Ordinary Differential Equations (For Ordinary People) - Waterman, G.
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DESCRIPTIONMath 321 Oregon Institute of Technology.
Ordinary Differential Equations(for ordinary people)Gregg WatermanOregon Institute of Technologyc2014 Gregg WatermanThis work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.The essence of the license is thatYou are free: to Share to copy, distribute and transmit the work to Remix to adapt the workUnder the following conditions: Attribution You must attribute the work in the manner specified by the author (but not in anyway that suggests that they endorse you or your use of the work). Please contact the author firstname.lastname@example.org to determine how best to make any attribution. Noncommercial You may not use this work for commercial purposes. Share Alike If you alter, transform, or build upon this work, you may distribute the resulting workonly under the same or similar license to this one.With the understanding that: Waiver Any of the above conditions can be waived if you get permission from the copyright holder. Public Domain Where the work or any of its elements is in the public domain under applicable law,that status is in no way affected by the license. Other Rights In no way are any of the following rights affected by the license: Your fair dealing or fair use rights, or other applicable copyright exceptions and limitations; The authors moral rights; Rights other persons may have either in the work itself or in how the work is used, such as publicityor privacy rights. Notice For any reuse or distribution, you must make clear to others the license terms of this work. Thebest way to do this is with a link to the web page below.To view a full copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letter to CreativeCommons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.Contents1 Functions and Derivatives, Variables and Parameters 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Functions and Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Derivatives and Differential Equations . . . . . . . . . . . . . . . . . . . . . 71.4 Parameters and Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Chapter 1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Chapter 1 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Introduction to Differential Equations 162.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Differential Equations and Their Solutions . . . . . . . . . . . . . . . . . . . 172.3 Classification of Differential Equations . . . . . . . . . . . . . . . . . . . . . 212.4 Initial Conditions and Boundary Values . . . . . . . . . . . . . . . . . . . . . 242.5 Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.6 Chapter 2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.7 Chapter 2 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Solving First Order Equations 333.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.2 Solving By Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . 353.3 Solving With Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . 383.4 Applications of First Order ODEs . . . . . . . . . . . . . . . . . . . . . . . . 413.5 Some Theoretical Concerns . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.6 Chapter 3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.7 Chapter 3 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 Qualitative and Numerical Methods: First Order 534.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.2 Solution Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.3 Direction Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.4 Phase Portraits and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . 594.5 Numerical Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.6 Chapter 4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.7 Chapter 4 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 695 Second Order Linear Constant Coefficient ODEs 755.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.2 An Initial Exploration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.3 Homogeneous Equations With Constant Coefficients . . . . . . . . . . . . . . 815.4 Particular Solutions, Part One . . . . . . . . . . . . . . . . . . . . . . . . . . 865.5 Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.6 Particular Solutions, Part Two . . . . . . . . . . . . . . . . . . . . . . . . . . 915.7 Linear Independence of Solutions, Reduction of Order . . . . . . . . . . . . . 945.8 Chapter 5 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 98i6 Applications of Second Order Differential Equations 1006.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.2 Free, Undamped Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1036.3 Forced, Undamped Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . 1066.4 Damped Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.5 Forced, Damped Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1096.6 RLC Electrical Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.7 Chapter 6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126.8 Chapter 6 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 1167 Boundary Value Problems 1227.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1227.2 Deflection of Horizontal Beams . . . . . . . . . . . . . . . . . . . . . . . . . 1237.3 Deflection of Vertical Columns . . . . . . . . . . . . . . . . . . . . . . . . . . 1267.4 Eigenfunctions and Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . 1307.5 The Heat Equation in One Dimension . . . . . . . . . . . . . . . . . . . . . . 1337.6 Chapter 7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1367.7 Chapter 7 Practice Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 137A Review of Calculus and Algebra 141A.1 Review of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141A.2 Review of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146A.3 Solving Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 147A.4 Series and Eulers Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151B Solutions to Exercises 153B.1 Chapter 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153B.2 Chapter 2 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155B.3 Chapter 3 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156B.4 Chapter 4 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159B.5 Chapter 5 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161B.6 Chapter 6 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162B.7 Chapter 7 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164B.8 Solutions for Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165Index167ii1 Functions and Derivatives, Variables and Parameters1.1 IntroductionLearning Outcome:1. Understand functions and their derivatives, variables and parameters.Performance Criteria:(a) Determine the independent and dependent variables for functionsmodeling physical and biological situations. Give the domain(s)of the independent variable(s).(b) For a given physical or biological situation, sketch a graph showingthe qualitative behavior of the dependent variable over the domain(or part of the domain, in the case of time) of the independentvariable.(c) Interpret derivatives in physical situations.(d) Find functions whose derivatives are given constant multiples ofthe original functions.(e) Identify parameters and variables in functions or differential equa-tions.Much of science and engineering is concerned with understanding the relationships be-tween measurable, changing quantities. Whenever possible we try to make these relationshipsprecise and compact by expressing them as equations relating variables; often such equationsdefine functions. In this chapter we take a look at ideas you should be familiar with, buthopefully you will now see them in a deeper and more illuminating way.11.2 Functions and VariablesPerformance Criteria:1. (a) Determine the independent and dependent variables for functionsmodeling physical and biological situations. Give the domain ofthe independent variable(s).(b) For a given physical or biological situation, sketch a graph showingthe qualitative behavior of the dependent variable over the domain(or part of the domain, in the case of time) of the independentvariable.As scientists and engineers, we are interested in relationships between measurable physicalquantities, like position, time, temperature, numbers or amounts of things, etc. The physicalquantities of interest are usually changing, so are called variables. When one physicalquantity (variable) depends on one or more other quantities (variables), the first quantity issaid to be a function of the other variable(s). Example 1.2(a): Suppose that a mass is hanging ona spring that is attached to a ceiling, as shown to theright. If we lift the mass, or pull it down, and let it go, itwill begin to oscillate up and down. Its height (relativeto some fixed reference, like its height before we lifted itor pulled it down) varies as time goes on from when westart it in motion. We say that height is a function oftime.springmass Example 1.2(b): Consider a beam that extends hori-zontally ten feet out from the side of a building, as shownto the right. The beam will deflect (sag) some, with thedistance below horizontal being greater the farther outon the beam one looks. The amount of deflection is afunction of how far out a point is on the beam.deflection isexaggerated Example 1.2(c): Suppose that we have a tank containing 100 gallons of water with10 pounds of salt dissolved in the water, as shown below and to the left. At some timewe begin pumping a 0.3 pounds salt per gallon (of water) solution into the tank at twogallons per minute, mixing it thoroughly with the solution in the tank. At the sametime the solution in the tank is being drained out at two gallons per minute as well.See the diagram below and to the right.100 galwater10 lbs salt0.3 lb/gal at2 gal/min2 gal/min100 galsolutionA lbs salt2Because the rates of flow in and out of the tank are the same, the volume in the tankremains constant at 100 gallons. The initial concentration of salt in the tank is 10pounds/100 gallons = 0.1 pounds per gallon. Because the incoming solution has ahigher concentration, the amount of salt in the tank will change as time goes on. (Theamount will increase, since the concentration of the incoming solution is higher thanthe concentration of the solution in the tank.) We can say that the amount of salt inthe tank is a function of time. Example 1.2(d): Consider the equation y = 12x2. For any number other than zerothat we select for x, there is a corresponding value of y that can be determined bysubstituting the x value and computing the resulting value of y. y depends on x, ory is a function of x Example 1.2(e): A drumhead with a radius of 5 inches is struck by a drumstick.The drum head vibrates up and down, with the height of the drumhead at a pointdetermined by the location of that point on the drumhead and how long it has beensince the drumhead was struck. The height of the drumhead is a function of the two-dimensional location on the drumhead and time. Example 1.2(f): Different points on the surface of a cube of metal one foot on a sideare exposed to different temperatures, with the temperature at each surface point heldconstant. The cube eventually attains a temperature equilibrium, where each point onthe interior of the cube reaches some constant temperature. The temperature at anypoint in the cube is a function of the three-dimensional location of the point.In each of the above examples, one quantity (variable) is dependent on one or moreother quantities (variables). The variable that depends on the other variable(s) is calledthe dependent variable, and the variable(s) that its values depend on is (are) called theindependent variable(s). Example 1.2(g): Give the dependent and independent variable(s) for each of Examples1.2(a) - (f).Example 1.2(a): The dependent variable is the height of the mass, and the independentvariable is time.Example 1.2(b): The dependent variable is the deflection of the beam at each point, andthe independent variable is the distance of each point from the wall in which the beam isembedded.Example 1.2(c): The dependent variable is the amount of salt in the tank, and the indepen-dent variable is time.Example 1.2(d): The dependent variable is y, and the independent variable is x.Example 1.2(e): The dependent variable is the height at each point on the drumhead, andthe independent variables are the location (in two-dimensional coordinates) of the point onthe drum head, and time. Thus there are three independent variables, and the two spacevariables would most likely be given using polar coordinates, since the head of the drum iscircular.Example 1.2(f): The dependent variable is the temperature at each point in the cube, andthe independent variables are the three coordinates giving the position of the point, in threedimensions. 3NOTE: Throughout this text, the symbol will be used to indicate the end of the answer to an examplethat is posed in a question and answer format.When studying phenomena like those given in Examples 1.2(a), (b), (c), (e) and (f),the first thing we do after determining the variables is establish coordinate systems for theindependent variables and, perhaps, the dependent variable: When position is an independent variable, we must establish a one (for the spring),two (for the drumhead) or three (for the cube of metal) dimensional coordinate system,with an origin (zero point) at some convenient location. If time is an independent variable, we must establish a time coordinate system bydetermining when time zero is. It may not be clear that there is a coordinate system for the temperature in the cubeof metal, or the amount of salt in the tank. For the temperature, the decision whetherto measure it in degrees Fahrenheit or degrees Celsius is actually the establishing of acoordinate system, with a zero point and a scale (both of which differ depending onwhich temperature scale is used). The choice of zero for the amount of salt in the tank will be the same regardless of howit is measured, but the scale can change, depending on the units of measurement.Once weve established the coordinate system(s) for the independent variable(s), weshould determine the domain of our function, which means the values of the independentvariable(s) for which the dependent variable will have values. Lets look at some examples. Example 1.2(h): For Example 1.2(a), suppose that we pull the mass down and thenlet it go at a time we call time zero (the origin of our time coordinate system. Timet is the independent variable, and the values of it for which we are considering theheight of the mass are t 0. Example 1.2(i): For Example 1.2(b), we will use a position coordinate system con-sisting of a horizontal number line at the top of where the beam emerges from the wall(so along the dashed line in the picture), with origin at the wall and positive values (infeet) in the direction of the beam. Letting x represent the position along the beam,the domain is 0 x 10.The four functions described in Examples 1.2(a)-(d) are functions of a single variable;the functions in Examples 1.2(e) and (f) are examples of functions of more than onevariable. The differential equations associated with functions of one variable are calledordinary differential equations, and the differential equations associated with functionsof more than one variable are (out of necessity) partial differential equations. In thisclass we will study only ordinary differential equations.The function in Example 1.2(d) is a mathematical function, whereas the functions fromExamples 1.2(a)-(c), (e) and (f) are not. (We might call them physical functions.) Inyour previous courses you have studied a variety of types of mathematical functions, includ-ing polynomial, rational, exponential, logarithmic, and trigonometric functions. The mainreason that scientists and engineers are interested in mathematics is that many physical sit-uations can be mathematically modeled with mathematical functions or equations. This4means that we can find a mathematical function that reasonably well describes the relation-ship between physical quantities. For Example 1.2(a), if we let y represent the height ofthe mass the equation that models the situation is y = A cos(bt), where A and b areconstants that depend on the spring and how far the mass is lifted or pulled down beforereleasing it. We will see that the deflection of the beam in Example 1.2(b) can be modeledwith a fourth degree polynomial function, and the amount of salt in the tank of Example1.2(c) can be modeled with an exponential function.Of course one tool we use to better understand a function is its graph. Suppose forExample 1.2(a) we started the mass in motion by lifting it 1.5 inches and releasing it (withno upward or downward force). Then the equation giving the height y at any time t wouldbe of the form y = 1.5 cos(bt), where b depends on the spring. Suppose that b = 5.2 (withappropriate units). Then the graph would look like this:0.6 1.2 1.81.50.01.5time(seconds)height(inches)For functions of one variable we always put the independent variable (often it will be time)on the horizontal axis, and the dependent variable on the vertical axis. We can see from thegraph that the mass starts at a height of 1.5 inches above its equilibrium position (y = 0).It then moves downward for the first 0.6 seconds of its motion, then back upward. It isback at its starting position every 1.2 seconds, the period of its motion. This periodicup-and-down motion can be seen from the graph. (Remember that the period T is thetime at which bT = 2pi, so T = 2pib.)Note that even if we didnt know the value of b in the equation y = 1.5 cos(bt) we couldstill create the given graph, we just wouldnt be able to put a scale on the time (horizontal)axis. In fact, we could even create the graph without an equation, using our intuition ofwhat we would expect to happen. Lets do that for the Example 1.2(c). Example 1.2(j): A tank contains 100 gallons of water with 10 pounds of salt dissolvedin it. At time zero a 0.3 pounds per gallon begins flowing into the tank at 2 gallons perminute and, at the same time, thoroughly mixed solution is pumped out at 2 gallonsper minute. (See Example 1.2(c).) Sketch a graph of the amount of salt in the tank asa function of time.The initial amount of salt in the tank is 10 pounds. Weknow that as time goes on the concentration of salt in thetank will approach that of the incoming solution, 0.3 poundsper gallon. This means that the amount of salt in the tankwill approach 0.3 lbs/gal 100 gal = 30 pounds, resultingin the graph shown to the right, where A represents theamount of salt, in pounds, and t represents the time, inminutes. 3010tA5NOTE: We have been using the notation sin(bt) or cos(bt) to indicate the sine or cosine of the quantitybt. It gets to be a bit tiresome writing in the parentheses every time we have such an expression, so we willjust write sin bt or cos bt instead.61.3 Derivatives and Differential EquationsPerformance Criteria:1. (c) Interpret derivatives in physical situations.(d) Find functions whose derivatives are given constant multiples ofthe original functions.Mathematical models for physical situations can often be developed from various prin-ciples and laws of physics. The physical principles do not usually lead us directly to thefunctions that model physical situations, but to equations involving derivatives of thosefunctions. Equations containing derivatives are called differential equations. More onthis later. In this section we will review the concept of a derivative and see an example of asimple differential equation, along with how it arises.When you hear the word derivative, you may think of a process you learned in a firstterm calculus class. In this course it will be important that you can carry out the processof finding a derivative; if you need review or practice, see Appendix B. In this sectionour concern is not the mechanics of finding derivatives, but instead we wish to recall whatderivatives are and what they mean.To reiterate what was said in the previous section, a function is just a quantity thatdepends on one or more other quantities, one in our case. Again, we refer to the firstquantity (the function) as the dependent variable and the second quantity, that it dependson, is the independent variable. If we were to call the independent variable x and thedependent variable y, then you should recall the notationdydxfor the derivative. Thisnotation can be loosely interpreted as change in y per unit of change in x. Technicallyspeaking, any derivative of a function is really the derivative of the dependent variable (whichIS the function) with respect to the the independent variable. We sometimes use the notationy instead ofdydx. Obviously it is easier to write y, but that notation does not indicatewhat the independent variable is and it does not suggest a ratio, or rate.Lets consider a couple examples of the meaning of the derivative in physical situations. Example 1.3(a): Suppose again that we take a mass hanging from a ceiling on aspring, lift it and let it go, and suppose the equation of motion is y = 1.5 cos 5.2 t. Thederivative of this function isdydt= 7.8 sin 5.2t, a new function of the independentvariable. This functions value at any time t can be interpreted as how fast the theheight of the mass is changing with respect to time, at that particular time. If the heightunits are inches and the time units seconds, then the units of the derivative are inchesseconds=inches per second, indicating that the derivative of the function y at a given time isthe velocity of the mass at that time. For example, the derivative at time 0.5 secondsisy(0.5) = 7.8 sin[(5.2)(0.5)] = 4.02,telling us that the mass is moving downward (indicated by the negative sign) at aboutfour inches per second at one half second after being set in motion.7 Example 1.3(b): Now recall the beam of Example 1.2(b), sticking out from a wallthat it is embedded in. If x represents a horizontal position along the beam andy represents the deflection (sag) of the beam at that horizontal position, then thederivativedydxis the change in deflection per unit of horizontal change, which is justthe slope of the beam at that particular point.Well now take a break from actual physical situations to ask some questions aboutderivatives, in a mathematical sense. After doing so, well see that such questions relatedirectly to certain real-life situations. Example 1.3(c): Find a function whose derivative is seven times the function itself.Note that the derivative of y = ekt, where k is a constant, is y = kekt. This showsthat exponential functions are essentially their own derivatives, with perhaps a constantmultiplier. If k was seven, the original function would be y = e7t and the derivative wouldbe y = 7e7t = 7y, seven times the original function y. Example 1.3(d): Find a function whose second derivative is sixteen times the functionitself.Here we should again be expecting an exponential function, but it will get multiplied twicebecause of the chain rule. Note that if y = e4t, then y = 4e4t and y = 16e4t = 16y,so y = e4t is the function we are looking for. But in fact it is not the function, but onlyone such function. The function y = e4t is another such function, as is y = 5e4t. (Youshould verify this last claim for yourself, and note that we could use any value where thefive is.) Example 1.3(e): Find a function whose derivative is 16 times the function itself.The previous example shows that the desired function is not an exponential function, asthe only likely candidates were shown to have second derivatives that are positive sixteentimes the original function. What we want to note here is that if we take the derivative ofsine or cosine twice, we end up back at sine or cosine, respectively, but with opposite sign.However, each time we take the derivative of a sine or cosine of kx, the chain rule givesus a factor of k on the outside of the trig function. Thus we see thaty = sin 4x = y = 4 cos 4x = y = 16 sin 4x = 16yy = cos 4x = y = 4 sin 4x = y = 16 cos 4x = 16yThis shows that y = sin 4x and y = cos 4x are functions whose derivatives are 16 timesthe original functions themselves. Consider Example 1.3(e) above. The words the second derivative is 16 times theoriginal function can be written symbolically asd2ydt2= 16y, since the function is thedependent variable y. This is a differential equation, an equation containing a derivative.(Differential equations can contain derivatives of any order. The order of the differentialequation is the highest order derivative occurring in the differential equation, so this is asecond order differential equation.)8Now consider the following physical situation: One end of a spring is attached to a ceiling,as shown to the left below. We then hang an object with mass m (we will refer to boththe object itself and its mass as the mass - one must note from the context which weare talking about) on the spring, extending it by a length l to where the mass hangs inequilibrium. This is shown in the center picture below. There are two forces acting on themass, a downward force of mg, where g is the acceleration due to gravity, and an upwardforce of kl, where k is the spring constant, a measure of how hard the spring pullsback. The spring constant is a property of the particular spring. When the mass hangs inequilibrium these two force are equal in magnitude to each other, but in opposite directions.F = klF = mgl0yly (pos)F = k(l y)F = mgWe will put a scale (in appropriate length units, like inches) beside the mass, with the zeroat the point even with the top of the mass at rest and with the positive direction being up.If we then lift the mass up to a position y0, where y0 < l, and release it, it will oscillateup and down. If we assume (for now) that there is no resistance, it will oscillate betweeny0 and y0 forever; such motion is known as simple harmonic motion. Consider themass when it is at some position y in this oscillation, as shown above and to the right.There will be an upward force of k(ly) due to the spring and a downward force of mg dueto gravity. Remembering that force is mass times acceleration and that acceleration is thesecond derivative of position with respect to time, the net force is thenF = ma = md2ydt2= k(l y)mg = kl ky mg = ky,since kl = mg.Extracting the equation md2ydt2= ky from the above and dividing both sides by m givesd2ydt2= kmy. If the values of k and m are such thatkm= 16, this equation becomesd2ydt2= 16y, the equation describing the situation from Example 3(e)! If we could find afunction y that satisfies the differential equation (more about what this means in the nextchapter), it will model the motion of the mass as it oscillates up and down. This shows thatwhat seems like an irrelevant mathematical question about derivatives (posed in Example3(e)) is actually very relevant for a practical application.91.4 Parameters and VariablesPerformance Criterion:1. (e) Identify parameters and variables in functions or differential equa-tions.If you have not recently read the explanation of the spring-mass system at the end ofthe last section, you should probably skim over it again before reading this section. Recallthat for the spring-mass system, the independent variable is time and the dependent variableis the height of the mass. Assuming no resistance, once the mass is set in motion, it willexhibit periodic oscillation. It should be intuitively clear that changing either the amount ofthe mass or the stiffness of the spring (expressed by the spring constant k) will change theperiod of oscillation. The mass m and the spring constant k are what we call parameters,and they should not be confused with the variables, which are time and the height of themass. Parameters will show up in three places: As characteristics of the physical systems themselves, quantified by numerical values. As constants within differential equations. As constants in the solutions to differential equations.Lets illustrate these three manifestations of parameters using our spring-mass system. Asmentioned above, the two physical parameters are the mass of the object hanging on thespring, and the stiffness of the spring, given by the spring constant. If the mass was 0.5 kgand the spring constant was 8 N/m (Newtons per meter) the differential equation would be0.5d2ydt2= 8y.Here we see the two parameters showing up in the differential equation. If we multiply bothsides by two and subtract the right side from both sides we obtaind2ydt2+ 16y = 0,where the 16 is the new parameter km, which we often rename as 2. In this case2 = 16 1sec2. The most general solution to this equation isy = C1 sint+ C2 cost;the variables are t and y, and C1, C2 and are parameters. The parameter dependson the mass and spring constant( =km)and the parameters C1 and C2 dependon how the mass is set in motion, what we will call initial conditions. In Chapter 2 wewill see the significance of C1 and C2 and how they are determined.We now consider the horizontal beam of Example 1.2(b). One might guess that someparameters that determine the amount of deflection of the beam would be the material thebeam is made of, the thickness and shape of the beam (square, I-beam, etc.), the lengthof the beam, and perhaps other things.10 Example 1.4(a): The differential equation, and its solution, for the beam of Example1.2(b) areEId4ydx4= w and y =w24EIx4 + c3x3 + c2x2 + c1x+ c0,where E is Youngs modulus of elasticity of the material the beam is made of, I is thecross-sectional moment of inertia of the beam about the neutral axis, and w is theweight per unit of length. Give the variables and parameters for both the differentialequation and the solution.We can see from the derivative in the differential equation that the independent variable isx and the dependent variable is y. The remaining letters all represent parameters: themodulus of elasticity E, the cross-sectional moment of inertia I, and the weight per unitof length w. In the solution we see these parameters again, along with four others, c0, c1,c2 and c3.The last four parameters in the solution will depend on the length of the beam and howit is supported, in this case by being embedded in the wall at its left end and having nosupport at the right end. These things are what are called boundary conditions. Welldiscuss them some more in Chapter 2, and look at specific situations using them in Chapter7. Here are a few comments about initial conditions and boundary conditions: Ordinary differential equations in which time is the independent variable will have initialconditions. Ordinary differential equations in which position along a line is the independent variablewill have boundary conditions. Most partial differential equations have both position and time as independent variables,so they usually have both initial conditions and boundary conditions.In some sense, parameters are variables that change from situation to situation, but oncethe situation is determined the values of the parameters are constant. At that point, the onlythings that change are the variables. An example you have probably encountered multipletimes can be used to illustrate what we mean. Suppose that a projectile is launched straightup in the air. The height at any time is dependent on the amount of time after launch, sothe independent variable is time and the dependent variable is height. Letting t representtime and y represent height, the equation for height isy = gt2 + v0t+ h0,where g is the acceleration due to gravity, v0 is the initial velocity, and h0 is the initialheight; these three quantities can be changed (g can only be changed significantly by beingon another planet!), but once the object is launched they cannot be changed.111.5 Chapter 1 Summary For our purposes, a function is a dependent variable that depends on one or moreindependent variables. (For most of this course we are concerned only with functions ofone independent variable.) The values of the independent variable for which values of the dependent variable areobtained are called the domain of the function. The graph of a function gives us a quick way to determine the general behavior of thefunction. The derivative of a function gives the rate of change of the dependent variable withrespect to the independent variable. Exponential functions are essentially their own derivatives (of any order). Sine and co-sine are essentially their own second derivatives. Essentially means that the functionand the derivative differ only by a factor of (multiplication by) a constant. Differential equations are equations containing derivatives of a function. When thefunction is a function of one variable, the differential equation is an ordinary differentialequation; when the function is a function of more than one variable, the differentialequation is a partial differential equation. Parameters are values that change from situation to situation, but that do not changeonce the situation is set. Variables are values that change once the situation is set.121.6 Chapter 1 Practice ExercisesSection 1.21. Some material contains a radioactive substance that decays over time, so the amount ofthe radioactive substance is decreasing. (It doesnt just go away - it turns into anothersubstance that is not radioactive in a series of steps. For example, uranium eventuallyturns into lead when it decays.)(a) Give the dependent and independent variables.(b) Sketch a graph of the amount A of radioactive substance versus time t. Labeleach axis with its variable - this will be expected for all graphs.2. A student holds a one foot plastic ruler flat on the top of a table, with half of the rulersticking out. They then tweak the end of the ruler, causing it to vibrate up and down.(This is roughly a combination of Examples 1.2(a) and (b).)(a) Give the dependent and independent variables. (Hint: There are two independentvariables.(b) Give the domains of the independent variables.3. Consider the drumhead described in Example 1.2(e). Suppose that the position of anypoint on the drumhead is given in polar coordinates (r, ), with r measured in inchesand in radians. Suppose also that time is measured in seconds, with time zero beingwhen the head of the drum is struck by a drumstick. Give the domains of each of thesethree independent variables.4. Consider the cube of metal described in Example 1.2(f). Suppose that we position thecube in the first octant (where each of x, y and z is positive), with one vertex (corner)of the cube at the origin and each edge from that vertex aligned with one of the threecoordinate axes. Each point in the cube then has some coordinates (x, y, z). Give thedomains of each of these three independent variables.5. Suppose that at some time zero a rock is thrown up in the air.(a) Sketch a graph of the height of the rock versus time, until it hits the ground. In thiscase, label each axis with its variable and some appropriate units of your choice.(b) What is the shape of the graph?(c) What sort of function might be used to model the height of the rock at any timet?(d) Discuss how the domain for this scenario compares to that for Exercise 1(b) -include an explanation of why they differ.136. When a solid object with some initial temperature T0 is placed in a medium (like air orwater) with a constant temperature Tm, the object will get cooler or warmer (dependingon whether T0 is greater or less than Tm), with its temperature T approachingTm. The rate at which the temperature of the object changes is proportional to thedifference between its temperature T and the temperature Tm of the medium, so itcools or warms rapidly while its temperature is far from Tm, but then the cooling orwarming slows as the temperature of the object approaches Tm.(a) Suppose that an object with initial temperature T0 = 80 F is place in a waterbath that is held at Tm = 40 F. Sketch a graph of the temperature as a functionof time. You should be able to indicate two important values on the vertical axis.(b) Repeat (a) for Tm = 40F and T0 = 30F.(c) Repeat (a) for Tm = 40F and T0 = 40F.7. (a) Suppose that a mass on a spring hangs motionless in its equilibrium position. Atsome time zero it is set in motion by giving it a sharp blow downward, and there isno resistance after that. Sketch the graph of the height of the mass as a functionof time.(b) Suppose now that the mass is set in motion by pulling it downward and simplyreleasing it, and suppose also that the mass is hanging in an oil bath that resistsits motion. Sketch the graph of the mass as a function of time.Section 1.31. Find the derivative of each function without using your calculator. You MAY use thecourse formula sheet. Give your answers using correct derivative notation.(a) y = 2 sin 3x (b) y = 4e0.5t (c) x = t2 + 5t 4(d) y = 3.4 cos(1.3t 0.9) (e) y = te3t (f) x = 4e2t sin(3t+ 5)2. Find the second derivatives of the functions from parts (a)-(c) of Exercise 1. Give youranswers using correct derivative notation.3. The temperature T of an object (in degrees Fahrenheit) depends on time t, measuredin minutes, anddTdt= 2.7 when t = 7. Interpret the derivative in a sentence, usingeither increasing or decreasing.4. The amount A of salt in a tank depends on the time t. If A is measured in poundsand t is measured in minutes, interpret the fact thatdAdt= 1.3 when t = 12.5.Again, use increasing or decreasing.145. The height of a mass on a spring at time t is given by y, where t is in seconds andy is in inches.(a) Interpret the fact thatdydt= 5 when t = 2.(b) Interpret the fact thatd2ydt2= 3 when t = 2.(c) Is the mass speeding up or slowing down at time t = 2? Explain.6. (a) Find a function y(x) whose derivative is 3 times the original function. Is theremore than one such function? If so, give another.(b) Find a function y(t) whose second derivative is 9 times the original function.Is there more than one such function? If so, give another.(c) Find a function x(t) whose second derivative is 9 times the original function. Isthere more than one such function? If so, give another.(d) Find a function y(x) whose second derivative is 5 times the original function.Is there more than one such function? If so, give another.Section 1.41. As mentioned previously, when a solid object with some initial temperature T0 isplaced in a medium (like air or water) with a constant temperature Tm, the objectstemperature T will approach Tm as time goes on. The rate at which the temperatureof the object changes is proportional to the difference between its temperature T andthe temperature Tm of the medium, giving us the differential equationdTdt= k(Tm T ),where k is a constant dependent on the material the object is made from.(a) Keeping in mind that parameters are quantities that vary from situation to situationbut do not change once the situation is fixed, give all of the parameters.(b) Give the independent variable(s).(c) Give the dependent variable.2. Suppose that a mass on a spring hangs motionless in its equilibrium position. At sometime zero it is set in motion by pulling it downward and simply releasing it, and supposealso that the mass is hanging in an oil bath that resists its motion. The independentvariable is time, and the dependent variable is the height of the mass. Give as manyparameters as you can think of for this situation - there are four that occur to me.3. When dealing with certain electrical circuits we obtain the differential equation andsolutionLdidt+Ri = E and i =ER+(i0 ER)eRLtGive the independent variable, dependent variable, and all the parameters.152 Introduction to Differential Equations2.1 IntroductionLearning Outcome:2. Understand differential equations and initial value problems, and thenature of their solutions.Performance Criteria:(a) Determine the independent and dependent variables for a givendifferential equation.(b) Determine whether a function is a solution to an ordinary differ-ential equation (ODE); determine values of parameters for whicha function is a solution to an ODE.(c) Classify differential equations as ordinary or partial; classify or-dinary differential equations as linear or non-linear, homogenous.Give the order of a differential equation.(d) Identify the functions a0(x), a1(x), ..., an(x) and f(x) for alinear ordinary differential equation.(e) Write a first order ordinary differential equation in the formdydx= F (x, y) and identify the function F . Classify first-orderODEs as separable or autonomous.(f) Identify initial value problems and boundary value problems. De-termine initial or boundary conditions.(g) Determine whether a function satisfies an initial value problem(IVP); determine values of constants for which a function satisfiesan IVP.In this chapter we expand on the ideas of Chapter 1 to look more closely at ordinarydifferential equations, and we will see what we mean by a solution to a differential equationor an initial value problem. It is important to understand these fundamental concepts beforemoving on to learning techniques for solving differential equations.We will also learn various classifications of differential equations. This is important inthat different types of equations have different methods for solving them.162.2 Differential Equations and Their SolutionsPerformance Criteria:2. (a) Determine the independent and dependent variables for a givendifferential equation.(b) Determine whether a function is a solution to an ordinary differ-ential equation (ODE); determine values of parameters for whicha function is a solution to an ODE.An equation that contains one or more derivatives is called a differential equation.Here are some examples that we will be considering:Equation 1:dydx+ 3y = 0 Equation 2: y + 3y + 2y = 0Equation 3: y + 9y = 26e2t Equation 4: 15.3d4ydx4= 1.4Equation 5:dydx=xyEquation 6:2ux2+2uy2=utNote that equations 1, 2, 3 and 5 contain not only derivatives of the function y, but thefunction itself as well. (We can really think of the function as the zeroth derivative.)The first five of these equations are all ordinary differential equations, meaning thatthey contain ordinary derivatives. The last one contains partial derivatives and is called apartial differential equation. (Some of you may have not yet taken a course in which youlearn about partial derivatives.) We often use the abbreviations ODE for ordinary differentialequation and PDE for partial differential equation.The order of a differential equation is the order of the highest derivative in the equation.Equations 1 and 5 above are first order, Equations 2, 3 and 6 are second order, and Equation4 is fourth order. In this course we will focus entirely on ordinary differential equations, andmost of the equations we will work with will be first or second order.When looking at a differential equation, it is often possible to determine the independentand dependent variables of interest. Derivatives are always of the dependent variable, andwith respect to the independent variable (or one of the independent variables in the case ofa function of more than one variable). So for Equation 1, the dependent variable is y andthe independent variable is x. Example 2.2(a): Give the dependent and independent variables for the rest of theequations.For Equations 4 and 5 the dependent variable is y and the independent variable is x. Forequation 3 the dependent variable is y, and since the derivative is an ordinary derivativethere must be only one independent variable, and it has to be t, the only other variablevisible in the equation. The dependent variable in Equation 2 is y, and it is not possibleto determine the independent variable in that case. Lastly, u is the dependent variablein Equation 6. There are three independent variables, x, y and t, which is why partial17derivatives are required. Any situation with more than one independent variable willresult in a partial differential equation. Is x = 5 a solution to 4x 2 = 10? One way to answer this question is to substitutefive for x in the left hand side of the equation and see if it simplifies to become the righthand side. If it does, then five is a solution to the equation:4(5) 2 = 20 2 = 18 6= 10, so x = 5 is not a solutionOn the other hand, x = 3 is a solution to 4x 2 = 10:4(3) 2 = 12 2 = 10, so x = 3 is a solutionWhat the above shows us is that a solution to an algebraic equation is a number that, whensubstituted for the unknown value, makes the equation true. We should recall that someequations have more than one solution. For example, both 3 and 3 are solutions to theequation x2 9 = 0.In the case of a differential equation, a solution to the equation is NOT a number; it is afunction.Solution to a Differential EquationA solution to a differential equation is a function for which thefunction and its relevant derivatives can be substituted into the equationto obtain a true statement.There are some differential equations whose solutions are relations rather than functions;well solve a few of those, but for all of the applications we will consider, the solutions to theODEs modeling the situations will be functions.When asked to verify that, or determine whether, a function is a solution to an ODE,you need to show some work supporting whatever your conclusion is. The following exampleshows one way to do this. Example 2.2(b): Show that y = 5 cos 4t is a solution to d2ydt2= 16y.We compute the left hand side (LHS) and right hand side (RHS) separately:dydt= 20 sin 4t = LHS = d2ydt2= 80 cos 4tRHS = 16(5 cos 4t) = 80 cos 4tBecause LHS = RHS, y = 5 cos 4t is a solution tod2ydt2= 16y. For the above example the left hand side was just one derivative. When the left handside is more complicated, a standard method of verifying a solution is to first calculate anyderivatives that appear on the left hand side of the equation, then substitute them into theleft hand side. If the right hand side is fairly simple, we might be able to simplify the leftside directly to the right hand side, as done in the next example.18 Example 2.2(c): Determine whether y = Ce2t, where C is any constant, is asolution to the differential equation y + 3y + 2y = 0.First we see that y = Ce2t(2) = 2Ce2t and y = 2Ce2t(2) = 4Ce2t, soLHS = 4Ce2t + 3(2Ce2t) + 2(Ce2t) = 4Ce2t 6Ce2t + 2Ce2t = 0 = RHS.Therefore y = Ce2t is a solution to y + 3y + 2y = 0. This last example shows that a differential equation can have an infinite number of solutions(since C can be any real number), and well see the same thing in the next example as well. Example 2.2(d): Verify that y = C1 sin 3t + C2 cos 3t, where C1 and C2 are anyconstants, is a solution to y + 9y = 0.First we see that y = 3C1 cos 3t3C2 sin 3t and y = 9C1 sin 3t9C2 cos 3t. ThereforeLHS = (9C1 sin 3t 9C2 cos 3t) + 9(C1 sin 3t+ C2 cos 3t) = 0 = RHS,so y = C1 sin 3t+ C2 cos 3t is a solution to y + 9y = 0. In this last example the function y = C1 sin 3t+ C2 cos 3t is a solution regardless of thevalues of the parameters C1 and C2. Because C1 and C2 can take any values, we say theyare arbitrary constants. We will often use the lower case c and upper case C for arbitraryconstants, sometimes with subscripts like above. We call all the functions obtained by lettingthe constants take different values a family of solutions for the differential equation. Thesolution to every first order equation will contain a constant that can take on infinitely manyvalues, and solutions to second order equations contain two arbitrary constants, as in theabove example. This may seem to contradict the result of Example 2.2(c), but the mostgeneral solution in that case is y = C1e2t + C2et; the solution verified in that exampleis for the case in which C2 = 0. The fact that C1 and C2 are subscripted differentlymeans that they are probably, but not necessarily, different constants. Other letters willoccasionally be used as constants.In this next example you will see a situation where a function is a solution only when theparameter takes a certain value; in this case the constant (parameter) is NOT arbitrary. Example 2.2(e): Determine any values of C for which y = Ce2t is a solution tothe differential equation y + 9y = 26e2t.The derivatives of the given function are y = 2Ce2t and y = 4Ce2t. Substitutingthe second derivative into the left hand side of the ODE givesLHS = 4Ce2t + 9Ce2t = 13Ce2t.y = Ce2t is a solution only if LHS = RHS, which requires that 13C = 26. ThereforeC = 2. The equation y + 9y = 0 is what we will call the homogenous equation associatedwith the equation y + 9y = 26e2t. (More on this later.) y + 9y = 0 is a secondorder homogenous equation, and Example 2.2(d) shows that the solution to the second order19homogenous equation has not one, but two, arbitrary constants. The function y = 2e3t iswhat we call a particular solution to the non-homogenous equation y+9y = 26e2t.A particular solution is one for which the values of constants cannot be arbitrary: Theconstant in this case must be two.A family of solutions that has one arbitrary constant, like the family from Example 2.2(c),is often referred to as a one-parameter family of solutions. The parameter is the constantC. The family y = C1 sin 3t+ C2 cos 3t from Example 2.2(d) is a two-parameter familyof solutions, with the parameters being C1 and C2. Solutions containing all possiblearbitrary constants will be called general solutions.This section has contained a lot of information! Lets summarize the important points: A solution to a differential equation is a function for which the function and itsrelevant derivatives can be substituted into the equation to obtain a true statement. Solutions to first order differential equations contain one arbitrary constant, and so-lutions to second order differential equations contain two arbitrary constants. All thesolutions obtained by letting constants take all possible values are called families ofsolutions. A solution to a differential equation that contains constants that are not arbitrary iscalled a particular solution to the differential equation. A family that encompasses all possible solutions of a differential equation is called ageneral solution to the differential equation.In Sections 2.4 and 2.5 we will see that if we impose an additional condition or two on adifferential equation the values of the arbitrary constants can be determined.202.3 Classification of Differential EquationsPerformance Criteria:2. (c) Classify differential equations as ordinary or partial; classify or-dinary differential equations as linear or non-linear, homogenousor non-homogeneous. Give the order of a differential equation.(d) Identify the functions a0(x), a1(x), ..., an(x) and f(x) for alinear ordinary differential equation.(e) Write a first order ordinary differential equation in the formdydx= F (x, y) and identify the function F . Classify first-orderODEs as separable or autonomous.There are many different classifications and types of differential equations; we will focuson just a few classifications here. Lets consider the following examples, most of which wesaw in the previous section.Equation 1:dydx+ 3y = 0 Equation 2: x2y + xy + x2y = 0Equation 3: y + 9y = 26e2t Equation 4: 15.3d4ydx4= 1.4Equation 5:dydx=xyEquation 6:2ux2+2uy2=utHere are the classifications well be interested in: Ordinary differential equations (ODEs) versus partial differential equations (PDEs).We have already discussed this; Equations 1 - 5 are ODEs and Equation 6 is a PDE. Itis worth mentioning here that a solution to an ODE is a function of just one variable,whereas a solution to a PDE is a function of more than one variable. The solution toEquation 6 is a function u of the three variables x, y and t. Differential equations are classified by order, which is the highest derivative occurringin the equation. Equations 1 and 5 are first order, Equations 2, 3 and 6 are secondorder (PDEs are classified by order the same way that ODEs are), and Equation 4 isfourth order. An ODE that can be written in the forman(x)dnydxn+ an1(x)dn1ydxn1+ + a2(x)d2ydx2+ a1(x)dydx+ a0(x)y = f(x) (1)is called a linear differential equation. Equations 1 through 4 are linear ODEs: Equation 1 is first order linear, with a1(x) = 1, a0(x) = 3 and f(x) = 0.21 Equation 2 is second order linear, with a2(x) = x2, a1(x) = x, a0(x) = x2 andf(x) = 0. This particular equation is known as a Bessel equation of order zero(where order does not refer to the order of the ODE - how confusing!). It isobtained when working with a PDE called the wave equation, used for things likemodeling the vibration of a drumhead. Equation 3 is second order linear with a2(t) = 1, a1(t) = 0, a0(t) = 9 andf(t) = 26e2t. Note the variable is t, rather than x, because the independentvariable in this case is t. Equation 4 is fourth order linear with a4(x) = 1, a3(x) = a2(x) = a1(x) = a0(x) =0 and f(x) = 1.4. A linear equation, so an ODE of the form (1) above, is called homogeneous if f(x) = 0.Equations 1 and 2 are homogeneous, Equations 3 and 4 are non-homogeneous. Onemust be a bit careful, because there is another meaning of homogenous associated withODEs! The difference between the two uses must be determined by the context in whichthey are used. The definition just given is the only one well be using. If we multiply both sides of Equation 5 by y we get y dydx= x, which is not of the form(1). Any effort to get the coefficient ofdydxto be a function of x will fail, so Equation5 is non-linear. Note that if the original equation had instead beendydx=yx, wecould multiply both sides by x and subtract y to get xdydx y = 0. This equationIS linear, with a1(x) = x, a0(x) = 1 and f(x) = 0.Suppose that we have a first order ODE with independent variable x and dependent variabley. Such an equation can always be written in the formdydx= F (x, y), where F is simplya function of the two variables x and y. Consider for example the Equation B below; itcan be written asdydx= x+ 2xy, so F (x, y) = x+ 2xy for that equation.A.dydx xy= 0 B. y 2xy = x C. y + 2y = y2 D. 5dydx 3y = sin x Example 2.3(a): Determine the functions F (x, y) for Equations A, C and D above.Each of the equations can be solved fordydxto getA.dydx=xyC.dydx= y2 2y D. dydx=35y +15sin xWe can see now that the functions F for the three equations areA. F (x, y) =xyC. F (x, y) = y2 2y D. F (x, y) = 35y +15sin x We can now define two other categories of first order differential equations.22 When F is the product of a function of x and a function of y, written compactlyas F (x, y) = g(x)h(y), the ODE is called separable. When F is really just a function of y (so F (x, y) = f(y)) the ODE is calledautonomous. Note that by letting g(x) = 1, any autonomous equation is alsoseparable (but not vice-versa!). Example 2.3(b): Determine whether any of the the equationsA.dydx xy= 0 B. y 2xy = x C. y + 2y = y2 D. 5dydx 3y = sin xare separable or autonomous.The only one of the equations that can be written in the formdydx= f(y) is C, so it isautonomous (and therefore separable as well). For Equation A, F (x, y) =xy= x 1y, so it isseparable, with g(x) = x and h(y) =1y. For Equation B, F (x, y) = x+2xy = x(1+2y),so it is also separable, with g(x) = x again and h(y) = 1+2y. In the case of Equation DF (x, y) = 35y+ 15sin x, which is clearly not just a function of y, so it is not autonomous.We also see that it is not possible to write F (x, y) in the form g(x)h(y) either, so it isalso not separable. We will see the significance of separable and autonomous equations later. For now weshould note a bit of algebra that can be performed with a separable equation. To beginwith, we need to think of the derivativedydxas being the quotient of the two differentialsdy and dx. Treating each like we would a variable, when we are working with a separableequation we can get all the x stuff on one side of the equation and the y stuff on theother side:dydx xy= 0dydx=xydy =xydxy dy = xdxSeparable equations are often easy to find solutions for, as well do in Chapter 3, and com-putations like the above will be part of the process.232.4 Initial Conditions and Boundary ValuesPerformance Criterion:2. (f) Identify initial value problems and boundary value problems. De-termine initial or boundary conditions.In Section 2.2 we learned that ODEs have families of solutions, with each member of afamily obtained by fixing a value for an arbitrary constant (parameter) in the case of a firstorder ODE and two constants for a second order. The number of parameters in a generalsolution that describes an entire family is equal to the order of the ODE. When solvingactual problems we will determine the values of such arbitrary constants from additionalgiven parameters called initial values (also called initial conditions) or boundary values(boundary conditions).Recall Examples 1.2(a): Example 1.2(a): Suppose that a mass is hanging ona spring that is attached to a ceiling, as shown to theright. If we lift the mass, or pull it down, and let it go,it will begin to oscillate up and down. The height y ofthe mass (relative to some fixed reference, like its heightbefore we lifted it or pulled it down) is a function of thetime t that has elapsed since we set the mass in motion.springmassFor this example, the height of the mass at any time t depends on the amount of themass and the stiffness of the spring, but it also depends on how we set the mass in motion.The mass could be set in motion by lifting it some amount, like one inch, then letting it go.It could instead be set in motion by just hitting it downward from its position at rest, insuch a way that its velocity is three inches per second at the instant we hit it. These actionscan be described by what we call initial conditions for the function y: Example 2.4(a): Consider the mass on the spring, set in motion by lifting it one inchand letting it go. Give the height and velocity of the mass at the time it is let go, usingfunction notation.Taking up to be positive, at time zero (the moment we set the mass in motion) the height ofthe mass is one inch, so we write y(0) = 1. Since we simply release the mass at time zero,the velocity at time zero is zero. Recalling that velocity is the first derivative of position,we can describe this by y(0) = 0. The two mathematical statementsy(0) = 1, y(0) = 0are called initial conditions for the function describing the height of the mass at any timet. We will be dealing with initial conditions throughout this course, so try to get a goodunderstanding of them now.24 Example 2.4(b): Consider the mass on the spring, this time setting it in motion byhitting it downward at three inches per second from its position at rest. Give the initialconditions for the height function y.Because we are forcing the mass from its position at rest, its initial height is zero. This isgiven using function notation by y(0) = 0. The fact that it has downward velocity of threeinches per second at time zero gives us the initial condition y(0) = 3. Example 2.4(c): Suppose that the mass is set in motion by pulling it down two inches,then giving it an upward velocity of five inches per second to begin. Give the initialconditions for the height function y.The initial conditions are y(0) = 2 and y(0) = 5. Now consider Example 1.2(b): Example 1.2(b): Consider a beam that extends hor-izontally out from the side of a building, as shown tothe right. The beam will deflect some amount y ateach point on the beam, with the amount of deflectiondepending on the distance x from the wall.deflection isexaggeratedFor this situation the deflection of the beam does not depend on time, so there are noinitial conditions. However, there are conditions imposed on the beam, depending on whetherand how it is supported at each end. We will save most of the details of this for later, but letsconsider the left end of the beam for the time being. If we let y = 0 represent the height ofthe dashed line and if x represents the distance along the beam from the wall, it should beclear that the deflection is zero where the beam meets the wall, or y(0) = 0. Additionally,assuming that the left end of the beam is embedded in the wall horizontally, the slope of thebeam at the wall is zero as well. But slope is derivative, so we have y(0) = 0 as well. Itturns out that at the right end of the beam we also know that y and y are both zero.If the right end of the beam was ten feet from the wall, we would write these conditions asy(10) = 0, y(10) = 0.We can think of the ends of the beam as the boundaries of the beam; the conditionsy(0) = 0, y(0) = 0, y(10) = 0, y(10) = 0are called the boundary conditions for the beam. Example 2.4(d): Consider a twenty foot beamthat is embedded in walls at both ends, as shownto the right. The beam will deflect downwardsome in the middle; the deflection is exaggeratedin the picture. Give the boundary conditions forthe beam.The boundary conditions are y(0) = 0, y(0) = 0, y(20) = 0 and y(20) = 0. 25Initial conditions arise when our function depends on time and boundary conditions arisewhen our function depends on position in space (or in the above examples, position along aline). We will work primarily with initial conditions, but you will see boundary conditionslater in the course (see Chapter 6). There are situations where a function depends onboth position and time; for such situations there are both initial conditions and boundaryconditions. We will not see these, because they are described by partial differential equations.Partial differential equations are also required when working with boundary conditions only,when the function of interest is a function of more than one space variable. Such functionswould arise when dealing with sheets or solids, rather than beams, which can be thought ofas one-dimensional lines.262.5 Initial Value ProblemsPerformance Criterion:2. (g) Determine whether a function satisfies an initial value problem(IVP); determine values of constants for which a function satisfiesan IVP.Consider again the differential equationdydx+3y = 0, for which any function of the formy = Ce3x is a solution. Suppose we impose the additional condition that y = 4 whenx = 0. This is called an initial condition, and we often write such a condition in the formy(0) = 4. Substituting these values into y = Ce3x gives 4 = Ce3(0), leading to C = 4.When we combine a differential equation with one or more initial values, we have whatis called an initial value problem (IVP). The solution to an initial value problem is afunction or equation that satisfies both the differential equation and the initial value(s).Thus y = 4e3x is a solution to the IVPdydx+ 3y = 0 , y(0) = 4 Example 2.5(a): Verify that y = 72e5t+ 52sin t 12cos t is the solution to the initialvalue problemdydt+ 5y = 13 sin t, y(0) = 3dydt= 352e5t + 52cos t+ 12sin t, soLHS = 352e5t + 52cos t + 12sin t+ 5(72e5t + 52sin t 12cos t)= 352e5t + 52cos t + 12sin t+ 352e5t + 252sin t 52cos t= 262sin t= 13 sin t= RHSThis shows that the function satisfies the differential equation.We must now show that the function satisfies the initial condition. When t = 0,y = 72e5(0) + 52sin 0 12cos 0 = 72+ 0 12= 62= 3,so the function satisfies the initial condition also. Therefore it is a solution to the IVP. We should observe in the above example that the process of checking the initial condition iseasier than checking the differential equation; this is often the case.27 Example 2.5(b): Determine whether y = 5e3t 2e3t is a solution to the initialvalue problemy 9y = 0 , y(0) = 3 , y(0) = 8This time lets check the initial conditions first. We see that y(0) = 52 = 3, so the firstinitial condition is met. We next compute y(t) = 15e3t 6e3t, so y(0) = 15 6 6=8, So the second initial condition is not met. Therefore the function is NOT a solutionto the IVP. 282.6 Chapter 2 Summary A solution to a differential equation is a function for which the function and its relevantderivatives can be substituted into the equation to obtain a true statement. When a solution contains arbitrary constants we call it a family of solutions. A familythat includes all possible solutions to a differential equation is called a general solution;a solution that contains no arbitrary constants is called a particular solution. General solutions to first order differential equations contain one arbitrary constant, andgeneral solutions to second order differential equations contain two arbitrary constants. To verify that a function is a solution to a differential equation, we substitute thefunction and its derivative into the left side and see if the result is the right side, ORwe substitute the function and its derivatives into both sides and see if the results areequal. Values of arbitrary constants are determined by initial (or boundary) conditions. Fora first order equation, one initial condition is needed to determine the one constant.For a second order equation, two initial conditions are needed to determine the twoconstants. A differential equation together with either initial values or boundary values is calledan initial value problem or a boundary value problem. We classify ordinary (and partial) differential equations: The order of a differential equation is the order of the highest derivative in theequation. In addition to classifying ordinary differential equations by order, we also classifythem as linear or non-linear, homogeneous or non-homogeneous. First order ordinary differential equations can also be classified as separable (ornot), and autonomous (or not). Recognizing the classification(s) of an ordinary differential equation are important forknowing how to solve the equation.292.7 Chapter 2 Practice ExercisesSection 2.21. For each of the following differential equations, determine the independent and depen-dent variables when possible. (You should always be able to identify the dependentvariable.)(a)dydx 2y = 0 (b) y y = 0 (c) 2ut2= 3(2ux21+2ux22+2ux23)(d) y + 9y = 26e2t (e)ut= 0.52ux2(f)d2xdt2 5dxdt+ 6x = 10 sin t(g) Ld2udx2+ g sin u = 0 (h) EId4ydx4= w (i) utt c2(urr + 2rur) = 02. (a) For part (b) of the previous exercise you should not have been able to identify theindependent variable. Given that the solution is y = 3ex 5ex, what is theindependent variable?(b) The differential equation y6y+9y = 0 has solution y = C1e3t+C2te3t. Whatare the independent and dependent variables?3. Is y = sin 2t a solution todydt+ 2y = 0?4. Is y = 3ex 5ex a solution to y y = 0?5. (a) Verify that y = 5e3x is also a solution to dydx+ 3y = 0.(b) Verify that y = Ce3x, where C is any constant, is a solution to the samedifferential equation.6. (a) Verify that y = C1 sin(3t) + C2 cos(3t), where C1 and C2 are any constants, isa solution to y + 9y = 0.(b) Verify that y = 2e2t is a solution to the differential equation y+9y = 26e2t andshow that y = 3e2t is NOT a solution to the same differential equation.(c) Verify that y = C1 sin(3t) + C2 cos(3t) + 2e2t is a solution to the differentialequation y + 9y = 26e2t.7. Verify that x = C1e3t +C2e2t + sin t+ cos t, where C1 and C2 are any constants, isa solution to the differential equationd2xdt2 5dxdt+ 6x = 10 sin t.8. Consider the differential equation y 6y + 9y = 0.(a) Verify that y = ce3t is a solution to the differential equation.(b) Verify that y = cte3t is a solution to the differential equation.309. Consider the differential equationdydx y = 4e3x.(a) Is there any value of c for which y = cex a solution to the equation? If so, whatis the value?(b) Is there any value of c for which y = ce3x a solution to the equation? If so, whatis the value?(c) Recall that a solution to a differential equation that cannot have an arbitraryconstant in it is called a particular solution to the equation. Give a particularsolution to the differential equationdydx y = 4e3x.(d) Is there any value of c for which y = cex a solution to the equationdydx y = 0?If so, what is the value?10. This last exercise has little or no conceptual value; it basically just involves goingthrough some mildly tedious calculations. However, I would expect you to be able to dothis kind of thing successfully (without using a calculator) if someone put a gun to yourhead and said that your life depended on it! So here it is: Verify that y = e3t sin 2t isa solution to y 6y + 13y = 0.Section 2.31. For Exercise 1 of the previous section, list the letters of all the ordinary differentialequations.2. Give the order of each of the following ordinary differential equations. Assume that anyletters not used in derivatives represent constants.(a)dydt 2y = 0 (b) y y = 0 (c) 1ydydx+ y = 1(d) y + 9y = 26e2t (e)1xdydx+ y = 1 (f)d2xdt2 5dxdt+ 6x = 10 sin t(g) Ld2udx2+ g sin u = 0 (h) EId4ydx4= w (i)dydx+ xy = 13. For each of the first order equations from Exercise 2, give the function F if the theequation was written in the formdydx= F (x, y). (Use the appropriate variables for theequation.)4. For each of the ODEs from Exercise 2 that are linear, give the values of the functionsf , a0, a1, a2, ... (Include the independent variable, like a1(x) = x2, for example.) Ifthe independent variable cannot be determined, use x.5. For each of the first order equations from Exercise 2 that are separable, give the functionsg and h, using the appropriate independent variable.6. Which of the first order equations from Exercise 2 are autonomous?31Section 2.51. Verify that y = 1x+ 3 is a solution to the initial value problemx2dydx= 1, y(1) = 22. Determine whether y = 2 sin(3t) + e2t is a solution to the initial value problemy + 9y = 13e2t, y(0) = 1, y(0) = 4Note that the most general solution to the differential equation y + 9y = 13e2t is y =C1 sin(3t) + C2 cos(3t) + e2t, but to satisfy the given initial conditions it must be the casethat C2 = 0.3. For each of the following, determine whether the given function is a solution to theinitial value problem that is given after it. If it is not, tell why not.(a) Function: y = 14ex+ 12e2x+ 34e3x IVP: yy2y = e3x, y(0) = 32, y(0) = 1(b) Function: y = 3ex + 12sin x IVP: y + y = sin x, y(0) = 3(c) Function: y = 52ex2 12IVP:dydx 2xy = x, y(0) = 2(d) Function: x = 2 sin 2t+ 3 cos 2t IVP:d2xdt2+ 4x = 0, x(0) = 3, x(0) = 44. It can be shown that y = A sin5 t + B cos5 t is the general solution to thedifferential equation y + 5y = 0. Find the solution to the initial value problemy + 5y = 0 , y(0) = 3 , y(0) = 23by inserting the t and y values into the function and its first derivative and solvingto get A and B.323 Solving First Order Equations3.1 IntroductionLearning Outcome:3. Solve first order differential equations and initial value problems; setup and solve first order differential equations modeling physical prob-lems.Performance Criteria:(a) Solve first order ODEs and IVPs by separation of variables.(b) Demonstrate the algebra involved in solving a relation in x andy for y; in particular, change ln y = f(x) to y = g(x), showingall steps clearly.(c) Use an integrating factor to solve a first order linear ODE or IVP.(d) Solve an applied problem using separation of variables or an in-tegrating factor.(e) Give an ODE or IVP that models a given physical situation in-volving growth or decay, mixing, Newtons Law of Cooling or anRL circuit.(f) Sketch the graph of the solution to a mixing or Newtons Law ofCooling problem, indicating the initial value and the steady-stateasymptote.(g) Identify the transient and steady-state parts of the solution to afirst order ODE.In the first two chapters you found out what what ordinary differential equations (ODEs)and initial value problems (IVPs) are, and what it means for a function to be a solution toan ODE or IVP. You then saw how to determine whether a function is a solution to an ODEor IVP, and we looked at a few real world situations where ODEs and IVPs arise fromphysical principles.Our goal for the rest of the course is to solve ODEs and IVPs and to see how the ODEs,IVPs and their solutions apply to real situations. We can solve ODEs (and PDEs) in threeways: Analytically, which means paper and pencil methods that give exact algebraic solu-tions. Qualitatively, which means determining the general behavior of solutions without actu-ally finding function values. Results of qualitative methods are often expressed graphi-cally. Numerical methods which result in values of solutions only at discrete points in timeor space. Results of numerical methods are often expressed graphically or as tables ofvalues.33In this chapter you will learn how to find solutions analytically for first order ODEs andIVPs. You will see two methods, separation of variables and integrating factors. Separation of variables is the simpler of the two methods, but it only works for separableODEs, which you learned about in Section 2.3. It is a useful method to look at becausewhen it works it is fairly simple to execute, and it provides a good opportunity to reviewintegration, which we will need for the other method as well. Solving with integrating factors is a method that can be used to solve any linear firstorder ODE, whether it is separable or not. The method of solution is more complicatedthan separation of variables, but not necessarily any more difficult to execute once youlearn it.After learning these two methods we will again look at applications, but only for first orderODEs and IVPs.343.2 Solving By Separation of VariablesPerformance Criteria:3. (a) Solve first order ODEs and IVPs by separation of variables.(b) Demonstrate the algebra involved in solving a relation in x andy for y; in particular, change ln y = f(x) to y = g(x), showingall steps clearly.So far you have learned how to determine whether a function is a solution to a differentialequation or to an initial value problem. But the following question remains: How do we findsolutions to differential equations? We will spend much of the course learning some methodsfor finding solutions. If the ODE is separable, we can apply the simplest method for solvingdifferential equations, called separation of variables. The bad news is that separationof variables only works for separable (so necessarily also first order) equations; the goodnews is that those sorts of equations actually occur in some real life situations. Lets lookat an example. Example 3.2(a): Solve the differential equation y cos 3xy= 0.Note that we can write the ODE as dydx= cos 3x 1y= g(x)h(y), where g(x) = cos 3x andh(y) = 1y. Therefore the ODE is separable; lets separate the variables and solve:dydx=cos 3xyChange to dydxnotation and get the term with thederivative alone on one side.dy =cos 3xydx Multiply both sides by dx.y dy = cos 3xdx Do some algebra to get all the x stuff on one sideand the y stuff on the other. At this point thevariables have been separated.y dy =cos 3xdx Integrate both sides.12y2 + C1 =13sin 3x+ C2 Compute the integrals.12y2 = 13sin 3x+ C Subtract C1 from both sides and let C = C2 C1.DO NOT solve for y unless asked to. The resulting solution for the above example is not a function, but is instead a relation. Insome cases we will wish to solve for y as a function of x (or whatever other variables wemight be using), but you should only do so when asked to.In the next example you will see a simple, but a very useful, type of differential equations.35 Example 3.2(b): Solve the differential equation dydt+ 0.5y = 0 by separation ofvariables, and solve the result for y.First lets solve the ODE by separation of variables:dydt+ 0.5y = 0dydt= 0.5ydy = 0.5y dtdyy= 0.5 dtdyy=0.5 dtln |y|+ C1 = 0.5t + C2ln |y| = 0.5t + C3where C3 = C2C1. We now solve for |y|, using the facts that the inverse of the naturallogarithm is the exponential function with base e and if |x| = u, then x = u (thedefinition of absolute value):ln |y| = 0.5t+ C3eln |y| = e0.5t+C3 take e to the power of each side|y| = e0.5t eC3 inverse of natural log and xaxb = xa+b|y| = C4e0.5t eC3 is just another constant, which we call C4y = C4e0.5t the definition of absolute valuey = Ce0.5t absorb the into C4, calling the result C The last step above might seem a bit fishy, but it is valid. In most cases we have initialvalues, which then determine the constant C, including its sign: Example 3.2(c): Solve the initial value problem dydt+ 0.5y = 0, y(0) = 7.3.We already solved the differential equation in the previous example, so we just need to findthe value of the constant by substituting the initial values into the solution y = Ce0.5t:7.3 = Ce0.5(0)7.3 = CThe solution to the IVP is y = 7.3e0.5t. 36Dont assume that the the constant is always the initial value! Example 3.2(d): Solve the initial value problem y cos 3xy= 0, y(0) = 4.We already solved the differential equation in Example 3.2(a), so we just need to find thevalue of the constant. Substituting x = 0 and y = 4 into the solution from that examplewe get12(4)2 = 13sin 3(0) + C8 = 0 + CC = 8The solution to the IVP is 12y2 = 13sin 3x+ 8. 373.3 Solving With Integrating FactorsPerformance Criterion:3. (c) Use an integrating factor to solve a first order linear ODE or IVP.Lets begin with an example that demonstrates the limitation of separation of variables. Example 3.3(a): Solve dydx 3y = e4x.Note that if we try to separate the variables we getdydx 3y = e4xdydx= 3y + e4xdy = (3y + e4x) dxHere we see that there is no way to get the 3y term back over to the left side with dy.(This is because 3y + e4x cannot be written in the form g(x)h(y).) Therefore thisequation cannot be solved by separation of variables. The following derivative computation provides the key for solving equations like the oneabove. Example 3.3(b): Suppose that y = y(x) is some function of x. Find the derivativeof ye3x (with respect to x).Because both y and e3x are functions of x, we must use the product rule:ddx(ye3x) = yddx(e3x) + e3xddx(y) = 3ye3x + e3x dydx= e3x(dydx 3y)Notice now that if we were to multiply the left side of the ODE of Example 3.3(a) by e3x wewould get the result of Example 3.3(b). This indicates an idea for solving the ODE:dydx 3y = e4xe3xdydx 3e3xy = e4xe3x Multiply both sides by e3xd(ye3x)dx= e4xe3x From Example 3.3(b)d(ye3x) = ex dx Combine exponentials, multiply both sides by dxd(ye3x) =ex dx Integrate both sides38ye3x = ex + C Carry out the integrationsye3xe3x = exe3x + Ce3x Multiply both SIDES by e3xy = e4x + Ce3x Apply properties of exponentsThus the solution todydx 3y = e4x is y = e4x + Ce3x.The method just shown for finding the solution todydx 3y = e4x probably seems a bitmysterious, to say the least! This is called the integrating factor method, which we nowsummarize. Note that it only applies to linear first order ODEs, which can always be putinto the formdydx+ p(x)y = q(x).Solving a 1st Order Linear ODE Using An Integrating FactorTo solve a first order ODE in the formdydx+ p(x)y = q(x),1) Compute u =p(x) dx. The integrating factor is eu.2) Multiply both sides of the equation by the integrating factor eu.The left side of the equation then becomesd(yeu)dx.3) Multiply both sides of the equation by dx and integrate both sides.The left side will become yeu.4) Solve for y by multiplying both sides by eu.Note that after integrating both sides of the equation there will be a constant added tothe right side. This constant will be multiplied by eu in the solution. For the equationdydx 3y = e4x, p(x) = 3 sop(x) dx = 3dx = 3x and eu = e3x.Any first order linear ODE can be solved using the integrating factor method, as longas euq(x) can be integrated; sometimes you can use this method or separation of variablesand they both will work. Now lets take a look at executing the above steps with anotherexample. Example 3.3(c): Solve dydx+yx= x2 for x > 0 by the integrating factor method.First we note that p(x) =1xand q(x) = x2. Because x > 0, |x| = x andu =1xdx = ln x. Therefore eu = elnx = x. We now carry out steps (2) through (4)above:39dydx+yx= x2 original equationxdydx+ y = x3 multipy both sides by eu, which in this case is xd(xy)dx= x3 use the product rule in reverse to collapse the left sided(xy) =x3 dx multiply both sides by dx and integratexy = 14x4 + C include a single constant of integration on the right sidey = 14x3 +Cxdivide both sides by eu = x 403.4 Applications of First Order ODEsPerformance Criteria:3. (d) Solve an applied problem using separation of variables or anintegrating factor.(e) Give an ODE or IVP that models a given physical situationinvolving growth or decay, mixing, Newtons Law of Cooling oran RL circuit.(f) Sketch the graph of the solution to a mixing or Newtons Lawof Cooling problem, indicating the initial value and the steady-state asymptote.(g) Identify the transient and steady-state parts of the solution toa first order ODE.Radioactive Decay and Population GrowthIn general, we can assume that the rate at which a quantity of radioactive material decaysis proportional to the amount present. For example, 20% of the material might decay inany 600 year period. If there were 1000 pounds initially, 200 pounds would decay over600 years, but if there were only 100 pounds initially, only 20 pounds would decay over600 years. If we let A represent the amount of material at any time t, then the rate atwhich the material decays is given by the derivativedAdt. The above discussion tells us thatthere is some constant of proportionality k for whichdAdt= kAWe will find that k is negative because the amount is decreasing.Similarly, suppose that N represents the number of individuals (which could be peopleor any other animals) in a region, and assume the that the population is growing. If therewere no constraints like famine, disease and such, the population should grow continuously.The derivativedNdtwould represent the rate of change of population with respect to time.When the population is small we would expect a small change in population over a fixedtime period, but when the population is large wed expect a greater increase in populationover the same time period, because there is a larger population having offspring. However,wed again expect the rate of change to be proportional to the population itself, resulting inthe differential equationdNdt= kN . This is of course the same equation as above but inthis case we can expect k to be positive, because here there is growth rather than decay.Clearly the differential equations for both radioactive decay and population growth arethe same, and both can easily be solved by separation of variables, or even just by guessing,as long as we remember that the solution must contain an arbitrary constant! The arbitraryconstant is in addition to the constant k; the additional constant is introduced by the factthat we must essentially integrate once to determine the function N = N(t).41 Example 3.4(a): Five hundred rainbow trout are introduced into a previously barren(no fish in it) lake. Three years later, biologists estimate that there are 1730 rainbowtrout in the lake. Assuming that the population satisfies the ODEdNdt= kN , deter-mine the function N = N(t) that gives the number N of rainbow trout in the lakeat time t.The ODEdNdt= kN says that we are looking for a function N(t) whose derivative isk times the function itself, and N = ekt clearly is such a function. This function containsno constant of integration, but we recall that N = Cekt is also a solution, for any valueof C. The general solution is then N = Cekt. The fact that there are 500 fish in thelake at time zero gives us 500 = Ce0, so C = 500 and N = 500ekt.To find k we substitute N = 1730 when t = 3 into our solution to get 1730 = 500e3k,and solve to find k = 0.414. The equation for the number of rainbow trout at any timet is then N = 500e0.414t. Heres how one would solve the differential equation from the last example by separationof variables, with some of the steps combined:dNdt= kNdNN= k dt multiply by dt and divide by NlnN = kt+ C integrate both sides - absolute value is notneeded because N must be greater than zeroN = Cekt exponentiate both sides and apply xaxb = xa + bMixing ProblemsLets go straight to an example: Example 3.4(b): A tank contains 80 gallons of water with 10 pounds of saltdissolved in it. Fluid with a 0.3 pounds per gallon salt concentration is being pumpedinto the tank at a rate of 7 gallons per minute. The fluid is continually mixed and,at the same time, the fluid is being drained from the tank at a rate of 7 gallons perminute. Letting A represent the amount of salt in the tank, in pounds, Sketch a graphof A as a function of time t. Label any values you can on the A axis.In the next two examples we will set up and solve the initial value problem for this situationanalytically, arriving at an equation that will give us the amount of salt at any time t.Before doing that it would be good to have some idea of what the behavior of the solutionwould be. We did this previously, in Example 1.2(j), but lets repeat the reasoning here.42The initial amount of salt in the tank is 10 pounds. Weknow that as time goes on the concentration of salt in thetank will approach that of the incoming solution, 0.3 poundsper gallon. This means that the amount of salt in the tankwill approach 0.3 lbs/gal 100 gal = 30 pounds, resultingin the graph shown to the right, where A represents theamount of salt, in pounds, and t represents the time, inminutes. 3010tANow lets set up an IVP and solve it: Example 3.4(c): Letting A represent the amount of salt in the tank, in pounds, givean initial value problem describing this situation.Taking the concentration of salt in the incoming fluid times the rate at which the fluid iscoming in, we get that salt is entering the tank at a rate of(0.3 lbsgal)(7 galmin) = 2.1 lbsmin.Now let A = A(t) be the amount (in pounds) of salt in the tank at any time t minutes.Then the concentration of salt in the tank is A80, so by the same sort of calculation thatwe just did the rate at which salt is leaving the tank is 7A80= 0.0875A lbsmin. The net rateof change of salt in the tank is the amount coming in minus the amount going out, or(2.1 0.0875A) lbsmin. But this quantity, being a rate of change, is also a derivative. Namelyit isdAdt, giving us the differential equationdAdt= 2.1 0.0875AWe also have the initial value A(0) = 10 pounds, so we have the initial value problemdAdt= 2.1 0.0875A , A(0) = 10 We will see differential equations like the one above in several contexts, so lets look athow to solve it by separation of variables. (One can solve it using an integrating factor aswell.) Doing this from scratch requires a u-substitution. This form of ODE comes up oftenenough that we should use the following formula instead, obtained by doing the substitution:1ax+ bdx =1aln | ax+ b |+ C (1)We can now use this result to solve the above IVP.43 Example 3.4(d): Solve the IVP from Example 3.4(b).Multiplying both sides of the ODE by dt and dividing by the quantity 2.10.0875A givesusdA2.1 0.0875A = dtTo use equation (1) we note that our left side is like the left side of (1), but with x = A,a = 0.0875 and b = 2.1. The formula then tells us that the left side integratesto 10.0875ln(2.1 0.0875A) + C. (We dont need absolute value because the quantity2.1 0.0875A is the rate at which the amount of salt is changing, and this is positive dueto the concentration of the incoming solution being higher than the initial concentration inthe tank.) Thus when we integrate both sides and combine the constants we have 10.0875ln(2.1 0.0875A) = t+ Cln(2.1 0.0875A) = 0.0875t+ C2.1 0.0875A = e0.0875t+C2.1 0.0857A = Ce0.0875t0.0875A = 2.1 + Ce0.0875tA = 24 + Ce0.0875tApplying the initial value A(0) = 10 we get 10 = 24+C, so C = 14 and the solutionto the IVP is A = 24 14e0.0875t. One should verify that this solution is reasonable by graphing it and seeing that the resultlooks like the graph obtained in Example 3.4(b).Newtons Law of CoolingNewtons Law of CoolingSuppose that a solid object with initial temperature T0 is placed ina medium with a constant temperature Tm, and let T = T (t) bethe temperature of the object at any time t after it is placed in themedium. The rate of change of the temperature T with respect to timeis proportional to the difference between the temperatures of the mediumand that of the object. That is,dTdt= k(T Tm)for some constant k > 0. Together with T (0) = T0, this gives us aninitial value problem for the temperature of the object.The medium that the object is placed in might be something like air, water, etc., andTm stands for temperature of the medium, and we will always consider situations for44which this temperature is constant. It is also sometimes called the ambient temperature.Note that if the temperature of the medium is greater than the temperature of the object,the rate of change of temperature must be positive, which is why k must be positive. Alittle thought will tell you that k must be positive if the temperature of the medium is lessthan the temperature of the object as well. Example 3.4(e): Suppose that a solid object with initial temperature 88F is placedin a medium with ambient temperature 50F, and after one hour the object has atemperature of 65F. Determine the equation for the temperature T as a function ofthe time t.Because the objects initial temperature 88F is higherthan the ambient temperature of 50F, it will cool afterbeing placed in the medium. Newtons Law of Cooling tellsus that it cools more rapidly at first, when the temperaturedifference between the object and the medium is large. Asthe object cools to temperatures closer to the ambient tem-perature, the rate of cooling decreases. This is shown bythe graph to the right.8850tTNow the initial value problem for this situation isdTdt= k(T 50) , T (0) = 88The differential equation becomesdTT 50 = k dt, and the method of Example 3.4(d)gives us T = 50 + Cekt before applying the initial condition. The initial condition givesus T = 50 + 38ekt. To determine k we apply T (1) = 65 to get k = 0.930, so thefinal solution is T = 50 + 38e0.930t. Electric CircuitsERLWe will work with two kinds of electrical circuits in this class.The first is a circuit that consists of a voltage source, a resis-tor and an inductor. The voltage source can be constant, orit can be variable, usually in an oscillating manner (so-calledalternating current). The resistor has a characteristic calledresistance, which is measured in units called ohms. The in-ductors characteristic is called inductance, which is measuredin henries. Although resistance and inductance could be vari-able, they always be constants in our considerations. We willuse E = E(t) for the voltage, R for the resistance and L forthe inductance. To the right is a schematic diagram of such acircuit.There is another electrical quantity that we will now consider, current. Basically, currentis something flowing around the circuit. We will use the symbol i for current, and it ismeasured in units called amperes. (Amps, for short.) We will usually think of our circuits45as having a switch as well, which is open (off) until time zero, when it is closed (turnedon). From that point on the current is (usually) changing, and is a function of time: i = i(t).In technical jargon, the voltage E(t) is called the input function or driving function.The current function i(t) is called the output function or response function.RL CircuitConsider an electric circuit as described above, with an applied volt-age E(t) volts (possibly a function of time) and constant resistance ofR ohms and constant inductance of L henries. The current i = i(t),in amperes, satisfies the differential equationLdidt+Ri = E(t)Methods for solving ODEs arising from the above are straightforward, so we wont look atexamples of how to solve them. Instead we will look first at the case where the voltage sourceE(t) is a constant function, and then well consider the situation where it is sinusoidal.Suppose that L = 14henry, R = 15 ohms, and E = 12 volts; in this case the ODE is14didt+ 15i = 12. If we separate the variables we obtaindi12 15i = 4 dt ordi48 60i = dt.Lets look at the first of these along with the separated equations arising in Examples 3.4(d)and 3.4(e):di12 15i = 4 dtdA2.1 0.0875A = dtdT70 T = k dtNote that these are all of the formdxa bx = c dt, where a, b and c are constants. Thisillustrates the fact thatphysical situations that seem to have nothing incommon lead to the same differential equation.We will see this principle in action again when we study second order ODEs.Transient and Steady-State SolutionsNow let us consider an RL circuit where again L = 14henry and R = 15 ohms, but forwhich E(t) = sin 3t. In this case we would not be able to separate the variables to obtainone of the above forms, so wed solve the equation using an integrating factor. In doing so,we would obtain the solutioni =463sin 3t+ Ce60twhere C is a constant that would be determined by an initial condition. Note that thesolution has two parts: The part 463sin 3t, which is periodic and goes on forever in the same way. This partof the solution is called the steady-state part or steady-state solution.46 The part Ce60t, which approaches zero as time goes on, so it dies out. Such asolution or part of a solution is called the transient solution or transient part ofthe solution.In practice, a machine with moving parts or an electrical circuit are often referred to assystems. When a system is turned on, it often exhibits a certain behavior as it startsup, which is often called the transient response of the system. Then it will settle in to asteady state behavior or response. Note that the ideas of transient and steady state solutionsonly make sense when the independent variable is time.For the situation where E has the constant value of 12 volts, the solution to the differentiaequation isi =1215sin 3t+ Ce60t,where C is again determined by the initial current in the circuit. We can see that here thesteady-state solution is i = 1215, where 12 is the voltage and 15 is the resistance; this saysthat in the long run the circuit will exhibit Ohms Law V = IR. This is because thecurrent approaches a constant value, and the inductor only affects the circuit when there ischange in the current, causing flux in the coil of the inductor.473.5 Some Theoretical ConcernsExistence and Uniqueness3.6 Chapter 3 Summary We can solve ODEs (and PDEs) in three ways: Analytically, which means paper and pencil methods that give exact algebraicsolutions. Qualitatively, which means determining the general behavior of solutions withoutactually finding function values. Results of qualitative methods are often expressedgraphically. Numerical methods which result in values of solutions only at discrete points intime or space. Results of numerical methods are often expressed graphically or astables of values. The two most commonly applicable methods for solving first order ODEs analyticallyare separation of variables and the integrating factor method. Separation of variables only works for equations that can be written dydx= g(x)h(y),and for which the antiderivativesg(x) dx anddyh(y)can be determined. The integrating factor method only applies to linear first order ODEs. Such ODEs canbe put in the formdydx+ p(x)y = q(x), and to carry out the integrating factor methodthe antiderivatives u =p(x) dx andeu q(x) dx must exist. Some applications of first order ODEs are population growth and radioactive decay,mixing problems, Newtons Law of Cooling problems, and RL electric circuits. Very different physical situations often result in the same differential equation. Suppose that the independent variable for an ODE is time, so the solution is a functionof time. Any part of the solution that goes to zero as time goes to infinity is called thetransient part of the solution, and any part of the solution that is constant or periodicis called the steady state part of the solution. It is not necessary that all parts of solutions exhibit transient or steady state behavior,but it is often the case that they do.483.7 Chapter 3 Practice ExercisesSection 3.21. Use separation of variables to solve each of the following differential equations. Dontsolve for y.(a)dydx= x sec y (d) y = y2x+ 3(b) dx+ x3y dy = 0 (e) x2 dy = ey dx(c) x2 + y4dydx= 0 (f) y =5x+ 3y2. Solve each of the following initial value problems. DO NOT solve for y, and giveconstants in exact form.(a) y = xy, y(1) = 3 (c)dydxeyx= 3, y(0) = 2(b) xdxdt+ 5t = 3, x(2) = 4 (d) y = y4 cos t, y(0) = 23. Some of the following initial value problems can be solved by separation of variables.Solve the ones that CAN be solved by that method. DO solve for y and give constantsin exact form again.(a)dydx 3y = 0 , y(0) = 4 (b) xdydx y = x , y(1) = 2(c) y 4xy = 0 , y(0) = 2 (d) y 2x = xy , y(2) = 5(e)dydx y = e3x , y(0) = 4 (f) dydx=y 1x+ 3, y(1) = 34. (a) Solve the initial value problem y 2xey = ey , y(0) = 0.(b) Solve the initial value problem y 2xey = ey , y(1) = 3. Give the exactform for the unknown constant.5. (a) Solve the differential equation y + 2ty = 0. You should get ln y = t2 + C.(b) We now want to get y as a function of t. e both sides of the equation and usethe fact that elnu = u. Use also the facts that xa+b = xaxb and e raised to aconstant power is yet another constant. You should now have a family of solutionsto the differential equation.(c) Use the initial condition y(0) = 7 to determine the value of the arbitrary constant.You now have a solution to the initial value problem.49Section 3.31. Solve the IVPdydx 3y = e4x, y(0) = 1.2. Use an integrating factor to solve y + 2y = 0.4e2t. (Note that y is now a functionof t.) Solve for y.3. (a) Use an integrating factor to solvedydx 12y = 0.(b) Solve the same ODE by separation of variables. Solve for y and compare with youranswer to (a) (and take any action that might be suggested by this comparison!).4. Solve the IVP y + 2y = 0.4e2t, y(0) = 3. (See Exercise 2.)5. Solve the IVPdydx 12y = 0, y(0) = 32.6. (a) Solve the ODE y 5y = 3 cos 2t. Use your formula sheet to avoid some verymessy integration.(b) Solve the IVP y 5y = 3 cos 2t, y(0) = 4.7. (a) Solve the ODEdydt+ 3y = t2 + 5t 1.(a) Solve the IVPdydt+ 3y = t2 + 5t 1, y(0) = 2.8. In this exercise you will see another method for solving the ODE from Exercise 6. Thismethod will be used later when we solve second order ODEs.(a) The equation y5y = 0 is the homogeneous equation associated with y5y =3 cos 2t. Substitute y = Cert into the homogenous equation to determine whatvalue r must have in order for y = Cert to be a solution. For that value of r,y = Cert is called the homogeneous solution to y 5y = 3 cos 2t.(b) Find the values of A and B for which y = A sin 2t + B cos 2t is a solution toy 5y = 3 cos 2t. Do this as follows:i. Find y and substitute it and y into the differential equations to get anequation involving sines and cosines of 2t.ii. Combine the like terms on the left side of the equation to get only one sine termand one cosine term.iii. You will need to note that on the right side of your equation 3 cos 2t is thesame as 3 cos 2t + 0 sin 2t. Equate the coefficient of cos 2t on the left sidewith the coefficient of cos 2t on the right side to get an equation involvingthe unknowns A and B. Then repeat for sine to get another equation withA and B.iv. Solve two equations for the two unknowns A and B. The resulting y =A sin 2t+B cos 2t is called the particular solution to y 5y = 3 cos 2t.(c) Write the sum of the homogeneous and particular solutions. This is known as thegeneral solution, and should match what you found in Exercise 6(a).9. The IVPs from Exercises 3(b) and 3(e) of Section 2.2 couldnt be solved by separationof variables, but they can be done with integrating factors. You will do them here.50(a) Solve the IVPdydx y = e3x , y(0) = 4.(b) Solve the IVP xdydx y = x , y(1) = 2. Begin by multiplying through by 1x.Section 3.41. When a person takes a medication, the amount in their body decreases exponentially,in the same way that a radioactive element decays. Suppose that a person takes 80grams of some medication, and that we somehow know (???) that 12 hours later theystill have 23 grams in their system.(a) Give the initial value problem for this situation, using A for amount in grams andt for time in hours.(b) Solve the IVP. Your answer should still contain an unknown constant k.(c) Determine k and give the amount function A.2. An underground storage tank contains 1000 gallons of water with 87 pounds of contam-inant in it. At some time we will call time zero, clean water is pumped into the tankat a rate of 5 gallons per minute and the thoroughly mixed solution is pumped out atthe same rate.(a) Set up an IVP for this situation, using A for the amount of contaminant, inpounds, and t for time, in hours.(b) Solve the IVP.(c) Determine when the amount of contaminant has decreased to five pounds.3. (a) Suppose that a solid object is placed in a medium with ambient temperature 70F.Solve the differential equation. The constant k will be unknown for now, andthere will be another constant that arises as well.(b) Suppose that the initial temperature of the object is 32F. Solve for the constantthat arose in solving the ODE.(c) Now suppose that after one hour the object has a temperature of 58F. Use thisinformation to determine the constant k. Give units with your answer.(d) What is the steady-state solution? What is the transient solution?4. (a) Suppose that the voltage in resistor-inductor series circuit is supplied by a 12 voltbattery, so E(t) = 12. The inductance of the circuit is 12henry, and theresistance is 10 ohms. At time zero the current in the circuit is zero. Find thecurrent function i(t) by solving the initial value problem just described.(b) Now suppose that the voltage is variable, with equation E(t) = 10 sin 2t, and theinitial current is zero. Solve the IVP.(c) Give the transient and steady-state parts of your solution. (Make it clear of coursewhich is which!)5. In general, the solution to the differential equation for Newtons Law of Cooling isT (t) = Tm + (T0 Tm)ekt ,where T0 is the initial temperature.51(a) What happens if T0 = Tm?(b) What is the steady-state solution?(c) What is the transient solution?Section 3.4 Solutions1. (a) T = 70 Cekt (b) C = 38, so T = 70 38ekt(c) k = ln 196(d) Steady-state: 70 Transient: 38e ln 196 t2. (a) i = 65 65e20t or i = 65(1 e20t)(b) i = 5101(10 cos 2t+ sin 2t) 50101e20t or i = 50101cos 2t + 5101sin 2t 50101e20t(c) The transient part is 50101e20t and the steady-state part is 5101(10 cos 2t+sin 2t).3. (a) If T0 = Tm the temperature will not change because the initial temperature ofthe object will be the same as the temperature of the medium; the solution will becompletely steady-state.(b) The steady-state solution is Tm.(c) The transient solution is (T0 Tm)ekt524 Qualitative and Numerical Methods: First Order4.1 IntroductionLearning Outcome:4. Use qualitative methods to determine the general behavior of a so-lution to a first order IVP. Use a numerical method to generate anumerical approximation of the solution to a first order IVP.Performance Criteria:(a) Sketch solution curves to an ODE for different initial values.Given a set of solution curves for a first order ODE, select theone having a given initial value.(b) Sketch a small portion of the direction field for a first orderODE.(c) Given the direction field and an initial value for a first orderIVP, sketch the solution curve.(d) Determine whether an ODE is autonomous.(e) Create a one-dimensional phase portrait for an autonomousODE.(f) Determine critical points/equilibrium solutions of an au-tonomous ODE, and identify each as stable, unstable or semi-stable.(g) Sketch solution curves of an autonomous ODE for various initialvalues.(h) Given a first order IVP and a step size, generate a numericalsequence of solution values.(i) Given the direction field for a first order ODE and an initialvalue, sketch the solution generated by Eulers method.All of our efforts so far in this course have been focused on solving first order ODEsor IVPs by what we call analytic methods; this has resulted in what we call analyticsolutions, meaning equations of functions or relations that give exact solutions to ODEs orIVPs. When can find such a solution, we say that we can solve the ODE or IVP analytically.Now the fact of the matter is that there are many differential equations of interest toscientists, engineers and mathematicians that cannot be solved analytically! So what are weto do? Well, there are two approaches that can be taken when working with such situations: Qualitative Methods Sometimes it is sufficient to know how the solution to an ODEor IVP will behave, without needing to know what the actual solution is. For example,when investigating a mixing situation we might only need to know the initial value and53the limiting value that the solution approaches. Maybe we just need to know whetherthe solution will have a nonzero steady-state part. We will see that the general behaviorof solutions to an ODE can sometimes be determined from the ODE, without actuallyhaving to find a solution. We will often express this behavior graphically, but withoutlabeling values on one or both axes. Numerical Methods Suppose that we have an exact solution y(t) to an IVP, wherey represents something like the temperature of an object or the current in an electricalcircuit. Then for any time t we can compute a value of y that gives us the temperatureof the object or current at that particular time. There are methods that can give usvery good approximations of y for various times t, even when we dont have anexplicit formula for finding its value.In this chapter we will get an introduction to qualitative and numerical methods for ODEs.544.2 Solution CurvesPerformance Criterion:4. (a) Sketch solution curves to an ODE for different initial values.Given a set of solution curves for a first order ODE, select theone having a given initial value.Consider this situation, which we considered previously: A tank contains 80 gallons ofwater with 10 pounds of salt dissolved in it. Fluid with a 0.3 pounds per gallon salt con-centration is being pumped into the tank at a rate of 7 gallons per minute. The fluid iscontinually mixed and, at the same time, the fluid is being drained from the tank at a rateof 7 gallons per minute.A quick computation reveals that the initial concentration of the solution in the tankis 0.125 pounds per gallon, less than the concentration of the fluid that is replacing it.Therefore the concentration of the fluid in the tank will increase, but it can never exceed theconcentration of the incoming fluid. If all of the fluid in the tank had the concentration ofthe incoming fluid, there would be (0.3)(80) = 24 pounds of salt. If we were then to graphthe amount of salt in the tank as a function of time we would get the solid curve graphedbelow. The limit of the amount of salt in the tank is 24 pounds.604020AmountofSalt(lbs)TimeNow suppose that the tank had A0 = 60 pounds of salt initially, giving an initialconcentration of 0.5 pounds per gallon, higher than the concentration of the incoming fluid.Therefore the amount of salt in the tank will decrease, with a limit of 24 pounds again. Thedashed line on the graph above shows the amount of salt as a function of time, for the initialamount of 40 pounds. The dotted curve is the solution curve for the initial amount of 24pounds, and the dotted line is both the asymptote for the other solutions, as well as thesolution when the initial amount of salt in the tank is 24 pounds.55As we saw before, the situation with the tank can be modeled with a differential equation.The various curves graphed on the previous page are called solution curves for the ODE.Each curve is associated with a particular initial value. Notice that none of the solutionscurves cross each other; this is not always the case, but will be for most of the ODEs thatwell look at. For an initial salt amount of 15 pounds, the solution curve will lie midwaybetween the curves for initial amounts of 10 and 20 pounds, without crossing either.In the exercises you will use your calculator or a graphing utility to plot solutions curvesfor ODEs.564.3 Direction FieldsPerformance Criteria:4. (b) Sketch a small portion of the direction field for a first orderODE.(c) Given the direction field and an initial value for a first orderIVP, sketch the solution curve.Consider again differential equationdydx+ y = x. We can rearrange the equation to getdydx= x y. Now remember that by solving the differential equation we mean findinga function y = y(x) that makes the equation true. Recall also that when considering thegraph of a function, the derivative of the function at some point the slope of the tangentline to the graph of the function at that point. So the second equation gives us a formulafor finding the slope of the tangent line to the unknown function y(x) at any point (x, y).To be more specific, consider the point (3, 1). Theequationdydx= x ytells us that at that point the slope of the tangent line tothe solution curve will bedydx(3,1)= 3 1 = 2So the tangent line through the point (3, 1) has slope 2.The dotted line below is that line; we will just keep thesmall part of it that actually goes through the point.4-22 Example 4.3(a): Find slopes for the remaining grid points, for dydx= x y.It is often easiest to determine slopes not by going point topoint, but to find all points where the slope is the same.For this equation, the slope will be zero at every point wherex = y. Similarly, the slope will be one at all the points wherex is one unit larger than y; for the above grid those are thepoints (4, 3), (3, 2), (2, 1) and (1, 0). Similarly, the slopewill be two at the points where x is two greater than y:(0,2, (1,1), (2, 0), (3, 1) and (4, 2). The slope linesfor slopes zero, one and two are plotted on the grid shownto the right. The remaining slopes can be seen on the nextpage. 4-2257The result of what we have been doing is something called a direction field or slopefield; the completed direction field can be seen below and to the left. Direction fields are away of studying the behavior of solutions to differential equations without actually solvingthe equations analytically. The slope lines that we have drawn in on the direction field arenot all that are possible; such an slope line exists for every point in the plane. Given adirection field and an initial value, we can sketch solution curves by drawing curves startingat the initial value point that are tangent to the imagined slope lines at all points that acurve goes through. Below and to the right you can see the solution curves correspondingto the initial values y(0) = 2, y(0) = 1 and y(0) = 3.4-224-22584.4 Phase Portraits and StabilityPerformance Criteria:4. (d) Determine whether an ODE is autonomous.(e) Create a one-dimensional phase portrait for an autonomousODE.(f) Determine critical points/equilibrium solutions of an au-tonomous ODE, and identify each as stable, unstable or semi-stable.(g) Sketch solution curves of an autonomous ODE for various initialvalues.Recall that a first order ODE is called autonomous if it can be written in the formdydx= f(y). That is, when we get the derivative alone on the left side of the equation, theright hand side is a function of only the dependent variable;dydt= y2 2y is an example ofan autonomous ODE. Note thatdydt= 0 whenever y = 0, so if we had an initial conditionof y(0) = 0, then the value of y would never change because the rate of change withrespect to time is zero. Therefore the solution to the IVPdydt= y2 2y, y(0) = 0is the constant function y = 0. We have solved the IVP without doing any calculations!Now suppose that, for the same ODE, y(0) = 2. We see that when y = 2 we again havedydt= 0, so the value of y will again not change. This means that y = 2 is the solutionto the IVPdydt= y2 2y, y(0) = 2The graph below and to the right shows thetwo solution curves that we have obtained so farfor the ODEdydt= y2 2y, for the initial val-ues y(0) = 0 and y(0) = 2. We will callconstant solutions like those two equilibrium so-lutions; the word equilibrium essentially meaningunchanging. The question that should occur toyou is What happens for other initial values ofy? With a little thought we should be able tofigure that out. There are three key observationswe can make that will help answer the question:0120 1 2 3y(0) = 2y(0) = 0ty The direction field depends only on y alone, so for any given value of y the sloperemains constant.59 The right hand side of the ODE can be factored to y(y 2). From that we see thatif y < 0 both y and y 2 will be negative, so dydt= y(y 2) will be positive.(A negative times a negative is positive! Youve got to like the simplicity of that bitof mathematics!) Therefore any solution with an initial value less than zero will beincreasing. When 0 < y < 2, y is positive and y 2 is negative, so dydtis negative.Finally, when y > 2 we havedydt> 0 because both y and y 2 are positive wheny > 2. Whether positive or negative, the value of dydtapproaches zero as y gets nearerto either zero or two. Therefore the direction field lines become flatter (closer tohorizontal) for values of y close to zero and two.From these observations we can deduce that the direction field fordydt= y2 2y has theappearance shown below and to the left. The direction field with solutions corresponding tofour different initial conditions is shown in the center below.0120 1 2 3 xy0120 1 2 3y(0) = 2.5y(0) = 2y(0) = 1y(0) = .5xy20yWe will summarize the information in the three bullets above with something called aphase portrait, shown above and to the right. (Technically it is a one-dimensional phaseportrait. Those of you taking the second term of this course may see two-dimensional phaseportraits.) The vertical line indicates y values, with the two critical points zero and twoindicated. The critical points divide the line into three intervals, and the arrow in eachinterval indicates whether y is increasing or decreasing, in the particular interval, as timeincreases.Look again at the direction field in the middle above, with the four solutions curves drawnin. Note that solutions with an initial y value less than two (including less than zero) alltend toward the constant solution y = 0, as the phase portrait tells us they will. Whenthis occurs we say that the solution y = 0 is a stable equilibrium solution. You cansort of think that if we have an initial condition of zero the solution will be zero, and if webump off from zero a bit with our initial condition, the new solution obtained will tendback toward zero as time goes on.On the other hand you can see that if y starts with the value two it will remain two, butif it starts at any value close to, but not equal to, two the solution will diverge away fromtwo. (Again, look at the phase portrait to see how it indicates this.) For this reason we call60the solution y = 2 an unstable equilibrium solution. Other language you might hearis that y = 0 is a stable critical point on the phase diagram, and y = 2 is an unstablecritical point. We will indicate stable critical points with solid dots, and unstable criticalpoints with open circles. (We will also use open circles for semi-stable equilibria, which youwill see later.)Again, the entire analysis you have just seen is based on the fact that the ODE is au-tonomous. Two points of interest here are Autonomous first order ODEs are not just a curiosity - they occur naturally in manyapplications. Autonomous ODEs can be difficult or impossible to solve. However, an analysis like wejust did can make it very easy to determine how solutions to such an equation behave,in a qualitative sense.It is important to be able to recognize autonomous differential equations. Example 4.4(a): Determine which of the following first order ODEs are autonomous:dydx= y2 x y = 2y 1 dydt= t2 5t+ 1 xdxdt= x+ 1Clearly the first equation is not autonomous and the second is. The third and fourthequations might be a bit confusing, as the variables are no longer x and y. In the thirdequation the dependent variable is y and the independent variable is t. Because the righthand side is a function of only the independent variable t, the equation is not autonomous.In the fourth equation the dependent variable is x, the independent variable is t, andthe equation can be rewritten asdxdt=x+ 1x. The right hand side is then a function ofx alone, so the equation is autonomous. If we can determine the phase portrait for an autonomous ODE, we then have a prettygood idea what all solutions to the ODE look like, without having to go to the trouble ofcreating a direction field. The next example shows how this is done. Example 4.4(b): Sketch the phase portrait for dydt= y3 + 6y2 9y and use it tosketch solution curves for the initial conditions y(0) = 4, y(0) = 2 and y(0) = 1.Identify each equilibrium solution as stable or unstable.We factor the right hand side of the ODE, starting by factoring y out:dydt= y3 + 6y2 9y = y(y2 6y + 9) = y(y 3)2From this we can see that the equilibrium solutions are y = 0 and y = 3. Testing valuesof y in each of the three intervals (, 0), (0, 3) and 3) gives us the following: When y < 0, dydt> 0 When 0 < y < 3, dydt< 0 When y > 3, dydt< 0This gives us the phase portrait shown to the left at the top of the next page, which indicatesthat there are equilibrium solutions of y = 0 and y = 3. If y(0) < 0 the solution isincreasing, but will approach the equilibrium solution y = 0; if 0 < y(0) < 3 the solutiondecreases toward y = 0. If y > 3 the solution also decreases, but toward y = 3. These6130y10234 y(0) = 4y(0) = 2y(0) = 1tybehaviors are shown above and to the right, for the three solutions with the given initialvalues. The solution y = 0 is a stable equilibrium solution, and the solution y = 3 isneither stable nor unstable, but is what we call a semi-stable equilibrium solution. 624.5 Numerical SolutionsPerformance Criteria:4. (h) Given a first order IVP, a recursion formula, and a step size,generate a numerical sequence of solution values.(i) Given the direction field for a first order ODE and an initialvalue, sketch the solution generated by Eulers method.The point of numerical methods is that they can be used to solve IVPs that cannot besolved analytically. (We will sometimes use numerical methods with IVPs that we CANsolve analytically, so that we can compare numerical solutions with analytical solutions.)So numerical methods are needed when we want a function y = y(t) that we cant usealgebra and calculus methods to find an equation for. The use of numerical methods (bothfor solving IVPs and for other things) is a huge part of mathematics, science and engineering.Here you will be introduced to some methods for finding solutions to first order initial valueproblems. The simplest method is called Eulers method; the name comes from LeonhardEuler (pronounced oiler), a very prolific Swiss mathematician who lived in the 1700s. Youwill see that this is not a very good method, but we will study it because it is simple toimplement, it illustrates what are called iterative techniques, and the ideas behind it are thesame ones that make some better methods work.Before actually finding a numerical solution to an IVP, lets go over what a solution willconsist of, and the special notation well be using in this section. Suppose that we have anIVPdydt= f(t, y), y(t0) = y0;the solution to the IVP is some unknown function y = y(t). Consider the sequence of timest = 0, 0.2, 0.4, 0.6, 0.8, ... seconds, which well denote as t0, t1, t2, t3, t4, .... Note that thesetimes are spaced apart by the increment of 0.2 seconds - we could just as well have used anincrement of 0.1 seconds, which would give us insteadt0 = 0, t1 = 0.1, t2 = 0.2, t3 = 0.3, ...The increment used is denoted by the letter h, so for the first sequence of times listed aboveh = 0.2 and for the second sequence h = 0.1. We will refer to the value of h as the stepsize, meaning that time advances in steps of size h. Regardless of the value of h, wealways start with a known value of t0 (usually zero) and h, then we use the relationst1 = t0 + h, t2 = t1 + h = t0 + 2h, t3 = t2 + h = t0 + 3h, t4 = t3 + h = t0 + 4h, ...to determine the other time values.Lets continue this discussion for the first sequence of times, t0 = 0, t1 = 0.2, t2 = 0.4,t3 = 0.6, and so on. If we knew the exact solution y = y(t), we could compute the valuesof y(0.2), y(0.4), y(0.6), ... (we already know the value of y(0) = y0). Because we dontknow the exact solution, we instead come up with a sequence y1, y2, y3, ... wherey1 y(t1) = y(0.2), y2 y(t2) = y(0.4), y3 y(t3) = y(0.6), ...63Each value in the sequence must be determined before the following term can be found, sowe use the known value of y0 to find y1, then use that value of y1 to find y2, use thatto find y3, and so on, as far as we wish to have values for y. It should be noted that tofind any value in the y sequence we always use the previous value of y, but might alsouse other prior y values and perhaps some time values as well.Any method that proceeds sequentially like this is called an iterative method. To carryit out we need an iteration formula that tells us how to obtain a y value from previousy values and perhaps previous time values. The simplest such formula, as mentioned before,is the one for Eulers method:yn+1 = yn + hdydt(tn,yn)= yn + hf(tn, yn) (1)Lets think about what this tells us. It says that to get the next y value yn+1 we takethe current y value yn and add the product of the increment h and the derivativeevaluated at our current t and y values. But because the ODE can be put in the formdydt= f(t, y), we use the final expression above to compute each new y value.You may (understandably) be totally confused by now! Lets look at a specific example. Example 4.5(a): Use the Eulers method equation (1) to find y0, y1, y2, y3 and y4 forthe initial value problemdydt= t y, y(0) = 2,using a step size of h = 0.2.First we identify t0 = 0 and y0 = 2 (from the initial value), t1 = 0.2, t2 = 0.4, ... fromt0 and h, and f(t, y) = t y (from the ODE). By (1),y1 = y0 + hf(t0, y0) = 2 + (0.2)(0 2) = 1.6We then use this value and (1) again to obtainy2 = y1 + hf(t1, y1) = 1.6 + (0.2)(0.2 1.6) = 1.32Continuing, we gety3 = y2 + hf(t2, y2) = 1.32 + (0.2)(0.4 1.32) = 1.136,y4 = y3 + hf(t3, y3) = 1.136 + (0.2)(0.6 1.136) = 1.0288.More y values, if desired, could be obtained by continuing in the same way. You might be wondering why the formula yn+1 = yn + hdydt(tn,yn)works. Recall thedefinition of the derivativedydt= limh0y(t+ h) y(t)h.This says that as h approaches zero, the quantity on the right approaches the derivative.Thus, for small values of h,dydt y(t+ h) y(t)h.64Multiplying both sides by h and solving for y(t+ h) gives usy(t+ h) y(t) + hdydt,and if t = tn, then t+ h = tn+1; substituting these into the above gives usy(tn+1) y(tn) + h dydyt=tn.But y(tn+1) yn+1 and y(tn) yn, and our expression f(t, y) for the derivative maycontain both t and y, so we must evaluate the derivative at (tn, yn), giving us the finalresultyn+1 = yn + hdydt(tn,yn)= yn + hf(tn, yn). (1)This formula can also be derived using something called Taylor polynomials, whichmany of you have not yet seen. For this reason we will not go into it here but, if youare interested, you can see the derivation of Eulers method from a Taylor polynomial inmost any introductory differential equations book. Using Taylor polynomials has a coupleadvantages: Their use can lead to better formulas than the one for Eulers method. Their use can give us a bound for the error that occurs when using a numerical method.The distinction is a bit subtle and often blurred, but when we just use an iterative formula,as we will, we are performing a numerical method. When we try to analyze our method forhow accurate it is, we are doing something called numerical analysis.There is also a geometric explanation for how and why Eulers method works. Supposethat we have the initial value problemdydt= f(t, y), y(t0) = y0with solution y = y(t). Figure 4.5(a) on the next page shows the solution curve (which isreally unknown) through (t0, y0) and the points on that curve corresponding to two timest1 and t2 as described previously. We can think of generating the approximate numericalsolution consisting of a sequence y0, y1, y2, y3, ... as follows: We are given the initial value y0. We construct a line through (t0, y0) that is tangent to the solution curve - its slope isthe derivative at (t0, y0). See Figure 4.5(b). The point on that line with t-coordinatet1 is y1 (Figure 4.5(c)). We construct a line through (t1, y1) that is tangent to the solution curve passingthrough that point. Its slope is the derivative at (t1, y1), and the point on the line witht-coordinate t2 is y2. See Figure 4.5(d). Continue in this manner to obtain the values y3, y4, ....65tyt1ht2hy(t1)y(t2)(t0, y0)(t1, y(t1))(t2, y(t2))solution y = y(t)Figure 4.5(a)tyt1 t2(t0, y0)(t1, y(t1))(t2, y(t2))slope =dydt(t0,y0)Figure 4.5(b)ty(t0, y0)(t1, y(t1))t1 t2(t1, y1)Figure 4.5(c)ty(t0, y0)(t1, y(t1))(t2, y(t2))t1 t2(t1, y1)(t2, y2)slope =dydt(t1,y1)Figure 4.5(d)66We should make a couple observations/comments: It should be clear from Figure 4.5(b) that a smaller value of h would give us a valueof y1 that is closer to the true value y(t1). The error in the approximate solution does not necessarily grow as t gets larger.The following summarizes the idea of how Eulers method works.Eulers MethodTo solve the first order IVPdydt= f(t, y), y(t0) = y0 for some x in-crement h, use the recursive relationyn+1 = yn + hdydt(tn,yn)= yn + hf(tn, yn) (1)as follows:1) y0 = y(t0)2) Use the above relation with n = 0 to find y1.3) Use the relation with n = 1, remembering that t1 = t0 + h, tofind y2.4) Repeat as long as desired, remembering that tn = t0 + nh.There other methods besides Eulers method for generating approximate solutions toIVPs. The general method for applying them is the same as Eulers method, but the iterationformula (1) is different for them. You will see some of those other methods in the exercises.674.6 Chapter 4 Summary684.7 Chapter 4 Practice ExercisesSection 4.21. The general solution to the ODEdydx+ y = x is y = x1+Cex. Graph the solutioncurves for each of the following initial values, for 3 x 6. Use your calculator ifyou wish, then sketch the graphs.(a) y(0) = 2 (b) y(0) = 0 (c) y(0) = 1 (d) y(0) = 32. On your graph from Exercise 1, sketch what you think the graphs for initial conditionsy(0) = 2, y(0) = 1 and y(0) = 3 would look like. Then graph them on yourcalculator to check yourself.3. (a) The graph below shows some solutions todydx= xy2. Label each that you canwith its initial value y(0) = .(b) The solution to the ODE is y =2x2 + C. Use this to help identify the initial valuesfor portions of curves that you couldnt make that determination for in part (a).(All parts not labeled so far share initial values with those identified in part (a).)45-4-44. The graph of some solution curves for a differential equation are shown below and tothe left.(a) Which curve represents the solution for the initial condition y(3) = 1?(b) Sketch the solution curve for the initial condition y(1) = 3.5. Suppose that a group ofN0 individuals is put in an environment that can only supportKindividuals, and suppose that the growth rate of the population without any restrictionswould be r percent (in decimal form!) per year. Then the population N at any time tyears is given byN =N0KN0 + (K N0)ert69The value N0 is called the initial population, K is called the carrying capacity.Suppose that for some population the carrying capacity is 100 and the growth rate is20%. Graph the functions N for the initial populations below, for zero to forty years.Your graph will need to go up to at least 150 individuals.(a) N0 = 20 (b) N0 = 150 (c) N0 = 100 (d) N0 = 06. Think about the graph you got in the previous exercise, and make sure that you un-derstand (from a population growth point of view) why each curve looks the way itdoes.Section 4.3For Exercises 1-3, plot the direction field for the given ODE at integer coordinates over thevalues given for each variable.1.dydt=12xt, 1 t 2, 2 y 2.2.dydx= x2 2x. 0 x 4, 2 y 4.3.dydt= x+ t. 2 t 2, 2 y 2.4. On the direction field below and to the left, sketch the solution curves going throughthe given points.(a) (4, 5) (b) (2,2) (c) (6,4)- 6 - 4 - 2 0 2 4 6- 6- 4- 20246xyExercise 4- 3 - 2 - 1 0 1 2 3- 3- 2- 10132xyExercise 55. On the direction field above and to the right, sketch the solution curves going throughthe given points.(a) (2, 0) (b) (1.5,1) (c) (3, 0)Section 4.4701. Determine which of the following first order ODEs are autonomous.(a)dydt 2y = 0 (b) dydx+ xy = 1 (c)1ydydx+ y = 3(d) y = y2 7y + 10 (e) 1xdydx+ y = 1 (f)dxdt+ 6x2 = x3 + 9x2. For each of the ODEs in Exercise 1 that are autonomous, draw a phase portrait on a single separate graph, sketch in all equilibrium solution curves, and one solutioncurve with an initial value in each interval of the real line created by the values ofthe equilibrium solutions give all equilibrium solutions and, for each, tell what kind of equilibrium it is3. In (a) below, the graph of some solution curves to an ODE are given. (b) and (c) arephase portraits for two other ODEs.026yt(a)41y(b)520y(c)(a) For the solution curves in (a) above, identify each equilibrium solution and tellwhether it is a stable equilibrium, unstable equilibrium, or semi-stable equilibrium.(b) Repeat (a) for the phase portrait in (b) above.(c) Repeat (a) for the phase portrait in (c) above.4. (a) Draw the phase portrait for the ODE with solution curves shown in (a) of theprevious exercise.(b) Draw some solution curves for the ODE whose phase portrait is shown in (b) of theprevious exercise. Be sure to include curves for the equilibrium solutions and at leastone solution with initial value in each of the intervals created by the equilibriumpoints.5. (a) Give an ODE, in factored form, that could have the solution curve graph shown inExercise 3(a).(b) Give an ODE, in factored form, that could have the phase portrait shown in Exercise3(b).(c) Give an ODE, in factored form, that could have the phase portrait shown in Exercise3(c).71Section 4.5For Exercises 1 and 2 you will be considering the ODEdydt= y t.1. In this exercise you will use Eulers method to construct a numerical solution for aninitial value of y(0) = 2 and a step size of h = 0.2(a) Build a table with three columns, the first for n = 0, 1, 2, ..., the second for tn,and the third for yn. Complete the second column for t = 0 to t = 1, and finishthe top row by putting in the initial value y0 = y(0).(b) Use the recursion formula yn+1 = yn + hf(tn, yn) and the values in the first rowof your table to find y1. Put the result in your table. Then continue computingyn values up to time one. Do not round any of your answers and use alldigits of each yn to find yn+1.(c) Solve the IVP, using the integrating factor method to solve the ODE.(d) Add a fourth column to your table for the exact solution values y(t0), y(t1), y(t2),... Compute those values, rounding to four places past the decimal using youranswer to (c), and add them to your table.(e) The percent error is percent error =100 | approximate value actual value || actual value | .Find the percent error for each of your approximations, and add them to the tablein a fifth column. Round to the nearest hundredth of a percent.2. (a) Find a recursion formula for the given ODE and a step size h = 0.05 as follows:For the formula given in 1(b), insert 0.05 for h and apply the function f(t, y) =y t to tn and yn. Distribute the 0.05 and combine like terms. Your final resultshould be something of the form yn+1 = ayn + btn, for some constants a and b.(b) Repeat Exercise 1, but for an initial value of y(0) = 1.4, a step size of h = 0.05,and from t = 0 to t = 0.2, using your formula from part (a) to find eachsuccessive yn. Round both the yn and y(tn) values to four places pastthe decimal, and the percent error to two places past the decimal.3. Consider the IVPdydt+ ty = 0, y(0) = 2,(a) Using Eulers method, which is the recursion formula given in Exercise 1(b), findthe first four approximate solutions (including y0, which is really exact) to theIVP using a step size of h = 0.1. Compile your results in a table, like you havebeen doing.(b) Use separation of variables to solve the IVP. Use your answer to add a column toyour table for the exact values y(tn).So far you have approximated solutions to an ODE using an iterative method called Eulersmethod. Now you will use two other methods that give better approximations. For any firstorder equation, written in the formdydt= f(t, y), Eulers method uses the recursion formula72given in Exercise 1(b) to find each successive solution approximation from the previousapproximation.bEulers method is the simplest of a variety of methods called Runge-Kuttamethods. The derivation of these methods is beyond the scope of our course, but welluse two of them without necessarily understanding their derivations. The first methodwell look at is called the midpoint method, which has a fairly straightforward geometricinterpretation that well go over in class. Here is the formula for the method:73yn+1 = yn + hf(tn +h2, yn +h2f(tn, yn))(1)4. Lets use the midpoint method to obtain an approximate solution to the initial valueproblemdydt= y t, y(0) = 2,with a step size of h = 0.2, from t = 0 to t = 1. As in Exercise 2, you will firstderive a recursion formula specific to this problem.(a) Simplify the expression yn +h2f(tn, yn) for our particular ODE and step size, inthe manner you did for Exercise 2(a).(b) Insert your answer from (a) into the appropriate place in (1) and simplify (1) forour ODE. Go until you obtain something of the form yn+1 = ayn + btn + c forsome constants a, b and c.(c) Use your answer to (b) to fill out a table of values like you did for Exercise 1,up to time t = 1. As before, include a column at the right for exact values ofthe solution, rounded to four places past the decimal. (Remember that the exactsolution for this IVP is y = t+ 1 + et.)(d) How do the percent errors for this method compare with those for Eulers method?The most commonly used Runge-Kutta method uses a set of equations to obtain each suc-cessive approximation. These could all be combined into one equation like (1), but it wouldbe very cumbersome. Instead we useyn+1 = yn +16(k1 + 2k2 + 2k3 + k4) ,wherek1 = hf(tn, yn), k2 = hf(tn +h2, yn +12k1), k3 = hf(tn +h2, yn +12k2), k4 = hf(tn+1, yn+k3)The values k1, k2, k3 and k4 must be computed each time you iterate to find a new yn+1.5. Now well use the above method to approximate a solution to the same IVPdydt=y t, y(0) = 2.(a) Find simplified forms of k1 through k4 for our particular ODE, again with a stepsize of h = 0.2.(b) Build a table of values with columns for n, tn, yn, k1, ... , k4 and fill it outup through time t = 1. As before, round to four places past the decimal whennecessary.(c) Compare your approximate values with the exact values in your table for Exercises1 and 4.745 Second Order Linear Constant Coefficient ODEs5.1 IntroductionLearning Outcome:5. Solve second order linear, constant coefficient ODEs and IVPs. Un-derstand the nature of solutions to such ODEs and IVPs.Performance Criteria:(a) Evaluate a differential operator for a given function.(b) Solve a second order, linear, constant coefficient, homogeneousODE.(c) Find a particular solution to a second order linear, constant coef-ficient ODE using undetermined coefficients.(d) Solve a second order linear, constant coefficient IVP.(e) Given minimal, but sufficient, information about an ODE of theform ay + by + cy = f(t) and its general solution, determinewhether a function is a solution to ay + by + cy = f(t) oray + by + cy = 0.(f) Determine, from the definition, whether two functions are linearlyindependent.(g) Use the Wronskian to determine whether two solutions to a secondorder linear ODE are independent.(h) Given one solution to a second order ODE, use reduction of orderto find a second solution.Most of our study so far has focused on solving or analyzing first order ODEs and theirassociated initial value problems. We have seen how such equations model RL electriccircuits, mixing problems, growth of populations and decay of radioactive elements, and thetemperature of a solid object in some constant-temperature medium.There are other things we would like to model that require not first order ODEs, butsecond order. One example was given in Section 1.3; there we saw how such a system ismodeled by a second order ODE. In Chapter 6 we will investigate such systems in meredepth, along with certain electrical circuits. The same type of differential equations andinitial value problems model both situations. In this chapter we will see how to solve suchequations, without concerning ourselves initially with what such equations model.Recall that an ODE of the forman(x)dnydxn+ an1(x)dn1ydxn1+ + a2(x)d2ydx2+ a1(x)dydx+ a0(x)y = f(x)is called a linear differential equation, and when f(x) = 0 it is a homogeneous lineardifferential equation. The order of the equation is the highest derivative; in this case theequation has order n. The functions a0, a1, ..., an of the independent variable x are75the coefficients of the differential equation. In this chapter we will focus almost entirely onsecond order linear differential equations in which all the coefficients are constants and theindependent variable is time t, rather than x. An example of such an equation isy + 5y + 6y = 5 sin(3t) (1)We want to learn how to solve such ODEs and IVPs; it is helpful to keep in mind somegeneral principles, some of which we will address in more detail in Section 5.7: The solution to a second order ODE will contain TWO arbitrary constants. In order to determine the values of the two constants, we need TWO initial conditions. The method we will use to solve equations like (1) will consist of three steps, with oneadditional step when initial conditions are given:(1) We will first solve the homogeneous equation obtained by replacing f(t) (in thiscase, 5 sin(3t)) with zero. This will give us a function with two unknown constantsin it. This function is the homogenous solution, and we will denote it by yh.(2) We will find something called a particular solution, denoted by yp, for theequation (1). This is done using something called the method of undeterminedcoefficients.(3) The general solution to the equation (1) is y = yh+ yp, the sum of the homoge-nous solution and the particular solution.(4) If we have an IVP consisting of (1) and two initial conditions, we can then solvefor the two unknown constants. Note that this is done only AFTER adding thehomogenous and particular solutions.More generally than (1), we will be looking at initial value problems of the formay + by + cy = f(t), y(0) = y0, y(0) = y0, (2)with a, b and c being constants, a 6= 0. In addition, the initial values y0 and y0 areconstants as well. What the above things say is that ANY solution to the ODE in (2) willlook likey = C1g(t) + C2h(t) + yp(t), (3)where yh = C1g(t) + C2h(t) is the homogeneous solution to ay + by + cy = 0 and yp isthe particular solution to the ODE in (2). C1 and C2 are arbitrary constants, and g(t) and h(t) are both solutions (bythemselves) to ay + by + c = 0 that are somehow different from each other, in away that we will make more precise in Section 5.7. If we are given initial conditions, as in (2), values for C1 and C2 will be determinedby applying the initial conditions y(0) = y0, y(0) = y0 to the general solution (3).We will save theoretical concerns for the last section in the chapter. Section 5.2 willdevelop some intuitive understanding of what is going on, and Sections 5.3 through 5.6 willbe focused on the nuts and bolts of solving (2). In Chapter 6 we will then move on toapplications of such initial value problems, and developing a physical understanding of boththe initial value problems and their solutions.765.2 An Initial ExplorationPerformance Criterion:5. (a) Evaluate a differential operator for a given function.An ExampleIn this section we will begin by exploring a specific second order ODE in order to illustratesome ideas we will capitalize on in order to solve linear, constant coefficient, second orderODEs. The ODE that we will be considering isy + 9y = 5e2t, (1)and our goal is to find all possible solutions to this equation. Based on what we have seen sofar, we expect a function y = y(t) that contains one or more arbitrary constants that cantake any value. We begin with the observation that if y = C1e2t, then y = C2e2t andy = C3e2t. Of course we know that C2 = 2C1 and C3 = 4C1, but the point here isthat y and its derivatives are just constant multiples of e2t, so for the right constants wemight get y + 9y to equal 5e2t. Example 5.2(a): Determine whether there is a value A for which y = Ae2t is asolution to (1).First we observe thaty = Ae2t = y = 2Ae2t = y = 4e2t.Substituting these into the left side of (1) we gety + 9y = 4Ae2t + 9Ae2t = 13Ae2t,which will equal 5e2t if A = 513. Therefore y = 513e2t is a solution to (1). We used the letter A rather than a subscripted C in the above example because we wantto save C for arbitrary constants. A was of course NOT arbitrary, as it needs to have thevalue 513in order for y = Ae2t to be a solution.The next example shows that y = 51332t is not the only solution to (1). Example 5.2(b): Show that y = C sin 3t+ 513e2t is a solution to (1).Taking derivatives we gety = 3C cos 3t 101332t and y = 9C sin 2t+ 2013e2t.Substituting in to the left hand side of the ODE we gety + 9y = 9C sin 2t+ 2013e2t + 9(C sin 3t+ 513e2t)= 9C sin 2t+ 2013e2t + 9C sin 3t+ 4513e2t)= 6513e2t= 5e2tThus y = C sin 3t+ 513e2t is a solution to y + 9y = 5e2t. 77To examine further what is going on here it is convenient to develop some terminology andnotation.Differential OperatorsA function can be thought of as a mathematical machine that takes in a number and,in return, gives out a number. There are other mathematical machines that take in thingsother than numbers, like functions or vectors, usually giving out things like what they takein. These sorts of machines are often referred to as operators; a simple example of anoperator that you are quite familiar with is the derivative operator. When we take thederivative of a function, the result is another function. To indicate the action of a derivativeon a function y = y(t) we will write dydtasddt(y),which is like the function notation f(x), with ddttaking the place of f and y taking theplace of x. The derivative operator has a very special property that you should be familiarwith from calculus. If a and b are any constants and y1 and y2 are functions of t, thenddt(ay1 + by2) = ady1dt+ bdy2dt= addt(y1) + bddt(y2). (2)Operators that distribute over addition and pass through constants like this are calledlinear operators.You might guess that the second derivative, and other higher derivatives, are linear op-erators as well, and you would be correct. We are particularly interested in operators thatare created by multiplying a function and some of its derivatives by constants and addingthem all together. It is customary to denote such operators with the letter D, for differ-ential operator. An example would be D = 3d2dt2+ 5ddt 4, whose action on a functiony = y(t) is defined byD(y) = 3d2ydt2+ 5dydt 4y. (3)Lets look at a specific example of how this operator works. Example 5.2(c): For the operator D defined by (3), find D(y) for y = t2 3t.D(y) = 3d2dt2(t2 3t) + 5 ddt(t2 3t) 4(t2 3t)= 3(2) + 5(2t 3) 4(t2 3t)= 6 + 10t 15 8t2 + 12t= 8t2 + 22t 9 When we combine several mathematical objects by multiplying each by a constant andadding the results, we obtain what is called a linear combination of those objects. Theoperator D defined by (3) is a linear combination of a function and its first two derivatives.(We can think of the function itself as the zeroth derivative, making all the things being78combined derivatives.) You have seen linear combinations in other contexts; any polynomialfunction likef(x) = 3x4 7x3 + 13x2 x+ 5.83is a linear combination of 1, x, x2, x3 and x4. Those of you who have had a course inlinear algebra have seen linear combinations of vectors.Again, we can think of an operator as a machine that takes ina function and gives out some resulting function that is basedsomehow on the input function. This is illustrated to the rightfor Example 5.2(c). The next example demonstrates that a dif-ferential operator formed as a linear combination of derivativesis a linear operator.Dt2 3t8t2 + 22t 9 Example 5.2(c): Show that the operator D defined byD(y) = 3d2ydt2+ 5dydt 4yis a linear operator by showing that it satisfies (2).To determine whether D is linear we need to apply it to ay1 + by2:D(ay1 + by2) = 3d2dt2(ay1 + by2) + 5ddt(ay1 + by2) 4(ay1 + by2)= 3(ad2y1dt2+ bd2y2dt2)+ 5(ady1dt+ bdy2dt) 4ay1 4by2= 3ad2y1dt2+ 3bd2y2dt2+ 5ady1dt+ 5bdy2dt 4ay1 4by2= 3ad2y1dt2+ 5ady1dt 4ay1 + 3bd2y2dt2+ 5bdy2dt 4by2= a(3d2y1dt2+ 5dy1dt 4y1)+ b(3d2y2dt2+ 5dy2dt 4y2)= aD(y1) + bD(y2)Because D(ay1 + by2) = aD(y1) + bD(y2), D is a linear operator. At the second line above we have applied the fact that the first and second derivative arelinear operators. With a bit of thought it should be clear that this, along with the distributiveproperty, is what makes a linear combination of derivatives a linear operator.Back to the ExampleWe return now to considering the ODEy + 9y = 5e2t, (1)79for which we have shown that bothy = 513e2t and y = C sin 3t+ 513e2t (4)are solutions. If we now let D be the operator defined by D(y) = y + 9y, then (4) saysthatD( 513e2t) = 5e2t and D(C sin 3t+ 513e2t) = 5e2t.Looking a little more closely, we see thatD(C sin 3t) =d2dt2(C sin 3t) + 9(C sin 3t) = 9C sin 3t+ 9C sin 3t = 0.This explains why D applied to the sum of C sin 3t and 513e2t is a solution; by linearityof differential operators,D(y1 +513e2t) = D(y1) +D( 513e2t) = 0 + 5e2t = 5e2tfor any function y1 for which D(y1) = 0. This gives us the following:Let D be a linear differential operator with independent variable t. Ify2 is a solution to D(y2) = f(t) and y1 is a solution to D(y1) = 0,then D(y1 + y2) = f(t) also.The general form of equation that we will work with in this chapter isay + by + cy = f(t), (5)where a, b and c are constants. Our goal will be to find a general solution to thisequation, meaning a solution that encompasses all possible solutions. Such a solution willconsist of a solution to (5) that contains no arbitrary constants plus the sum of all possiblesolutions toay + by + cy = 0. (6)The solution to (5) without arbitrary constants is called the particular solution to theequation, and the sum of all possible solutions to (6) is called the homogeneous solution.We will denote the particular solution by yp and the homogenous solution by yh; thegeneral solution is then y = yh+ yp. For equations like (5) or (6) the homogeneous solutionwill always consist of two different (in a sense that we will discuss further in a later section)functions, each multiplied by an arbitrary constant and then added together.As an example, we will find that the ODEy + 9y = 5e2thas homogeneous solution yh = C1 sin 3t + C2 cos 3t and particular solution yp =513e2t,so the general solution isy = yh + yp = C1 sin 3t+ C2 cos 3t+513e2t.For reasons you will see in Section 5.6, we will find the homogeneous solution first usingmethods described in the next section. We will then find the particular solution in themanner that we did in Example 5.2(a).805.3 Homogeneous Equations With Constant CoefficientsPerformance Criterion:5. (b) Solve a second order, linear, constant coefficient, homogeneousODE.We begin by considering the second order linear, homogeneous ODEy + 5y + 6y = 0.It is easily shown that y = C1e3t+C2e2t is the general solution to the ODE. The obviousquestion to ask is how would one obtain that solution from the ODE? The answer is, byguessing! We are looking for some function y = y(t) for which we can multiply the functionand its first and second derivatives by constants, add the results and get zero. This indicatesthat y must essentially be equal to its first and second derivatives, which only occurs forexponential functions. Therefore we guess that the solution has the form y = ert, wherer is some constant. Example 5.3(a): Solve y + 5y + 6y = 0We substitute the guess y = ert into the ODE, resulting inr2ert + 5rert + 6ert = (r2 + 5r + 6)ert = 0Now the quantity ert is never zero, so it must be the case that r2 + 5r+ 6 = 0. Solvingthis gives us r = 3,2, so both y = e3t and y = e2t are solutions. Now it is nothard to show that if y1 and y2 are both solutions to a homogeneous ODE and C1 andC2 are constants, then y = C1y1 + C2y2 is also a solution. So in this case, the generalsolution is y = C1e3t + C2e2t. When using the above method to solveay + by + cy = 0 (1)we will arrive at (ar2 + br + c)ert = 0, leading to ar2 + br + c = 0. We will referto the equation ar2 + br + c = 0 as the auxiliary equation associated with the ODEay + by + cy = 0, and we will call solutions to this equation roots of the equation.When the auxiliary equation for a second order, linear, constant coefficienthomogeneous equation has two real roots r1 and r2, the solution to theODE isy = C1er1t + C2er2t.The most efficient method for finding the roots of the auxiliary equation r2 + 5r + 6 =0 from example 5.3(a) is to factor the left side, but we could have used the quadratic formular =bb2 4ac2a(2)81instead. We will at times want or need to use the quadratic formula, and at other timesfactoring or another method may be sufficient for finding the roots of the auxiliary equation.Regardless of what we might do in practice, an examination of (2) is useful for determiningthe various possibilities when obtaining the roots to the auxiliary equation: When b2 4ac > 0 there will be two real roots, as in example 5.3(a). When b2 4ac = 0 there will only be one real root, which is a rather special case. When b2 4ac < 0 the roots will be complex conjugates; that is, they will have theform r = k + i and r = k i, where i = 1. When b2 4ac < 0 and b = 0, the roots will be two purely imaginary numbersr = i and r = i. (This is just the previous case, with k = 0.)Each of the above cases results in different forms of the solution to the homogeneousequation (1), and the methods for solving the auxiliary equations are generally different aswell, although the quadratic formula could be used in all cases. Weve already taken care ofthe first case with Example 5.3(a).We now consider the second case above. Suppose that we were to try the method ofExample 5.3(a) for the ODE y + 6y + 9y = 0. We would only get one value for r,3. By the same reasoning as used before we then have the solution y = C1e3t + C2e3t;because these are like terms, this can be written y = Ce3t, where C = C1+C2. But (forreasons well go into in more depth later) the general solution to a second order ODE mustbe the sum of two different functions, each multiplied by arbitrary constants, so somethingis wrong! The following example shows that there is in fact another solution besides e3t. Example 5.3(b): Verify that y = te3t is also a solution to the ODE y+6y+9y = 0.Using the chain rule to compute the derivatives,y = t(e3t) + te3t = 3te3t + e3ty = 3(te3t) 3e3t = 3(3te3t + e3t) 3e3t = 9te3t 6e3tSubstituting into the ODE,y + 6y + 9y = (9te3t 6e3t) + 6(3te3t + e3t) + 9(te3t)= 9te3t 6e3t 18te3t + 6e3t + 9te3t= 0The general solution to the ODE y + 6y + 9y = 0 is then y = C1e3t +C2te3t. Anytime that our auxiliary equation has a repeated root r (both solutions are the same), oneof the solutions is ert, and the other solution is tert.When the auxiliary equation for a second order, linear, constant coefficienthomogeneous equation has only one real root r, the solution to the ODEisy = C1ert + C2tert.82We now consider the case where b2 4ac < 0 and b = 0, resulting in two purelyimaginary roots r = i. Well consider the homogeneous equation y + 9y = 0 that welooked at previously, which can be rewritten as y = 9y. Through our familiarity withderivatives and the chain rule, we guessed (correctly) that both y = sin 3t and y = cos 3t aresolutions, and it is not hard to show that y = C1 sin 3t + C2 cos 3t is a solution. Now howwould it work to try the method of Example 5.3(a) for this equation? Letting y = ert wehavey + 9y = r2ert + 9ert = (r2 + 9)ert = 0so we need to solve r2 + 9 = 0 r2 = 9. If we allow complex solutions, the solution tothis equation is r = 3i. To continue we will need the following:Eulers Formula: ei = cos + i sin To see where Eulers Formula comes from, see Appendix D. We will also use the two basictrig identities:cos() = cos , sin() = sin Combining the two solutions e3it and e3it, we gety = Ae3it +Be3it= A(cos 3t+ i sin 3t) +B[cos(3t) + i sin(3t)]= A cos 3t+ Ai sin 3t +B cos 3tBi sin 3t= (A+B) cos 3t+ (A B)i sin 3t= C1 cos 3t+ C2 sin 3t.Here A and B are constants, and C1 = A + B, C2 = (A B)i. Now it seems thatC2 should be an imaginary number, which is a bit disconcerting. However, it is possiblethat A and B are complex as well, in such a way that perhaps C2 turns out to be real!You will find that this method works, regardless. The following summarizes what we havejust seen.When the auxiliary equation for a second order, linear, constant coefficienthomogeneous equation has two purely imaginary roots r = i, thesolution to the ODE isy = C1 sint+ C2 cost.Lets use a specific example to examine the final situation, where the auxiliary equationhas two complex roots r = k i. Example 5.3(c): Solve y + 10y + 28y = 0Guess y = ert, so y = rert and y = r2ert. Theny + 10y + 28y = r2ert + 10rert + 28ert = (r2 + 10r + 28)ert = 083Again, ert cannot equal zero, so we solve r2 + 10r + 28 = 0; this is done in this caseusing the quadratic formula:r =10 102 4(28)2= 5 2i2Thereforey = Ae(5+2i2)t +Be(52i2)t= Ae5te2i2 t +Be5te2i2 t= e5t[(A cos(22 t) + Ai sin(22 t)) + (B cos(22 t) +Bi sin(22 t))]= e5t[(A cos(22 t) + Ai sin(22 t)) + (B cos(22 t) Bi sin(22 t))]= e5t[(A+B) cos(22 t) + (A B)i sin(22 t)]= e5t[C1 cos(22 t) + C2 sin(22 t)]The solution to y + 10y + 28y = 0 is y = e5t[C1 cos(22 t) + C2 sin(22 t)]. In general, we have the following.When the auxiliary equation for a second order, linear, constant coefficienthomogeneous equation has two complex roots r = k i, the solution tothe ODE isy = ekt(C1 sin t+ C2 cost).Example 5.3(c) and the discussion before it show how sine and cosine functions arisewhen r1 and r2 are complex numbers, but you need not show all those steps when solving.Unless asked to show more, we will just set up and solve the equation ar2 + br + c = 0,then give the solution that arises from the form of the roots. On the next page you will finda flowchart detailing the process of finding the solution to a second order, linear, constantcoefficient, homogeneous differential equation. You will need to memorize the forms of thesolution for the various forms of r1 and r2, but if you do enough exercises, you shouldknow them by the time you are done.84Solving Second Order, Linear, Constant Coefficient,Homogeneous ODEsDifferential Equationay + by + cy = 0Auxiliary Equationar2 + br + c = 0Isb2 4ac 0?yes noIs b = 0?yesnoSolve for r = iby isolating r2 andand taking the squareroot of both sidesSolution to the ODE isy = C1 sint+ C2 costSolution to the ODE isy = ekt(C1 sin t+ C2 cost)Use the quadraticformula to solvefor r = k iSolve by factoringor by using thequadratic formulaArethere two rootsr1, r2?yesnoSolutionto the ODE isy = C1er1t + C2er2tSolutionto the ODE isy = C1ert + C2tert855.4 Particular Solutions, Part OnePerformance Criteria:5. (c) Find a particular solution to a second order linear, constant coef-ficient ODE using undetermined coefficients.In the last section you learned how to solve ODEs of the form ay+by+cy = 0. We willsee these sorts of equations in two kinds of engineering applications, but they will generallynot be homogeneous; instead, they will be of the formay + by + cy = f(t),where f is some function that we will refer to as the forcing function. The reason forthis name will be apparent soon. More generally, we will be solving IVPs consisting of thiskind of ODE along with two initial values. The process for doing this will go like this:1) Solve the homogeneous ODE ay + by + cy = 0. Denote its solution by yh, calledthe homogenous solution.2) Find the particular solution yp of the ODE ay + by + cy = f(t).3) Construct the general solution y = yh + yp.4) Apply the initial conditions to the general solution to find the values of the arbitraryconstants.You have already seen how to obtain the homogeneous solution yh. The method we willuse for finding the particular solution is called themethod of undetermined coefficients.To do this we substitute a guess, which we will call the trial particular solution, into the ODE.The trial solution will have some unknown (undetermined, but soon to be determined!)constants. Their values can be determined by finding the result obtained when the trialsolution is substituted into the left hand side of the ODE, and then setting it equal tothe known right hand side of the ODE. Like terms are then equated, and the constantsdetermined. We already did this in Example 5.2(a), which we repeat here to recall what wedid. Example 5.4(a): Find a value A for which y = Ae2t is a solution to y+9y = 5e2t.First we observe thaty = Ae2t = y = 2Ae2t = y = 4e2t.Substituting these into the left side of (1) we gety + 9y = 4Ae2t + 9Ae2t = 13Ae2t.We now set this equal to the right hand side of the ODE to get 13Ae2t = 5e2t. The onlyway that this can be true is if the coefficients of e2t are equal: 13A = 5 A = 513.Therefore yp =513e2t is the particular solution to y + 9y = 5e2t. 86Using methods from the previous section, we would have already found the homogeneoussolution yh = C1 sin 3t + C2 cos 3t, so the general solution to y + 9y = 5e2t isy = yh + yp = C1 sin 3t+ C2 cos 3t+513e2t.Next we will examine the ODE y + 7y + 10y = 5t2 8. The associated homogeneousequation y + 7y + 10y = 0 has auxiliary equation r2 + 7r + 10 = 0, which factors to(r + 2)(r + 5) = 0, so the homogeneous solution is y = C1e2t + C2e5t. We now find theparticular solution. Example 5.4(b): Find the coefficients A, B and C for which yp = At2+Bt+C isa solution to y + 7y + 10y = 5t2 8.First we compute the needed derivatives:yp = At2 +Bt+ C = yp = 2At+B = yp = 2A.Next we substitute these values into the left side of the ODE and group by powers of t:yp + 7yp + 10yp = 2A+ 7(2At+B) + 10(At2 +Bt + C)= 2A+ 14At+ 7B + 10At2 + 10Bt+ 10C= 10At2 + (14A+ 10B)t+ (2A+ 7B + 10C)Setting this equal to the right hand side of the ODE gives us10At2 + (14A+ 10B)t+ (2A+ 7B + 10C) = 5t2 + 0t 8,and equating coefficients of powers of t (including the constant term) results in the threeequations10A = 5, 14A+ 10B = 0, 2A+ 7B + 10C = 8.From the first equation we see that A = 12. Substituting that value into the secondequation and solving for B results in B = 710. Finally, when we substitute the valueswe have found for A and B into the last equation and solve for C we get C = 41100.The particular solution to y + 7y + 10y = 5t2 8 is thenyp =12t2 710t 41100. When we combine the particular solution we just found with the already obtained homoge-neous solution, the general solution to y + 7y + 10y = 5t2 8 isy = yh + yp = C1e2t + C2e5t + 12t2 710t 41100.It is important to note that even though the right hand side 5t2 8 contains only at2 term and a constant term, we need to include a t term in our guess for the particularsolution. If we had instead guessed a particular solution of the form yp = At2 + B andsubstituted it into the ODE we would have obtained the two equations10A = 5 and 14A = 0,both of which cannot be true at the same time!87So far we have found a particular solution to ay + by + cy = f(t) when f(t) is anexponential function (Example 5.4(a)) and a polynomial (Example 5.4(b)). The only othertype of function we will consider for f(t) is a trigonometric function; suppose we have theODEy + 4y + 3y = 5 sin 2t. (1)Based on what we have seen so far, our first inclination might be that we should consider aparticular solution of the form yp = A sin 2t. Lets try it:yp = A sin 2t = yp = 2A cos 2t = yp = 4A sin 2tsoyp + 4yp + 3yp = 4A sin 2t+ 8A cos 2t+ 3A sin 2t= A sin 2t+ 8A cos 2t.Here we need A = 0 because there is no cosine term on the right hand side of (1), but thenwed have no sine term either!So what do we do? Well, the trick is to let yp have both sine and cosine terms. Example 5.4(c): Determine values for A and B so that yp = A sin 2t+B cos 2t isthe particular solution to y + 4y + 3y = 5 sin 2t.We see thatyp = 2A cos 2t 2B sin 2t, yp = 4A sin 2t 4B cos 2t.soyp + 4yp + 3yp = (4A sin 2t 4B cos 2t)+ 4(2A cos 2t 2B sin 2t) + 3(A sin 2t+B cos 2t)= (A 8B) sin 2t+ (8A B) cos 2tThus, in order for the left hand side to equal the right hand side, it must be the case that8A B = 0 because there is no cosine term in the right hand side, and A 8B = 5,so that the sine terms are equal. Solving the first equation for B and substituting intothe second equation results in A = 113. Substituting this back into the second equationgives B = 813. Our particular solution is then yp = 113 sin 2t 813 cos 2t .At this point we have the following when finding a particular solution to ay+by+cy = f(t): If f is a polynomial of degree n, thenyp = Antn + An1tn1 + + A2t2 + A1t+ A0Note that all powers of t less than or equal to the degree of f(t) are included. If f(t) = Cekt, then yp = Aekt. If f(t) = C1 sin kt + C2 cos kt, then yp = A sin kt + B cos kt. Even if one of C1 orC2 is zero, yp still contains both the sine and cosine terms.In Section 5.6 we will see that there is a bit more to be added, but for now the abovesummarizes what we have seen so far.885.5 Initial Value ProblemsPerformance Criteria:5. (d) Solve a second order linear, constant coefficient IVP.Now that we know how to find the general solution to ay+ by+ cy = f(t) we are readyto solve initial value problems whose ODEs are of this form. Here is how the process goes:Solving the IVP ay + by + cy = f(t), y(0) = y0, y(0) = y01) Find the homogeneous solution yh to ay + by + cy = 0.2) Use the method of undetermined coefficients to find the particularsolution yp of the ODE ay + by + cy = f(t).3) Construct the general solution y = yh + yp.4) Apply the initial conditions to the general solution to find the valuesof the arbitrary constants.We have already covered the first three steps of the above. It is very important toremember that the initial conditions are applied to the general solution to find thevalues of the arbitrary constants. A common mistake by students is to find the values of theconstants based on only the homogeneous solution - this is incorrect.Lets look at an example: Example 5.5(a): Solve the IVPy + 4y + 3y = 5 sin 2t, y(0) = 2, y(0) = 1We must first solve the homogeneous equation y + 4y + 3y = 0. The roots of theauxiliary equation are r1 = 1 and r2 = 3 (of course it doesnt matter which is which),so the homogeneous solution is yh = C1et + C2e3t. Because the right hand side is notof the same form as any part of the homogeneous solution, our trial particular solution isyp = A sin 2t+B cos 2t. This gives usyp = 2A cos 2t 2B sin 2t, yp = 4A sin 2t 4B cos 2t.soLHS = (4A sin 2t 4B cos 2t) + 4(2A cos 2t 2B sin 2t) + 3(A sin 2t+B cos 2t)= (A 8B) sin 2t+ (8AB) cos 2tThus, in order for the left hand side to equal the right hand side, it must be the case that8A B = 0 because there is no cosine term in the right hand side, and A 8B = 5,so that the sine terms are equal. Solving the first equation for B and substituting into the89second equation results in A = 113. Substituting this back into the second equation givesB = 813. Our particular solution is then yp = 113 sin 2t 813 cos 2t, and the generalsolution isy = yh + yp = C1et + C2e3t 113 sin 2t 813 cos 2tThe derivative of the general solution isy = C1et 3C2e3t 213 cos 2t + 1613 sin 2tApplying the initial conditions gives us the two equationsC1 + C2 813 = 2 and C1 3C2 213 = 1Adding these and solving for C2 gives us C2 = 2326 . Substituting this into either equationgives C1 =9126. The solution to the IVP is theny = 9126et 2326e3t 113sin 2t 813cos 2t It is often the case that finding the constants C1 and C2 comes down to solving a systemof two equations in two unknowns, as it did here.905.6 Particular Solutions, Part TwoPerformance Criteria:5. (c) Find a particular solution to a second order linear, constant coef-ficient ODE using undetermined coefficients.(e) Given minimal, but sufficient, information about an ODE of theform ay + by + cy = f(t) and its general solution, determinewhether a function is a solution to ay + by + cy = f(t) oray + by + cy = 0.At his point you have seen the entire process for solving initial value problems for secondorder, linear, constant coefficient differential equations. In this section we see one difficultythat can arise, and how to handle such situations. We begin with an example. Example 5.6(a): Determine the values of A and B for whichyp = A sin 3t+B cos 3tis the particular solution to the ODE y + 9y = 2 sin 3t.As usual, we begin by finding the derivatives of yp:yp = 3A cos 3t 3B sin 3t = yp = 9A sin 3t 9B cos 3t.We then haveyp + 9yp = 9A sin 3t 9B cos 3t+ 9(A sin 3t+B cos 3t) = 0,so there are no values of A and B for which y = A sin 3t + B cos 3t is a solution toy + 9y = 2 sin 3t. The problem here is that the homogeneous solution to y+9y = 2 sin 3t is yh = C1 sin 3t+C2 sin 3t. Thus we cannot hope to obtain 2 sin 3t when applying the operator D =d2dt2+9 toy = A sin 3t + B cos 3t, as the result is always zero. However, we will find that a differentguess for yp will give us the particular solution that we seek: Example 5.6(b): Determine the values of A and B for whichyp = At sin 3t+Bt cos 3tis the particular solution to the ODE y + 9y = 2 sin 3t.We use the product rule carefully to find the derivatives of yp:yp = 3At cos 3t+ A sin 3t 3Bt sin 3t+B cos 3tandyp = 9At sin 3t+ 3A cos 3t + 3A cos 3t 9Bt cos 3t 3B sin 3t 3B sin 3t.91Grouping the like terms of the second derivative gives usyp = 9At sin 3t 9Bt cos 3t 6B sin 3t+ 6A cos 3t.Substituting into the left side of the ODE gives usyp + 9yp = 9At sin 3t 9Bt cos 3t 6B sin 3t+ 6A cos 3t+ 9(At sin 3t+Bt cos 3t)= 6B sin 3t+ 6A cos 3t.In order for this to equal 2 sin 3t we must have A = 0 and B = 13, so the particularsolution to y + 9y = 2 sin 3t is yp = 13t cos 3t. We already knew the homogeneous solution, so the general solution to y+ 9y = 2 sin 3t isy = C1 sin 3t+ C2 cos 3t 13t cos 3t.We can now make an amendment to the listing at the end of Section 5.4 to get the overallsummary for guesses to use for particular solutions.Undetermined CoefficientsConsider the constant coefficient ODE ay + by + cy = f(t), andassume yh contains no terms that are constant multiples of f(t). Thetrial particular solution yp is chosen as follows. If f is a polynomial of degree n, thenyp = Antn + An1tn1 + + A2t2 + A1t+ A0 If f(t) = Cekt, then yp = Aekt. If f(t) = C1 sin kt+C2 cos kt, then yp = A sin kt+B cos kt. Evenif one of C1 or C2 is zero, yp still contains both the sine andcosine terms.When yh contains any term of the same form as f , yp will be as abovebut multiplied by the smallest power of t for which no terms of yp areof the same form as any terms of yh. Example 5.6(c): Find the trial particular solution to y + y 6y = 5t 3.The homogeneous solution is yh = C1e3t + C2e2t, so the trial particular solution isyp = At +B. Example 5.6(d): Find the trial particular solution to y + y 6y = 7 cos 5t.The homogenous solution is the same as in Example 5.6(c), so trial particular solution isyp = A sin 5t+B cos 5t. Example 5.6(e): Find the trial particular solution to y + y 6y = 4e2t.Again the homogeneous solution is yh = C1e3t + C2e2t. f(t) has the same form as oneof the terms of the homogeneous solution, the trial particular solution is yp = Ate2t. 92We conclude this section by further examining homogeneous and particular solutions toan ODEay + by + cy = f(t). (1)Lets denote the left side of (1) using the operator notation D(y). We have found that thehomogeneous solution consists of a linear combination of two functions g(t) and h(t) thatare both, by themselves, solutions to D(g) = 0 and D(h) = 0. By a linear combinationwe meanyh = C1g(t) + C2h(t),where C1 and C2 are ANY constants. When D is applied to the particular solutionyp the result is D(yp) = f(t). The general solution isy = C1g(t) + C2h(t) + yp(t).Applying D to the solution then givesD(C1g + C2h+ yp) = C1D(g) + C2D(h) +D(yp) = 0 + 0 + f(t) = f(t).Note the use of the fact that D is a linear operator in this computation. Example 5.6(f): The general solution to (1) is y = C1e2t+C2et+4 cos 5t. Whichof the following are solutions to ay + by + cy = 0?(a) y = 5et (b) y = 7e2t + 4 cos 5t (c) y = 7e2t + 5etThe homogeneous solution is the part containing the arbitrary constants, yh = C1e2t +C2et. It is a solution to ay + by + cy = 0 for all choices of C1 and C2, so thefunctions in (a) and (c) are both solutions. The function in (b) is a solution to (1), but notto ay + by + cy = 0 because D applied to 7e2t is zero, but when applied to theparticular solution yp = 4 cos 5t the result is f(t), not zero. Example 5.6(g): The general solution to (1) is y = C1e2t+C2et+4 cos 5t. Whichof the following are solutions to (1)?(a) y = 4 cos 5t (b) y = 8 cos 5t (c) y = 7e2t + 5et + 4 cos 5tWe again recognize that the homogeneous solution is yh = C1e2t + C2et and theparticular solution is yp = 4 cos 5t. Because the particular solution by itself is a solutionto (1), the function in (a) is a solution. Unlike the homogeneous solution, a constant in theparticular solution is not arbitrary, so the function in (b) is not a solution. The function in(c) is a solution, because it is simply the general solution with the arbitrary constants havingthe specific values C1 = 7 and C2 = 5. 935.7 Linear Independence of Solutions, Reduction of OrderPerformance Criteria:7. (e) Determine, using the definition, whether two functions are linearlyindependent.(f) Use the Wronskian to determine whether two solutions to a secondorder linear ODE are independent.(g) Given one solution to a second order ODE, use reduction of orderto find a second solution.We begin with two questions:(1) When solving the ODE y+3y+2y = 0, we assumed a solution of the form y = ert forsome constant r and found that r must equal 1 or 2. We then assumed thatevery solution to the ODE is of the form y = C1et + C2e2t. How do we know thatthis is the case?(2) When solving y + 2y + y = 0 we found only one solution, y = et. We thendemonstrated that y = tet is also a solution, and we assumed that the generalsolution to the ODE is y = C1et + C2tet. How might one know or find the secondsolution without it being given?In this section we will develop some language and see some theorems answer the first question,and well see a method called reduction of order that gives an answer to the secondquestion.Linearly Independent SolutionsLinearly Independent FunctionsTwo functions f and g are linearly dependent on an interval [a, b] ifthere exist two non-zero constants c1 and c2 for whichc1f(x) + c2g(x) = 0 for every x in [a, b]. (1)If (1) is true only when both c1 and c2 are zero, then f and g arelinearly independent on [a, b].The expression c1f(x) + c2g(x) above is called a linear combination of the functionsf and g. Example 5.6(a): Show that the two functions y = e2t and y = et are linearlyindependent for all values of t.Suppose that c1e2t + c2et = 0 for some c1 and c2 not equal to zero. Thene2t(c1 + c2et) = 0, but e2t is never zero so it must be the case that c1 + c2et = 0.This gives us et = c1/c2, which says that et is constant. Since that is not the case,y = e2t and y = et are linearly independent. 94There may be times that it is difficult to tell, using the definition above, whether twofunctions are linearly independent. In those cases we can use a new function created fromthe two functions, called the Wronskian, to determine whether the functions are linearlyindependent.The Wronskian of Two FunctionsThe Wronskian W of two functions f and g isW (x) = f(x)g(x) f (x)g(x).Those of you who have had linear algebra may recognize the Wronskian as the determinantof the 2 2 matrix [f(x) g(x)f (x) g(x)]Our interest is in determining whether two solutions to an ODE are linearly independent.Note that any linear second order homogeneous ODE with independent variable t can(almost always, and definitely in the case that p and q are constant) be written in theformy + p(t)y + q(t)y = 0 (2)The following tells how the Wronskian is used to determine whether two solutions to anequation of the form (2) are linearly independent.The Wronskian and Linearly Independent SolutionsTwo solutions f and g of (2) are linearly independent on the interval(a, b) if there exists some point x in the interval for which W (x) 6= 0. Example 5.6(b): Show that the two solutions y = sin 3t and y = cos 2t ofy + 9y = 0 are linearly independent for all values of t.The Wronskian of these two functions isW (t) = (sin 3t)(cos 3t) (cos 3t)(sin 3t) = sin2 3t cos2 3t = 1,which is clearly not zero for any value of t. Therefore y = sin 3t and y = cos 2t arelinearly independent for all values of t. We conclude with why we are interested in all of this.95General Solutions to y + p(t)y + q(t)y = 0If y1 and y2 are linearly independent solutions to y+p(t)y+q(t)y = 0,then the general solution isy = C1y1 + C2y2for arbitrary constants C1 and C2.Note that the above says that the general solution is a linear combination of the two solutionsy1 and y2.In the exercises you will show that y = et and y = tet are linearly independentsolutions to y + 2y + y = 0, so the above tells us that the general solution is theny = C1et + C2tet. Its now time to find out where the solution y = tet comes from.Reduction of OrderReduction of order is a method for finding a second solution to a second order differentialequation when one solution is already known. Our main interest in this is finding the secondsolution when we have repeated roots, so we will not go into the method in excessive detail.Perhaps the best way to introduce the method is through an example. The two key ideasare these: We will assume that if y1 = y1(t) is a solution, then the second solution has the formy2(x) = u(t)y1(t) for some function u. We then substitute y2 into the ODE, whichresults in a new ODE for u. The new ODE for u will contain u and u terms, but no u term. If we letv(t) = u(t) then v(t) = u(t), and making these two substitutions we get a firstorder equation in v. (This is where the name reduction of order comes from - wevereduced a second order equation to a first order equation.) We solve that for v, thensolve u(t) = v(t) to get u.Now lets get to that example! Example 5.6(c): Use the solution y1(t) = et and reduction of order to find a secondsolution to y + 3y + 2y = 0.We begin by assuming y2 = u(t)et. Then (using integration by parts),y2 = u(t)et + u(t)et and y2 = u(t)et 2u(t)et + u(t)et.Substituting into the ODE we gety2 + 3y2 + 2y2 = [u(t)et 2u(t)et + u(t)et] + 3[u(t)et + u(t)et] + 2u(t)et= u(t)et 2u(t)et + u(t)et 3u(t)et + 3u(t)et + 2u(t)et= u(t)et + u(t)et= et(u(t) + u(t))96Setting the result equal to zero (because we want y2 = u(t)et to be a solution toy + 3y + 2y = 0) and noting that et is never zero, we must have u(t) + u(t) = 0.Here we let v(t) = u(t), so v(t) = u(t) and this last ODE becomes v(t) + v(t) = 0.This equivalent to v(t) = v(t), so v(t) = C1et.We now replace v(t) with u(t) to obtain u(t) = C1et. The solution to this isu(t) = C2et + C3; for reasons to be given later, we can take C2 to be any non-zerovalue and C3 can have any value. Well take C2 = 1 and C3 = 0. Thereforey2(t) = u(t)et = etet = e2t. Disregarding the constant (because we will replace itwhen adding this solution to the one given), we have y2 = e2t. Forming the linear combination of the given solution and the one that we found usingit, we get y = aet + be2t for constants a and b. We now examine the way that theconstants C2 and C3 were handled in the above. Lets see what would have happened ifwe had not let C2 = 1 and C3 = 0. In that case we would have hady2 = u(t)y1(t) = (C2et + C3)et = C2e2t + C3et.When we then form a linear combination of y1 and y2 using constants A and B, wellgety = Aet+B(C2e2t+C3et) = Aet+BC2e2t+BC3et = (A+BC3)et+BC2e2t = aet+be2t,where a = A+BC3 and b = BC2. If we keep the constants C2 and C3, they essentiallyget absorbed into the constants for the linear combination of the two solutions.In Example 5.6(c) there was no need to use reduction of order to determine a secondsolution y2 = e2t from the first solution y1 = e2t; we could arrive at both solutions viathe auxiliary equation method. However, the above example demonstrates how the methodworks. In the exercises you will encounter exercises where you will again be asked to find asecond solution by this method when it is unnecessary, but you will also use it for situationswhere the second solution (and maybe the first as well) cant be obtained by methods we haveused so far. You will also use reduction of order to find the second solution to y+2y+y = 0,knowing the first solution y1 = et, which is obtained by the auxiliary equation method.975.8 Chapter 5 Practice ExercisesSection 5.21. Solve each of the following ODEs, assuming the independent variable is t.(a) y 3y 4y = 0 (b) 2y + 11y + 5y = 02. Solve each of the following ODEs, assuming the independent variable is t.(a) y + 6y + 9y = 0 (b) y + 4y + 4y = 03. Solve each ODE using the method described above. Assume the independent variableis t. Use exact values except for (i) - use decimals rounded to the hundredths placethere.(a) y 2y 3y = 0 (b) y + 2y + 10y = 0 (c) y + 10y + 25y = 0(d) y + 6y + 11y = 0 (e) y + 3y + 2y = 0 (f) y + 2y = 0(g) y + 2y + y = 0 (h) y + 4y + 29y = 0 (i) y + 3.1y + 4.5y = 0Section 5.31. In this exercise you will find the particular solutions for the ODEs from Examples5.3(b), (c) and (d). In each case, substitute the trial particular solution given in thecorresponding example into the ODE to determine the value(s) of any constant(s).(a) Find the particular solution to y + y 6y = 5t 3.(b) Find the particular solution to y + y 6y = 7 cos 5t.(c) Find the particular solution to y + y 6y = 4e2t.2. Suppose that when you were finding the particular solution to y+ y 6y = 4e2t youdidnt notice that 4e2t was of the same form as one of the terms of yh. Try a particularsolution of yp = Ae2t and see what happens. It will try to tell you that something iswrong!3. Solve each of the following IVPs by the method described in the box on the previouspage.(a) y + 9y = 4 sin t , y(0) = 2 , y(0) = 4(b) y + y 6y = 1 + 8t 6t2 , y(0) = 2 , y(0) = 3(c) y + 4y + 4y = 5e3t , y(0) = 0 , y(0) = 0(d) y + 7y + 10y = 6e2t , y(0) = 2 , y(0) = 11(e) y 10y + 25y = 30t+ 3 , y(0) = 2 , y(0) = 8(f) y + 4y = 3 sin 2t , y(0) = 12, y(0) = 524. The functions below are solutions to second order linear, constant coefficient initialvalue problems. Give the steady-state and transient parts of each.98(a) y = 23sin 3t+ 53cos 3t (b) y = e3t(4 sin t + 7 cos t) + 34cos 7t(c) y = 35sin 5t 65cos 5t+ 72e2t (d) y = 3te5t 7e5t + etSection 5.61. For each pair of functions, determine whether they are linearly independent, using thedefinition. If they arent, give nonzero constants c1 and c2 for which c1f(x)+c2g(x) =0 for all real numbers x.(a) f(x) = 3x2 5, g(x) = 2x+ 1 (b) f(x) = 4x+ 2, g(x) = 2x+ 1(c) f(x) = 3e5x, g(x) = 2e5x (d) f(x) = 3e5x, g(x) = e2x2. Use the facts that cos(x) = cos x and sin(x) = sin(x) for the following.(a) Give nonzero constants c1 and c2 such that c1 cosx + c2 cos(x) = 0. Arecosx and cos(x) linearly independent?(b) Repeat part (a) for sin x and sin(x).3. For each pair of functions in Exercise 1 that you determined were linearly independent,give the Wronskian and one value of x for which it is not zero.4. Find the Wronskian for y1 = ekt and y2 = tekt (where k is any nonzero constant)and give a value of t for which it is not zero. What does this tell us about the functionsy1 and y2?5. Use the Wronskian to determine whether ex and ex are linearly independent.6. Consider the ODE y + 8y + 15y = 0.(a) Given that one solution is y1 = e5t, use reduction of order to find another solution.(b) Use the auxiliary equation to find both solutions, to check your answer to (a).7. Using the auxiliary equation method with y + 2y+ y = 0, we get the single solutiony1 = et. Use reduction of order to obtain the second solution y2 = tet.8. Given that one solution to 2x2y+ xy 3y = 0 is y1 = 1x , find a second solution y2.9. Given that one solution to x2y+2xy 2y = 0 is y1 = x, find a second solution y2.996 Applications of Second Order Differential Equations6.1 IntroductionLearning Outcome:6. Understand the nature of solutions to second order linear, constantcoefficient ODEs and IVPs modeling spring-mass systems or RLC cir-cuits.Performance Criteria:(a) Translate given physical initial conditions for a mass on a springto equivalent mathematical statements. Set up and solve an IVPfor an undamped mass on a spring with no forcing function.(b) Sketch or identify the graph of the solution to an IVP for anundamped mass on a spring with no forcing function.(c) Write a function y = A sint + B cost in the alternate formy = C sin(t + ). From this, determine the amplitude, period,frequency, angular frequency and phase shift.(d) Determined whether a forced, undamped system will exhibit res-onance, beats, or neither.(f) Determine from the coefficients of a second order, constant coef-ficient homogeneous ODE whether the solution will be (a) under-damped, (b) critically damped, or (c) overdamped.(g) Sketch the graph of overdamped or underdamped solutions forgiven initial conditions without finding the solution to the differ-ential equation.(h) For a spring-mass system or electric circuit, understand the rela-tionships between the physical situation (presence and type of damping and/orforcing) the form of the differential equation, including the function f the the analytic and graphical nature of the solution (in par-ticular, the presence and appearance of transient and steady-state parts of the solution)Our practical applications so far have been focused on first order ODEs and their asso-ciated initial value problems. We have seen how such equations model RL electric circuits,mixing problems, growth of populations and decay of radioactive elements, and the temper-ature of a solid object in some constant-temperature medium.There are other things we would like to model that require not first order ODEs, butsecond order. One example was given in Chapter 1, a spring-mass system. In Section 1.3we saw how such a system is modeled by a second order ODE, like you solved in the last100chapter. In this chapter well add damping to that system, and well see that such a systemis completely analogous to an electrical circuit consisting of a resistor, inductor, capacitorand voltage source all connected in series. The same type of differential equations and initialvalue problems model both situations.We will continue to work with equations of the formy + 5y + 6y = 5 sin 3t, (1)which you learned how to solve in the last chapter. We want to know not only how to solvesuch ODEs and IVPs, but also to know what kind of behavior to expect from solutions,based on the original ODE or IVP. Here is a summary of some of the ideas you will runacross - some should be familiar already: The left hand side of the ODE represents some sort of system like a mass on a springor a simple electric circuit. The function on the right represents some kind of input tothe system; we will sometimes call that function a forcing function, since it forcesmovement (for a mass on a spring) or current (for an electrical circuit) in the system. The solution function y will consist of exponential or trigonometric functions, orproducts of the two. The solution may have several terms, some of which may dieout (go to zero) as time goes on, and others that may not. The parts that die outare called transient parts of the solution, and parts that are constant or periodic arecalled steady-state parts. You encountered these ideas previously, in Chapter 3.Most of our study so far has focused on solving or analyzing first order ODEs and theirassociated initial value problems. We have seen how such equations model RL electriccircuits, mixing problems, growth of populations and decay of radioactive elements, and thetemperature of a solid object in some constant-temperature medium.There are other things we would like to model that require not first order ODEs, butsecond order. One example was given in Chapter 1, a spring-mass system. In Section 1.3we saw how such a system is modeled by a second order ODE. In this chapter well adddamping to that system, and well see that such a system is completely analogous to anelectrical circuit consisting of a resistor, inductor, capacitor and voltage source all connectedin series. The same type of differential equations and initial value problems model bothsituations.Recall that an ODE of the forman(x)dnydxn+ an1(x)dn1ydxn1+ + a2(x)d2ydx2+ a1(x)dydx+ a0(x)y = f(x)is called a linear differential equation, and when f(x) = 0 it is a homogeneous lineardifferential equation. The order of the equation is the highest derivative; in this case theequation has order n. The functions a0, a1, ..., an are the coefficients of the differentialequation. In this chapter we will focus almost entirely on second order linear differentialequations in which all the coefficients are constants and the independent variable is time t,rather than x. An example of such an equation isy + 5y + 6y = 5 sin(3t) (1)We want to learn not only how to solve such ODEs and IVPs, but how to know what kindof behavior to expect from solutions, based on the original ODE or IVP. Here is a summaryof some of the ideas you will run across - some should be familiar already:101 The solution to a second order ODE will contain TWO arbitrary constants. In order to determine the values of the two constants, we need two initial conditions. The method we will use to solve equations like (1) will consist of two parts:(1) We will first solve the homogeneous equation obtained by replacing f(t) (in thiscase, 5 sin(3t)) with zero. This will give us a function with two unknown constantsin it. This function is the homogenous solution, and we will denote it by yh.(2) We will find something called a particular solution, denoted by yp, for theequation (1). This is done using something called the method of undeterminedcoefficients.(3) The general solution to the equation (1) is y = yh+ yp, the sum of the homoge-nous solution and the particular solution.(4) If we have an IVP consisting of (1) and two initial conditions, we can then solvefor the two unknown constants. Note that this is done only AFTER adding thehomogenous and particular solutions. The left hand side of the ODE represents some sort of system like a mass on a springor a simple electric circuit. The function on the right represents some kind of input tothe system; we will sometimes call that function a forcing function, since it forcesmovement (for a mass on a spring) or current (for an electrical circuit) in the system. The solution function y will consist of exponential or trigonometric functions, orproducts of the two. The solution may have several terms, some of which may dieout (go to zero) as time goes on, and others that may not. The parts that die outare called transient parts of the solution, and parts that are constant or periodic arecalled steady-state parts. You encountered these ideas previously, in Chapter 3.1026.2 Free, Undamped VibrationPerformance Criteria:6. (a) Translate given physical initial conditions for a mass on a springto equivalent mathematical statements. Set up and solve an IVPfor an undamped mass on a spring with no forcing function.(b) Sketch or identify the graph of the solution to an IVP for anundamped mass on a spring with no forcing function.(c) Write a function y = A sint + B cost in the alternate formy = C sin(t + ). From this, determine the amplitude, period,frequency, angular frequency and phase shift.Suppose that we have a mass hanging on a spring, with no damping. In Section 1.3 wederived the differential equationmd2ydt2+ ky = 0 (1)where y is the height (from equilibrium) of the mass at any time t after it is set in motionby some means, with positive being upward. m is the mass of the mass (the first of thesebeing a quantity, and the second being an object) and k is the spring constant. First letsremind ourselves of how we can solve such ODEs: Example 6.2(a): Solve Equation 1 for a mass of 10.3 and a spring constant of 28.7(both with appropriate units).The auxiliary equation corresponding to the ODE is 10.3r2 + 28.7 = 0. Subtracting 28.7from both sides and dividing by 10.3 gives r2 = 2.786.... If we then take the square rootof both sides and round to the nearest hundredth, we have r = 1.67i. Therefore thesolution to the ODE is y = C1 sin 1.67t+ C2 cos 1.67t.Intuitively it is reasonable that, once set in mo-tion, a mass on a spring with no damping will os-cillate forever with the same amplitude. The graphof such motion would look something like what isshown to the right; this is referred to as harmonicmotion. It may not be clear how the solution to theabove differential equation, which contains a sum ofsine and cosine functions, gives a graph like the oneshown. Well soon see a computation that makes itclear how this happens.y (in)t (sec)A mass hanging on a spring can be set in motion by doing one of three things: Moving the mass away from its equilibrium position and letting it go, giving the mass some initial velocity upward or downward from its initial position, moving the mass away from its initial position and giving it an initial velocity.103If we move the mass upward or downward, that will give y(0) equal to some value otherthan zero, and if we impart an initial velocity the value of y(0) will be nonzero. Example 6.2(b): Suppose that the mass from Example 6.2(a) is set in motion bylifting it up one inch and giving it an initial velocity of 2 inches per second downward,both at time zero. Give the initial conditions in function form and sketch a graph ofwhat you expect the motion to look like. Then determine the solution to the initialvalue problem and graph it to check your graph.The condition of raising the mass up by one inch isgiven by y(0) = 1, and the initial velocity of twoinches per second downward is given by y(0) = 2,with the negative indicating that the velocity is in thedownward direction. Because the mass starts aboveits equilibrium point, we expect the y-intercept of thegraph to be at positive one. The condition of beinggiven an initial velocity downward means that theslope of the tangent line to the graph at t = 0 willbe negative. We therefore expect a graph somethinglike that shown to the right.y (in)t (sec)Applying the first initial condition with the solution from Example 6.2(a) gives C2 = 1.Taking the derivative of the solution obtained in Example 6.2(a) givesy = 1.67C1 cos 1.67t 1.67C2 sin 1.67t.Substituting the initial velocity initial condition into this gives us C1 = 1.20 (the zeroindicates accuracy to the hundredths place), so the solution to the IVP isy = 1.20 sin 1.67t+ cos 1.67t.The graph of this function agrees with that shown above. It is sometimes useful to change an expression of the form 1.20 sin 1.67t+cos 1.67t intoa single sine function with a phase shift, and here is what we use to do it:A sint+B cost = C sin(t+ ), where C =A2 +B2and = tan1BAif A > 0, = pi + tan1BAif A < 0Note that when using a calculator to compute , it must be in radian mode! Example 6.2(c): Change the solution y = 1.20 sin 1.67t + cos 1.67t into the formy = C sin(t+ ).C =(1.67)2 + 12 = 1.95, = pi + tan1 BA= 2.60Thus the solution can be written as y = 1.95 sin(1.67t+ 2.60) 104There are some important parameters associated with a function y = C sin(t+ ):Amplitude, frequency, period and phase shiftFor a function y = C sin(t+ ), C is the amplitude is the angular frequency T = 2piis the period f = 1T=2piis the frequencyis the phase shift = 2pif Example 6.2(d): For the function y = 1.95 sin(1.67t + 2.60), give the amplitude,period, frequency and phase shift.The amplitude is 1.95, the period is T = 2pi1.67= 3.76, the frequency is f = 2pi= 0.64 andthe phase shift is 2.601.67= 1.56. With a bit of careful thought, the following should be clear: Only the amplitude andand phase shift are influenced by the initial conditions. The period, frequency and angularfrequency in this case are all determined by m and k, the parameters of the system.1056.3 Forced, Undamped VibrationPerformance Criterion:6. (d) Determined whether a forced, undamped system will exhibitresonance, beats, or neither.Suppose that we have either a mass on a spring with no damping, subject to a sinusoidal external force, or an inductor and a capacitor (no resistor) in series with a voltage source that is puttingout a sinusoidal current.How would we expect the functions describing the position of the mass or the charge on thecapacitor to behave? That is what you will investigate in the exercises for this section. Youmight want to take a guess as to what you would expect, before the mathematics of thesituation answers the question.1066.4 Damped VibrationPerformance Criteria:6. (e) Determine from the coefficients of a second order, constant coef-ficient homogeneous ODE whether the solution will be (a) under-damped, (b) critically damped, or (c) overdamped.(f) Sketch the graph of overdamped or underdamped solutions forgiven initial conditions without finding the solution to the differ-ential equation.Harmonic Motion AgainLet us first return to the spring-mass system of Section 1.3. Re-read that section before reading the following. Then supposethat we submerge the mass in an oil bath, as shown to the right(think about an oil-damped shock absorber). As the mass movesup and down there is now an additional force acting on it, theresistance of the oil. You can guess that this resistance will causethe amplitude of the vibration to decrease over time, eventuallycausing it to die out. We will make the assumption that theforce is directly proportional to the velocity but in the oppositedirection; that is, for some positive constant C, the force ofresistance is Cdydt.oilSuppose that we then have some variable external force acting on the mass as well. (Thinkforce exerted upward on a shock absorber by the road, as a car drives along.) If this force issome function f(t), then the net force on the mass at any time t isFnet = md2ydt2= f(t) Cdydt ky.If we rearrange this equation we getmd2ydt2+ Cdydt+ ky = f(t),the second order differential equation that models the motion of a spring-mass system withdamping and an external forcing function F . We will study such ODEs and their IVPs ingreat detail in Chapter 4. There we will see that essentially the same equation is used tomodel the behavior of an electric circuit consisting of a voltage source, inductor, capacitorand resistor, all connected in series.In the previous section you studied the motion of a mass on a spring with no damping. Inthis section we will add damping; imagine a mass on a spring, but the mass is placed in anoil bath. For practical purposes, think of a shock absorber consisting of a spring in a cylindercontaining oil. In this case the differential equation that models the motion of the mass ismd2ydt2+ dydt+ ky = 0 , (1)107where m and k are again the mass of the mass and the spring constant, respectively. isa positive damping constant.Our main objective here is to understand the behavior of the solution function y(t) ofequation (1), and how that behavior varies depending on the parameters m, k and . Youwill be looking at some exercises to do this. Realize that the values given for the parametersmay not be realistic - they were chosen in such a way as to make the mathematics reasonable.1086.5 Forced, Damped VibrationPerformance Criterion:6. (g) Identify the transient and steady-state parts of the solution to adamped system with forced vibration.Whether you have been doing the mechanical exercises or the electrical, so far you have beendealing with what we call free vibration. In the mechanical case this means a mass on aspring, possibly damped, with no external force acting after time zero. Electrically, this is aseries circuit containing a capacitor and inductor, maybe a resistor, but no current source.In both the mechanical and electrical situations, we are dealing with a second order linearhomogeneous differential equation with constant coefficients:md2ydt2+ dydt+ ky = 0 or Ld2qdt2+Rdxdt+1Cq = 0When we add an external force, it gives us a function of time on the right side of the aboveequations:md2ydt2+ dydt+ ky = f(t) or Ld2qdt2+Rdxdt+1Cq = f(t)The function f is usually called the forcing function. Such a function is often a sine orcosine function, representing perhaps the input of a motor (mechanical) or an alternatingcurrent (electrical). We will now investigate what happens in this situation.1096.6 RLC Electrical CircuitsIn Section 4.3 you studied the motion of a mass on a spring with no damping. The electricalanalog to this is a circuit consisting of a resistor, an inductor and a capacitor in series witheach other. We will refer to such a circuit as an RLC circuit. There is no voltage source- imagine that the voltage source has just been removed and replaced with a short circuit.There are three ways that something of interest can be happening in the circuit: There is no current in the circuit, but there is an initial charge on the capacitor, there is no charge on the capacitor, but there is an initial current, there is both an initial charge on the capacitor and an initial current.These are analogous to the three possible conditions for the mass on the spring. The dif-ferential equation that models An RLC circuit is derived from Kirchoffs voltage law whichtells us that the sum of the voltage drops across each of the three components is zero. ByOhms law the voltage drop across the resistor is V = iR, where i is the current and R isthe resistance of the resistor. The voltage drops across the inductor and the capacitor areLdidtand Cq,where q is the charge on the capacitor. From Kirchoffs voltage law we now haveLdidt+Ri+ Cq = 0But the current i is the rate at which charge is passing by a point in the circuit: i =dqdtThusdidt=ddt(dqdt)=d2qdt2and the above equation becomesLd2qdt2+Rdqdt+1Cq = 0 , (2),where L is the inductance of the inductor in henrys, R is the resistance of the resistor inohms, and C is the capacitance of the capacitor in farads. All are positive quantities; thiswill be important! Note that this is exactly the type of equation that we have been using tomodel spring-mass systems.1101116.7 Chapter 6 SummaryPerformance Criteria:6. (h) For a spring-mass system or electric circuit, understand the re-lationships between the physical situation (presence and type of damping and/orforcing) the form of the differential equation, including the functionf the the analytic and graphical nature of the solution (inparticular, the presence and appearance of transient andsteady-state parts of the solution)In the previous chapter we saw how to solve second order linear ODEs with constant co-efficients, and their associated initial value problems. In this chapter we have seen twoapplications, spring-mass systems and electric circuits, and investigated various combina-tions of initial conditions and forcing functions f for both undamped and damped systems.In this section we will attempt to summarize all that we have seen. In particular, we wantto recognize from an ODE, the solution to an ODE, or the graph of the solution to an ODEwhether it models a situation in which the system is undamped, under-damped, critically damped or over-damped with or without an external forcing function for which the solution has transient or steady-state parts, or both resulting in or exhibiting resonance or beatsThe type of differential equation that we are talking about here is one of the formay + by + cy = f(t) (1)where the coefficients a, b and c are constant parameters based on the physical propertiesof the system we are considering: In a spring-mass system a is the mass, b is the coefficient of damping, and c is thespring constant. In an electric circuit, a is the inductance, b is the resistance and c is the reciprocalof the capacitance.It is clear that without a mass and a spring there is no spring-mass system, so for thatsituation neither a nor c can be zero. For the electric circuit situation it is reasonable toconsider a system with only a resistance and inductance, but that can be treated as a firstorder ODE in a manner you have already seen. For an RLC circuit of the sort we wish toconsider, none of the values a, b or c are zero, although we will consider the case b = 0 asa theoretical possibility.112The Left Side of the ODEThe left hand side of the ODE describes the system itself; in the case of a spring-mass system,it is the spring, the mass, and the damping. In an electric circuit it is the resistor, inductorand capacitor. The system doesnt cause motion or current, it just shapes it by the wayit reacts to it. There will not be any motion or current unless there are nonzero initialconditions, a forcing function f , or both. What is of real concern to us on the left handside of the equation (1) is the role of b, which controls the damping. Here is a summary ofthat: When b = 0 the system is undamped. When b2 4ac < 0 the system is under-damped. Oscillation will occur, but anyoscillation due to the system itself will decay. (The solution will have a transient part.)Any steady-state behavior will be due to the forcing function f . When b24ac > 0 the system will be over-damped, and there will be no oscillation dueto the system itself. The system will again have a transient part, and any steady-statepart of the solution is again due to the external forcing function f . When b2 4ac = 0 the system is critically damped, and the solution will behavesimilarly to the over-damped situation. The transient part will have a tekt term(k > 0), but still decays over time because ekt decays faster than t grows.Any quick investigation of an ODE of the form (1) should perhaps begin by observing whethera damping term is present. If it is, computation of b2 4ac should follow to determinewhich of the last three cases above we are dealing with.Initial ConditionsIn order to solve a second order ODE, we must have two initial conditions. Im not wellqualified to talk about their meaning in the case of an RLC circuit, but for a spring-masssystem the meaning of the initial conditions is pretty straightforward. The initial positiontells us whether the mass is raised or pulled down at time zero. In either case, there ispotential energy due to either gravity (for y(0) > 0) or the spring (for y(0) < 0) that isconverted to kinetic energy of motion when the mass is let go. y(0) is the initial velocityimparted to the mass; it is negative if the initial velocity is downward, and positive if theinitial velocity is upward.The effect of initial conditions is not lasting, unless the system is undamped. For anydamped system, the initial conditions will lead to a transient part of the solution. For anundamped system, initial conditions will lead to a steady-state part of the solution.The Forcing Function fThe function f , on the right hand side of (1), is the forcing function that is imposed on thesystem. In the case of the spring-mass system it might be something like effect of bumpsin the road for a shock absorber, or perhaps the effect of some vibration added by a motor.For an electrical circuit, it is the voltage source that is supplying the circuit. Often f willbe, in reality, a periodic function made up of sine or cosine functions. For this reason it issufficient to understand the behavior of the system when f is a single trig function.113f provides input to the system over time, unlike the initial conditions, which only supplyinput right at time zero. It, of course, leads to the particular solution to (1). At this pointwe will only consider decaying exponential or trigonometric forcing functions. A decaying exponential function is itself transient, so in a mathematical sense whenf is such a function it leads to a transient part of the solution. (That part of thesolution is the particular solution.) For an undamped system such a forcing functionwill also act to cause a steady-state part of the solution as well, even in the absence ofinitial values. (In that case it acts, in a sense, like an initial velocity.) When f is a periodic function like a trig function, it will provide input to the systemforever. Because of this, it will usually lead to a periodic steady-state part of thesolution. The one exception is for an undamped system, where we find the followingbehaviors: When the frequency of the forcing function is significantly different from the naturalfrequency (sometimes called the resonant frequency) of the system, the resultis a steady-state solution with two parts, one with the resonant frequency and onewith the frequency of the forcing function. When the frequency of the forcing function is the same as the resonant frequencyof the system, the forcing function will cause part the solution to be trigonomet-ric functions with continually increasing amplitude. This is the condition we callresonance. When the frequency of the forcing function is close to the resonant frequency of thesystem, it will cause vibration with increasing amplitude when the forcing functionis in phase with the vibration. Eventually the forcing function will become out ofphase with the natural vibration, canceling it out. It will then get back in phase,then out, over and over. The result is the phenomenon called beats.Solutions to Second Order Initial Value ProblemsWhen we look at the equation of the solution toay + by + cy = f(t), (1)it will generally have a number of terms, which are parts connected to the rest of the solutionby addition or subtraction. Any terms that that have factors of the form ekt, with k > 0,will decay as time goes on, so those are transient parts of the solution. All of them togetherare what we will call the transient part of the solution; this is sometimes called the transientresponse.Any parts of the solution that are periodic or constant, taken together, constitute thesteady-state part of the solution, or the steady-state response of the system. These occurwhen there is a constant forcing function, or a periodic forcing function with period differentfrom the period of the homogeneous solution, if it is periodic. A solution will also havea steady-state part if the system is undamped. Solutions can have parts that are neithertransient nor steady-state. An example is something like y = t sin 2t, obtained when theforcing function f is periodic with the same period as the homogeneous solution. Thisfunction is not periodic because its amplitude is increasing over time.114When the system is undamped, the solution to the homogenous equation has the formy = c1 sint + C2 cost, so it is periodic. The frequency depends on the system andis called the resonant frequency of the system. (Specifically, it depends on the mass andspring constant for the spring-mass system, and on the inductance and capacitance in theelectrical case.) When the forcing function f is periodic with the frequency a situationcalled resonance will occur. The result is a solution whose equation contains terms withsine and/or cosine functions multiplied by a factor of t that causes growth in the solutionas time goes on. Graphically we see a solution that oscillates with increasing amplitude, asmentioned in the previous paragraph.Consider again an undamped system. When f is periodic with frequency that is closeto, but not the same as, the resonant frequency of the system, the solution exhibits a phe-nomenon called beats. The appearance of the solution equation is no different than for theundamped situation with periodic forcing function having a frequency that is significantlydifferent than the resonant frequency. Graphically, however, the solution having beats willshow oscillation at one frequency but with periodic changes in amplitude. Electrical folkscall this amplitude modulation. (AM, like AM radio!)1156.8 Chapter 6 Practice ExercisesSection 6.21. Suppose that the mass is moved upward by 2.5 cm and released.(a) Sketch the graph of y versus t, taking care with the fact that positive y isupward. Make the amplitude of the motion clear on your graph.(b) Express the initial conditions mathematically by giving values for y(0) and y(0).2. Repeat Exercise 1 for a mass that is set in motion by hitting it upward (as it hangs atequilibrium), giving it an initial speed of 3 inches per second.3. Repeat Exercise 1 for a mass that is set in motion by giving it an initial speed of 8 cmper second downward from a point 2 cm above equilibrium. The amplitude is not two- make it clear whether it is more or less than two.The differential equation that models the motion of a mass on a spring, without damping, ismd2ydt2+ ky = 0 , (1)where m is the mass of the mass (that sounds confusing, huh?) and k is the springconstant. Both are positive quantities; this will be important!4. A mass of 34kg is attached to a spring with a spring constant of 15 N/m (Newtonsper meter). It is set into motion by raising it 2.5 cm and releasing it. (See Exercise 1.)(a) Set up an initial value problem consisting of a differential equation and two initialconditions.(b) Solve the initial value problem to find the position function y.(c) Graph your solution (use your calculator!) and compare with your answer to 1(a).They should agree; if they dont, figure out what is wrong and fix it!5. (a) Consider again a mass of 34kg is attached to a spring with a spring constant of15 N/m (Newtons per meter), as in Exercise 4. Solve the initial value problemobtained with the initial conditions of Exercise 3, where the initial speed was 8 cmdownward from a point 2 cm above equilibrium.(b) Graph your solution and compare with your answer to 3(a).(c) Put your solution in the form y = A sin(t+). Check it by graphing it and yoursolution to part (a) together; they should be the same!6. Any given situation involving undamped motion is essentially the same as the one youjust encountered, with some minor variations. For this reason it is convenient to simplysolve a generic problem like (1) once, and simply work with the solution. For reasonswhich (I hope) will become apparent, we letkm= 2 , where is some positivereal number. (Mathematically it would be OK for to be negative, but for practicalpurposes it is better if it is positive.)116(a) Solve (1) by first dividing through by m and using 2 forkm.(b) The constant is the angular frequency of the periodic motion of the mass.What is the angular frequency for Exercises 4 and 5?(c) We use the symbol T for the period of the motion, which is obtained fromT =2pi. What is the period of motion for Exercises 4 and 5? Give units withyour answer - they should be the appropriate time units.(d) The frequency of the motion is denoted by f , and it is given by1T=2pi. Findthe frequency for exercises 4 and 5.7. A mass of 410kg on a spring with spring constant 4 N/m is given an initial velocity of9 cm/sec upward from an initial position of 4 cm below equilibrium.(a) Find the equation of motion of the mass, y(t), in the form y = A sin(t+ ).(b) What are the period and frequency of the motion?Section 6.31. For this exercise and each of the following, work in decimals, rounding all values to thenearest tenth.(a) Solve the initial value problemx + 4.84x = 8 cos 5t , x(0) = x(0) = 0(b) Graph the solution on your calculator, sketch the graph. Be sure to get a viewingwindow that is appropriate. Put a scale on your graph.(c) Discuss the situation of transient and steady-state solutions. Why should we expectthis before even solving the differential equation?2. (a) Solve the initial value problemx + 4.84x = 8 cos 2.2t , x(0) = x(0) = 0(b) Graph the solution on your calculator, then sketch the graph.(c) The phenomenon you are observing here is called resonance. In either the me-chanical or electrical case, as the amplitude gets larger and larger, something willfail - the spring or one of the electrical components. What is it about the situationthat is causing this to happen?Note that the angular frequency of 2.2 that appears in the solution to the IVP comes fromthe homogeneous equation, so it depends only on the spring-mass or LC system, not on theforcing function. That frequency is sometimes called the natural frequency of the system.3. (a) Solve the initial value problemx + 4.84x = 8 cos 2t , x(0) = x(0) = 0(b) Graph your solution from t = 0 to t = 75. Sketch the graph.117(c) The phenomenon you are observing here is called beats - in electronics this isamplitude modulation. All I know about this is that the AM in AM radio standsfor amplitude modulation (FM is frequency modulation)! Ask your local EETinstructor for details. Look carefully at how this initial value problem compareswith the other two. What do you suppose it is that is causing the beats?(d) A trig identity can help us get a little better insight into the solution. You shouldbe able to write your solution in the form x(t) = A(cos0t cost). Use theidentitycosu cos v = 2 sin(v u2)sin(u+ v2)to rewrite your solution. The new form of the solution is trying to talk to you. Canyou see what it is trying to tell you?(e) Graph y = 19 sin(0.1t) and y = 19 sin(0.1t) together with the graph of thesolution, and sketch what you see. Can you now see what the solution to (d) istrying to tell you?Section 6.41. (a) Solve the initial value problem consisting of Equation (1) with m = 5, = 6 andk = 80, and initial conditions y(0) = 2, y(0) = 6. Give your answer in theform y = Ceat sin(t+) and all numbers in decimal form, rounded to the nearesttenth. (Note that 5.0007 rounded to the nearest tenth is 5.0, not 5! What is thedifference?)(b) Graph the solution to the IVP on your calculator. Adjust the viewing window toget about 3 cycles of the motion displayed fairly large. Sketch your graph.(c) Graph y = 2.3e0.6t and y = 2.3e0.6t together with the solution you graphedin (b). Add them to your sketch as dashed curves.What you have just seen is an example of what is called underdamped motion. There isdamping, but it is small enough to allow the mass to move up and down while the vibrationdecays.2. (a) Solve the initial value problem consisting of Equation (1) with m = 5, = 50 andk = 80, and initial conditions y(0) = 2, y(0) = 6. Give all numbers in decimalform, rounded to the nearest(b) Sketch the graph of your solution over a time period long enough to show what ishappening over time. Make your vertical scale such that the general shape of thegraph can be seen.(c) Compare and contrast your solution to this IVP with the solution to the IVP fromExercise 1, in terms of amplitude over time and oscillation.3. For your answer to 2(c) you should have noted that the solution of the IVP in Exercise1 oscillates as it decays, and the solution of the IVP in Exercise 2 does not. The initialconditions do not affect this (can you see why not?), so the only difference is the valueof .(a) Determine a value of that is the dividing line between equations have solutionsthat oscillate as they decay and those that do not oscillate. Explain how youdetermined it.118(b) Solve the IVP with m = 5, k = 80 and the value of you obtained in (a), alongwith the initial conditions from Exercise 2. Graph the solution on your calculatorand sketch the graph.(c) Suppose now that m = 5 and k = 80. How would you expect the solution to(1) to behave if was slightly smaller than the value obtained in (a), and why.How would you expect the solution to behave for slightly larger than the valueobtained in (a)? Answer these questions in complete sentences.For your answer to 3(c) you should have said that if is a little less than 40, the solution willbehave like that in Exercise 1, and if is a little more than 40 the solution will behave likethat in Exercise 2. The situation in Exercises 3(a) and 3(b) is called the critically dampedcase, meaning that if there is just a little less damping the behavior will be oscillatory, butif the damping has that critical value or greater, the motion will not be oscillatory.4. Describe a method for determining from the coefficients m, and k whether thesolution will be underdamped, overdamped, or critically damped.5. Sketch the graph of the solution to a 2nd-order, constant-coefficient homogeneous ODEwith the given damping conditions and initial conditions.(a) Underdamped, y(0) < 0, y(0) > 0.(b) Overdamped, y(0) < 0, y(0) < 0.(c) Overdamped, y(0) > 0, y(0) < 0. (I think there may be a couple differentappearances possible here - well investigate this more later.)(d) Undamped (not underdamped, but undamped), y(0) > 0, y(0) > 0.Section 6.51. Consider the second order initial value problemy + 2y + 10y = 9.4 sin t , y(0) = 5 , y(0) = 0In this exercise you will find the solution to this initial value problem, and you willinvestigate its behavior. Recall the process for solving such an equation: Find the homogeneous solution yh. It will contain two constants. Use undetermined coefficients (guessing) to find the particular solution yp to theequation. Add yc and yp to find the general solution to the equation. Apply the two initial conditions to determine the values of the unknown constants.Be sure to conclude by writing your final solution to the IVP.(a) Carry out the above process for the given IVP. Give all numbers as decimals,rounded to the tenths place.(b) Graph the solution on your calculator, for t = 0 to t = 10. Sketch your graph.(c) Look carefully at your solution (the equation itself, not the graph). Recall that thetransient part of a solution is any part that goes to zero as time goes on. Any partthat does not go to zero over time is called the steady-state part of the solution,or just the steady-state solution. Give the transient and steady-state parts of yoursolution, telling clearly which is which.119(d) Graph the steady-state solution together with the complete solution that you al-ready graphed. The complete solution should approach the steady state solution astime goes on. Add the graph of the steady-state solution to your graph from (b).Section 6.71. Here are some equations of the sort we have been discussing:(i) y + 3y + 2y = 0 (v) y + 9y = 0(ii) y + 4y = sin(9t) (vi) y + 5y + 7y = 7.4 sin(2.4t)(iii) y + 10y + 25y = 0 (vii) y + 3y + 5y = 0(iv) y + 25y = 3.1 sin(5t) (viii) y + 16y = 7 cos(3.8t) 4 sin(3.8t)(a) Which equations model an undamped system? Which model an under-dampedsystem? Critically damped? Over-damped?(b) Which equations will have solutions with a transient part? Which will have solu-tions with a steady-state part?(c) Which equations will have solutions that exhibit resonance? Which will have solu-tions exhibiting beats?2. Consider the following solutions to differential equations of the type we have beendiscussing (second order, constant coefficient).(i) y = C1et + C2e3t(ii) y = C1e2t + c2te2t(iii) y = C1 cos(5.1t) + C2 sin(5.1t)(iv) y = e1.2t[C1 cos(5t) + C2 sin(5t)](v) y = e0.4t[C1 cos(2t) + C2 sin(2t)] 1.3 cos(7t)(vi) y = C1 cos(3t) + C2 sin(3t) + 0.13 cos(8t) 1.46 sin(8t)(vii) y = C1 cos(3t) + C2 sin(3t) + 0.13t cos(3t) 1.46t sin(3t)(viii) y = A sin(0.1t) sin(6.1t)(a) Identify the transient and steady-state parts of each solution. (Some may not haveboth.)(b) Which solutions are for differential equations of the form ay + by + cy = 0?(c) Which solutions are for undamped systems? Which are for under-damped systems?Critically damped? Over-damped?1203. Below are some graphs of solutions to ODEs of the form ay + by + cy = f(t), whereeither, or both, of b or f(t) may be zero.yt(i)yt(ii)yt(iii)yt(iv)yt(v)yt(vi)yt(vii)(a) Which graphs are for solutions to undamped systems? Under-damped systems?Critically or over-damped systems? (You should not be able to tell the graphs forcritically damped or over-damped apart.)(b) Which graphs are for ODEs of the form ay + by + cy = 0?4. None of the ODEs in Exercise 1 have a solution equation given in Exercise 2, or solutiongraph given in Exercise 3. However, we CAN match up the FORMS of the ODEs,solution equations, and graphs of solutions. For example, equation (iii) from Exercise1 matches with solution (ii) from Exercise 2 and graph (i) from Exercise 3. Find eightother sets of three like this; one graph will have to be used more than once.1217 Boundary Value Problems7.1 IntroductionLearning Outcome:7. Set up and solve boundary value problems.Performance Criteria:(a) Solve a boundary value problem for the deflection of a horizontalbeam.(b) Give the boundary conditions for a horizontal beam.(c) Predict the shape of the deflection curve for a horizontal beamthat is supported in a given manner.(d) Give the boundary conditions for a vertical column.(e) Find the buckling modes (non-trivial solutions) for a vertical col-umn.(f) Find the critical loads for a vertical column.(g) Give the pinning conditions resulting in each of the bucklingmodes of a vertical column.(h) Determine eigenfunctions and eigenvalues for a derivative opera-tor.All of the applications that we have studied so far have involved some quantity that isa function of time; that is, time has been the independent variable. Arbitrary constantshave arisen in the process of solving the associated ODEs, and we have used given initialconditions to determine the values of the constants. In this chapter we look at deflection(bending) of horizontal beams and vertical columns, as a function of the distance along thebeam or column. The independent variable is then a one dimensional position variable.As discussed in Section 2.4, we use boundary conditions, rather than initial conditions, todetermine the values of the arbitrary constants.We will designate the variable x to denote the distance along the beam (or column)from one end or the other. Due to loading of the beam there will be some deflection y offof the straight line the beam would follow with no loading. The deflection will be differentat different points along the beam, so y = y(x). That is, the amount of deflection dependon where one is looking along the length of the beam. The situation will be significantlydifferent for horizontal and vertical columns; well take a look at both.1227.2 Deflection of Horizontal BeamsPerformance Criteria:7. (a) Solve a boundary value problem for the deflection of a horizontalbeam.(b) Give the boundary conditions for a horizontal beam.(c) Predict the shape of the deflection curve for a horizontal beamthat is supported in a given manner.In this section we will take a look at the differential equations associated with beams thatare suspended horizontally in some way. The beams themselves will not be horizontal overtheir entire lengths, because the force of gravity will cause some bending. The first thing tounderstand is the mathematical setup. Suppose that we have a beam of length 10 feet. Weput the cross-sectional center of its left end at the origin of an x-y coordinate plane, andthe cross-sectional center of its right end at the point (10, 0). The axis of symmetry of thebeam then runs along the x-axis from x = 0 to x = 10; see the figure below and to theleft.10yx10yxNow the beam will deflect (a fancy term for sag) in some way, due to any weight it issupporting, including its own weight. The shape it takes will depend on the manner inwhich it is supported (we will get into that soon), but one possibility is shown in the figureabove and to the right. The original axis of symmetry of the beam now follows the graph ofa function, which we will call y(x). Note that the domain of the function is just the interval[0, 10]. Our goal will be to find the mathematical equation of the function.Let us still consider a 10 foot beam, but we will repre-sent it with just the curve described by the axis of sym-metry. (From now on, when we talk about the beam, wereally mean the axis of symmetry of the beam.) Supposealso that the ends of the beam are what we call embed-ded. This means that they are not only supported atboth ends, but the ends are also held horizontal by beingclamped somehow. A good image to keep in mind is abeam that is stuck into two opposing walls of a structure.See the diagram to the right.10yxThe theory behind obtaining a differential equation to model a horizontal beam is beyondthe scope of this class. Suffice it to say that it involves ideas from the area of statics, like123the bending moment of the beam, and things involving the properties of the material fromwhich the beam is built. The differential equation itself is fourth order:EId4ydx4= w(x) (1)Here E is Youngs modulus of elasticity for the material from which the beam is made,I is the moment of inertia of a cross-section of the beam and w(x) is the load per unitlength of the beam. If the beam has uniform cross-section and the only weight that it issupporting is its own weight, then w(x) is a constant. We will stick with that condition.Because the ODE is fourth order, we will need four boundary conditions to determine allof the constants that will arise in solving it. We first recognize that because the two ends aresupported, there will be no deflection at either end. Therefore y(0) = y(10) = 0. This willbe the case for any horizontal beam that is supported at both ends. Next we consider thefact that the ends of the beam are embedded horizontally into a wall. The embedding causesboth ends to be horizontal right at the points where they leave the walls they are embeddedin, so the slope of the beam is zero at those points. Mathematically we express this byy(0) = y(10) = 0. When we put the ODE together with these boundary conditions we geta boundary value problem. Suppose that for our ten foot beam beam E = 10, I = 5 andw(x) = 100. The boundary value problem that we have is then50d4ydx4= 100, y(0) = 0, y(0) = 0, y(10) = 0, y(10) = 0 Example 7.2(a): Solve the boundary value problem just given.We begin by dividing both sides by 50 to getd4ydx4= 2. Our task now is to keep integratingboth sides until we find y = y(x). Integrating once givesd3ydx3= 2x+C1, and integratingagain givesd2ydx2= x2 + C1x + C2. Next we find thatdydx= 13x3 + 12C1x2 + C2x + C3,and applying the initial condition y(0) = 0 gives C3 = 0. Substituting this value andintegrating one more time we get y = 112x4 + 16C1x3 + 12C2x2 + C4, and applying theboundary condition y(0) = 0 results iny = 112x4 + 16C1x3 + 12C2x2. (2)We now apply the initial condition y(10) = 0 to get 0 = 100012+ 10006C1 +1002C2, andthe initial condition y(10) = 0 to get 0 = 10003+ 1002C1 + 10C2. To solve this systemwe multiply the first equation by 12 and the second by 6 to get the system to the leftbelow, which can be solved in the manner shown in the other steps:2000C1 + 600C2 = 10, 000300C1 + 60C2 = 2000= 20C1 + 6C2 = 10030C1 6C2 = 200 10C1 = 100C1 = 10Substituting this value for C1 and solving for C2 gives us C2 = 403 . Putting thesevalues into (2), the solution to the IVP is y = 112x4 53x3 203x2. 124Use your calculator to graph the solution from x = 0 to x = 10, using a y scalethat allows you to actually see the deflection of the beam. Does the result surprise you? (Itshould!) One annoying feature of the differential equation is that it is based on taking downto be the positive direction. To see what the actual shape of the beam will be, multiply yoursolution by 1, then graph it. Now the result should look something like the picture nearthe bottom of page 102.Suppose again that we have a 10 foot beam, but nowit is supported by a fulcrum at each end, and each end isfree to pivot around the fulcrum. We will call the beamsimply supported in this case. See the diagram to theright. In this case, we will have y(0) = y(10) = 0, justlike the embedded case. However, we can see that we willnot have y(0) = 0 or y(10) = 0.x = 0 x = 10So how do we get two more boundary conditions? Note that the downward force of gravityin the interior and the upward force of the supports at the ends bend the beam into a concaveup shape. (Remember concavity from Math 251? The upward concavity here means thatit must be the case that y(x) > 0 for values of x between, but not equal to 0 and 10.)However, there are no opposing forces to bend the beam right at its ends. Thus there is noconcavity right at the ends of the beam, resulting in the conditions y(0) = y(10) = 0.The final situation we will consider for now is a beamthat is embedded at the left end and free at the rightend, as shown. The left end of the beam is embedded, sowe know the values y(0) = y(0) = 0. We know neitherthe displacement nor the slope of the right end, but whatwe do know there is that there is no concavity there, soy(10) = 0. This gives us three boundary conditions, butof course we need four. The last condition comes fromsome theory we wont go into here, but it is y(10) = 0.x = 0 x = 10Let us now summarize the the possible boundary conditions for a horizontal beam: At an embedded end both y and y are zero. At a simply supported end y and y are zero. At a free end y and y are zero.I expect you be able to give any of those conditions - you should be able to figure all ofthem out each time you need them, without memorization, with the possible exception ofthe third derivative just discussed.1257.3 Deflection of Vertical ColumnsPerformance Criteria:7. (d) Give the boundary conditions for a vertical column.(e) Find the buckling modes (non-trivial solutions) for a vertical col-umn.(f) Find the critical loads for a vertical column.(g) Give the pinning conditions resulting in each of the bucklingmodes of a vertical column.In this section we will examine the behavior of avertical column when a load is applied directly down-ward on its top, as shown in the first picture to theright. We will begin by considering columns that arepinned; this means we will allow the top and bottomof the column to be at angles other than perpendicu-lar to the ground. We will require the top and bottomof the column to be vertically aligned with each other- later we will consider a situation where we will relaxthis condition. So, for example, when enough force isapplied downward the column will deflect horizontallyas shown in the second picture to the right.LoadForceWe will set up a coordinate system as shown belowand to the right, with x indicating the distance up-ward from the bottom of the beam and y = y(x) rep-resenting the horizontal deflection of the column atany point x. The differential equation governing thedeflection of the column isEId2ydx2= Py,where E and I are as in the last section; they are propertieswhich could vary along the column (with the variable x), butthis would be unusual. Well only consider columns where theydo not change. P is the (positive) force exerted downward onthe top of the column, and we will look at the effects of differentvalues of P , but for purposes of solving the ODE it is a constant.Since the ODE is second order we will need two conditions todetermine the solution. If the length of the column is denoted byL, we have the boundary conditions y(0) = 0 and y(L) = 0.If we rearrange the equation and combine it with the boundaryconditions we get the boundary value problem (BVP)EId2ydx2+ Py = 0, y(0) = 0, y(L) = 0.x = Ly(L) = 0y(0) = 0xy (pos)126We will see that solving this boundary value problem is somewhat different than solving thekind of BVPs we saw for horizontal beams, in the previous section. Lets go through thatnow, with specific values of E, I and L. The methods and ideas in this section aresignificantly different than the other things we have done in this course - youll need to putyour thinking cap on for this one! Example 7.3(a): Solve the boundary value problemEId2ydx2+ Py = 0, y(0) = 0, y(20) = 0, (1)with E = 800, I = 150 and P > 0.Substituting the values for E and I into the ODE and solving for the derivative gives usd2ydx2+ P120,000y = 0 (2)Using our methods from Chapter 5, this equation has auxiliary equation r2+ P120,000= 0 and,because P > 0, its roots are r = iP120,000. The solution to (2) is theny = C1 sinP120,000x+ C2 cosP120,000xApplying the boundary condition y(0) = 0 gives us C2 = 0, so the solution is theny = C sinP120,000x. (Ive omitted the subscript for simplicity.) Now here is where thingsstart to get interesting! The other boundary condition tells us that0 = C sinP120,000(20) (3)which, in turn, tells us that either C = 0 or sinP120,000(20) = 0. The first possibilitygives us the trivial solution y = 0 - it satisfies the differential equations and boundaryconditions, but it isnt particularly interesting! Considering the second possibility, sin =0 for = 0, pi, 2pi, 3pi, ..., npi, ... so, for C 6= 0, (3) will be true forP120, 000(20) = 0, pi, 2pi, 3pi, . . . , npi, . . . , (4)the first of which also gives us the trivial solution y = 0. Therefore, in theory at least(well talk later bout what all this means from a practical point of view), we can only havenonzero solutions the the BVP ifP120, 000=pi20,2pi20,3pi20, . . . ,npi20, . . .This gives us the nonzero solutionsy = C sin pi20x, y = C sin 2pi20x, y = C sin 3pi20x, . . . , y = C sin npi20x, . . .to the boundary value problem. 127This isnt really the end of the story, but we need to pause to catch our breath anddevelop some terminology before resuming. At this point what we know about the boundaryvalue problem (1) is that y = 0 is a solution, called the trivial solution (because it ismathematically uninteresting), and we only have nontrivial solutions for discrete values ofP ; those solutions are the ones the example concluded with. The non-trivial solutions arecalled buckling modes (for reasons youll soon see); the first non-trivial solution is calledthe first buckling mode, the second is the second buckling mode, and so on. The onlyvalues of P for which we can have nontrivial solutions are those that satisfy (4). Solvingfor P gives usP = 120000( pi20)2, 120000(2pi20)2, 120000(3pi20)2, . . . , 120000(npi20)2, . . .= 300pi2, 300(2pi)2, 300(3pi)2, . . . , 300(npi)2, . . .= 300pi2, 4(300pi2), 9(300pi2), . . . , n2(300pi2), . . .These values of P are called critical loads. Like the buckling modes, they are numbered,so 300pi2 is the first critical load, 4(300pi2) is the second critical load, etc. The wordload refers to the load held up by the column. Note that the second critical load is four(two squared) times the first critical load, the third critical load is nine (three squared) timesthe first critical load, and so on.What happens physically is this: When there is no load on the column it is perfectlystraight (the solution y = 0), and it remains that way until the first critical load is reached.At that point the column will deflect sideways, taking the shape of the curve y = C sin pi20x,shown at the left below. In reality, as the load increases beyond the first critical load, thedeflection will remain the same shape, but with increasing amplitude, until the column fails.201st buckling mode10202nd buckling mode13(20)23(20)203rd buckling modeIf we were able to prevent the middle point of the column, at x = 10, from deflecting,the column would be able to support the second critical load. Because the column is heldwith y(10) = 0, the deflection of the column will take the shape of a full period of thesine function, as shown in the middle picture above; this is the second buckling mode. Theact of preventing deflection is sometimes called pinning. If we pin the column at pointsone-third and two-thirds of the way along its length, the column would be able to supportthe third critical load, and the shape of the deflection would be given by the third bucklingmode, shown to the right above.128Lets revisit the boundary conditions y(0) = y(L) = 0, where L is the length ofthe column. We should first note that the only requirement we really had was that thetop and bottom of the column were vertically aligned. We could just as well have puty(0) = y(L) = 2, as shown in the diagram at the left below; however, the mathematicsinvolved are a bit simpler if we instead use y(0) = y(L) = 2, as we did. Physically wemust still insist that the top and bottom be aligned. Without this restriction, any horizontalshifting of the ceiling would result in hinging and a collapse. The beginning of such acollapse is indicated by the three diagrams below and to the right.Ly(0) = y(L) = 2xy (pos)LoadForceNow suppose the top and bottom were embedded, rather than being pinned. If the columnhad length L, we would then have the boundary conditionsy(0) = y(L) = 0. (5)This gives us two boundary conditions, which is what we should need in order to solve thesecond order ODEEId2ydx2+ Py = 0,and leaves us without any conditions on y(0) and y(L). In this situation, we couldconceivably allow the ceiling to drift laterally without collapse, because the columnbeing held in a vertical alignment at the bottom and top would provide enough rigidityto prevent collapse. The first three buckling modes for this situation are shown below; ifthe ceiling were prevented from drifting, the second mode would then become the first, thefourth would become the second, and so on. You will investigate this situation, along withthe corresponding critical loads, in the exercises.1st buckling mode 2nd buckling mode 3rd buckling mode1297.4 Eigenfunctions and EigenvaluesPerformance Criteria:7. (h) Determine eigenfunctions and eigenvalues for a derivative opera-tor.Consider again the differential equationEId2ydx2+ Py = 0Lets divide both sides of the differential equation by EI to get the ODEd2ydx2+PEIy = 0.Just to make things simpler we will replacePEIwith the symbol (lambda). SinceP,E and I are all positive quantities, is a positive real number also. This will beimportant soon.So we wish to solve the differential equation y + 2y = 0, which we have done severaltimes before (with specific numbers in place of ). We can rewrite this asd2dx2(y) = 2y (1)Remember that y is a function. The notationd2dt2(y) is used to indicate that the secondderivative with respect to x is what is sometimes called an operator, which means it is afunction that acts on functions, and the result of that action is another function. Equation(1) says that we are looking for a function for which the action of the second derivativeoperatord2dx2really just amounts to multiplying the function by the constant 2. Thissort of situation happens often enough, and is important enough, that mathematicians havedeveloped some language relating to it. We will introduce that language, as well as somethat leads up to it, in this section.OperatorsA function can be thought of as a mathematical machine that takes in a number and,in return, gives out a number. There are other mathematical machines that take in thingsother than numbers, like functions or vectors, usually giving out things like what they takein. As stated above, these sorts of machines are often referred to as operators. Example 7.4(a): As just mentioned, the second derivative d2dx2is an operator. Youshould be quite familiar with its action:d2dx2(5x3 + 7x2 2x+ 4) = 30x+ 14 130 Example 7.4(b): Of course the first derivative is an operator as well, and we cancreate new operators by forming something called a linear combination of derivatives.As an example, we can define an operatorD = 3d2dt2+ 5ddt 4by its action on a function y = y(t):D(y) = 3d2ydt2+ 5dydt 4y.So, for example, if y = e2t,D(e2t) = 3d2dt2(e2t) + 5ddt(e2t) 4e2t = 12e2t 10e2t 4e2t = 2e2tYou saw such operators when we studied second order linear ODEs. Example 7.4(c): You may know that when we multiply the matrix A =[ 4 63 5]timesthe vector u =we getAu =[ 4 63 5] =[(4)(1) + (6)(3)(3)(1) + (5)(3)]=[ 2218]Similarly, for the vector v =,Av =[ 4 63 5] =[ 21]We can think of A as an operator that acts on vectors with two components to createother vectors with two components. There are a couple of properties of the derivative operator that you should be familiarwith; in the following, f and g are functions, and c is a constant:ddx[f(x) + g(x)] =dfdx(x) +dgdx(x) andddx[cf(x)] = cdfdx(x)Similarly, if A is a matrix, u and v are vectors, and c is a scalar (constant),A(u+ v) = Au+ Av and A(cu) = cAuOperators with these two properties, of distributing over addition and passing throughconstants, are called linear operators. (This is where the linear in linear algebra comesfrom.) Many operators used in applications, including all of the ones above, are linearoperators.Eigenfunctions and EigenvaluesIn general, the result when an operator works on an object has no particular relationshipwith whatever it was that wasworked on, other than usually being the same general sort ofobject. However, for most operators there are a few objects that give a special result whenworked on, as well see in the following examples.131 Example 7.4(d): Consider again the second derivative d2dx2, and note thatd2dx2(sin 5x) = 25 sin 5x.The effect of the derivative on sin 5x is to simply multiply the function by 25. Example 7.4(e): Note that in Example 7.4(c), the result of A times v was simply1 times v.Example 7.4(d) above illustrates the following definition.Eigenfunctions and EigenvaluesLet A be an operator and y a nonzero function for which there is aconstant k such thatAy = ky.Then y is an eigenfunction of the operator A, with correspondingeigenvalue k. Example 7.4(f): Based on Example 7.4(d), sin 5x is an eigenfunction for the operatord2dx2, with corresponding eigenvalue 25. Example 7.4(g): Example 7.4(e), shows that v =is an eigenvector (insteadof eigenfunction) for the matrix A =[ 4 63 5], with eigenvalue 1. Example 7.4(h): Because the derivative of a constant is zero, which is also zero timesthe function, every constant function is an eigenfunction of the first derivative operatorddx, with corresponding eigenvalue zero. This emphasizes that even though the zerofunction isnt allowed as an eigenfunction, eigenfunctions are allowed to have eigenvaluesof zero.We finish with two final observations. First, for any constant C,d2dx2(C sin 5x) = 25C sin 5x = 25(C sin 5x).This indicates that any constant multiple of an eigenfunction is also an eigenfunction, withthe same eigenvalue. This holds for eigenvectors as well; we dont really think of suchmultiples as new eigenfunctions or eigenvectors. We also see thatd2dx2(cos 5x) = 25 cos 5x,showing that an operator can have more than one eigenfunction (beyond just constant mul-tiples) with the same eigenvalue. This also holds for matrices and eigenvectors.1327.5 The Heat Equation in One DimensionIn this section, we will consider the physically impossible but mathematically convenientsituation: We have a metal rod of length L (see picture below) that is perfectly insulatedalong its length, so that no heat can enter or escapealong its length, but for which heat can enter or leavethe ends. At any time t greater than zero and anyposition x along the rod, the function u(x, t) givesthex = 0 x = Ltemperature at that point x and time t. (We will think of the rod as being infinitelythin, so that the rod has only one point at each x position. If you are not happy withthis, an alternative is to think that if the rod had some thickness the temperature at everypoint in a cross-sectional slice at some x is the same, so we need not consider the othertwo space dimensions.) Suppose that at time zero there is some distribution of temperaturesalong the rod, given by a the function f(x) for 0 x L, and suppose also that the endsof the rod are held at temperature zero at all times t 0. The function f gives an initialcondition for each point x along the length of the rod, and the conditions that the endsare held at temperature zero are boundary conditions.What we would like to know is whether, and how, we can determine the temperature atany point x with 0 < x < L (we know the temperatures at x = 0 and x = L are alwayszero), at any time t > 0. Here the dependent variable u depends on the two independentvariables x and t. Some physical principles concerning heat give us a differential equationfor this situation and, due to there being two independent variables, it is a partial differentialequation. The equation (called the heat equation) and conditions given above can all bestated asut= k2ux2, u(0, t) = u(L, t) = 0, u(x, 0) = f(x) (1)Here the conditions u(0, t) = u(L, t) = 0 are boundary conditions and u(x, 0) = f(x) isessentially an initial condition (for every point along the rod). Thus we have a problem thatis a sort of combination initial value/boundary value problem. But we can really just thinkof it as a boundary value problem for this reason: If we were tothink of the Cartesian plane as representing position x alongthe horizontal axis and time t along the vertical axis we get apicture like the one to the right, where each point in the shadedregion represents a point x in the rod and some time t. Ourgoal is then to find the temperature at each of those points; inthis way we can think of trying to find function values in a regionthat is bounded by the line from zero to L on the x-axis and thetwo half-infinite lines from zero to infinity in the t directionat x = 0 and x = L. The condition u(x, 0) = f(x) can bethought of as a boundary condition along the bottom, and theconditions u(0, t) = u(L, t) = 0 are boundary conditions alongthe two sides.xtLRecall that if we have the function f(x, y) = x2y3, to get the partial derivativefxwesimply take the derivative of x2y3, treating y (and therefore y3) as a constant. Similarly,133we getfyby treating x as a constant, so we havefx= 2xy3 andfy= 3x2y2Lets try another. Example 7.5(a): Find uxandutfor u(x, t) = e2t sin 3x.When findinguxwe consider t to be a constant, so e2t is as well. The derivative isthenux= 3e2t cos 3t.When findingut, sin 3t is essentially a constant, sout= 2e2t sin 3t. Those whove had a multivariable calculus course will perhaps recall that the computation ofpartial derivatives can be significantly more complicated (and therefore difficult) than this,but if we understand the above example we are ready to understand how (1) is solved. Themethod for doing it is called separation of variables, which is similar in execution, up toa point, to the method of the same name we used to solve separable first order ODEs up toa point. After that point we must proceed differently.To begin, we assume that the function u(x, t) is actually a product of a function ofx alone and another function of t alone. There is no practical reason to think this mightbe the case, but the method works, so well use it! (This method was invented/discovered inthe 1700s by Daniel Bernoulli, part of a family of a number of accomplished mathematiciansand scientists. Daniel was also involved in the derivation of the ODE we used for horizontalbeams.) If we let g be the function of x and h be the function of t, then u(x, t) = g(x)h(t).Now remember that if we are taking the derivative of g(x)h(t) with respect to x, h(t) istreated as a constant, and when taking the derivative with respect to t, g(x) is treated asa constant, sout= g(x)h(t) and2ux2= g(x)h(t).The differential equation in (1) then becomes g(x)h(t) = kg(x)h(t). If we divide bothsides by kg(x)h(t) we geth(t)kh(t)=g(x)g(x). (2)Here is where we deviate from the procedure for solving first order separable equations. (2)needs to be true for all values of x and t, and this is likely only the case if both sidesof (2) are equal to some constant (again, well see that it works!) that we will call 2.For reasons we wont go into here, is positive. Setting each side equal to 2 andmultiplying by the denominators we getddth(t) = k2h(t) and d2dx2g(x) = 2g(x). (3)134In addition to this, we also have the boundary conditions g(0) = g(L) = 0 for the secondequation. The first equation in (3) tells us that h(t) is an eigenfunction for the firstderivative operator, with eigenvalue k2, and we know that h(t) is then any constantmultiple of ek2t. The second equation says that g(x) is an eigenfunction of the secondderivative operator with eigenvalue 2; because is positive, the eigenfunctions areconstant multiples of sint and cost. What we have done is called separation ofvariables, although it is a bit different from the method of the same name that we used tosolve some first order ODEs.Lets focus a bit more on the second ODE in (3) and its boundary values g(0) = G(L) = 0.The general solution to the ODE is g(x) = C1 sin t + C2 cost. Applying the conditiong(0) = 0 gives us C2 = 0, so well omit the subscript on C1 and say that the solutionis g(x) = C sint. (At this point this story should be starting to feel familiar!) We nowconsider the boundary condition g(L) = 0, which gives us 0 = C sinL. As before, whenconsidering vertical columns, we dont want to let C = 0, so we must have sinL = 0.This implies thatL = 0, pi, 2pi, 3pi, ... = = 0, piL,2piL,3piL, ...and the solutions to the boundary value problem (disregarding constants and the zero solu-tion arising from = 0) aresinpiLx, sin2piLx, sin3piLx, ...The solution h then becomes h(t) = ekpi2L2 , e4kpi2L2 , e9kpi2L2 , ..., so we get a sequence ofsolutions u(x, t) = g(x)h(t):u(x, t) = ekpi2L2 sin piLx, e4kpi2L2 sin 2piLx, e9kpi2L2 sin 3piLx, ... (4)Recall that when solving an ODE like y + 3y + 2y = 0 we assumed y = ert for someconstant r. From this we obtain y = et or y = e2t, but we saw that the sum ofconstant multiples of these two, y = C1et + C2e2t is the most general solution. By thesame reasoning, the most general solution to the PDE were looking at is an infinite sum ofthe solutions in (4):u(x, t) = C1e kpi2L2 sin piLx+ C2e 4kpi2L2 sin 2piLx+ C3e 9kpi2L2 sin 3piLx+ (5)This story goes on quite a bit longer, but lets end it with the following. In order to try tomeet the condition u(x, 0) = f(x) we must havef(x) = C1 sinpiLx+ C2 sin2piLx+ C3 sin3piLx+ (6)The right hand side of (6) is something called a Fourier series. This brings up the questionIn what way (or ways) do we interpret the equal sign in (6), and for whatfunctions f can such interpretation(s) be made?Attempts to answer this question gave birth to a large amount of mathematics over manyyears, starting with Joseph Fouriers work in the early 1800s, and with a major result provedas late as 1966. Perhaps some of you will investigate this subject more in later coursework.1357.6 Chapter 7 Summary1367.7 Chapter 7 Practice ExercisesSection 7.21. For each beam pictured below, list the boundary conditions. Assume that the height ofthe left end of each is zero.(a) (b)12 ft8 ft(c) (d)20 ft x = 0 x = 152. Suppose that you have a beam that is 8 feet long. For each of the scenarios given,determine first whether it would make any sense physically to have the beam supportedin the manner given. If not, explain why. If it does make sense, give the four boundaryconditions.(a) Left end embedded horizontally, right end simply supported.(b) Left end simply supported, right end free.(c) Both ends free.3. Find the deflection function y = y(x) for an eight foot beam that is embedded atboth ends, carrying a constant load of w(x) = 150 pounds per foot. Suppose also thatE = 30 and I = 80, in the appropriate units.(a) Give the appropriate boundary value problems (differential equation plus boundaryconditions).(b) Solve the differential equation and apply the boundary conditions in order to de-termine the constants. What is the final solution?(c) Graph your solution on the appropriate x interval, using a y scale that allowsyou to actually see the deflection of the beam. Remember to multiply the rightside of your solution from (b) by negative one so that it appears the same way thatthe beam will.137(d) Where do you believe the maximum deflection should occur? Find the deflectionthere - you need not give units with your answer, since I have been somewhat vagueabout the units of the constants E and I.4. (a) Sketch a graph of the deflection of a beam that is embedded at its left end and freeat its right end.(b) Suppose that the beam is 10 feet long, with values of w0, E and I of 100, 10and 5, respectively. Solve the boundary value problem.(c) Graph your solution and compare with your sketch in part (a). Of course theyshould be the same.(d) What is the maximum deflection of the beam, and where does it occur?5. Repeat parts (a)-(d) of Exercise 4 for a 10 foot beam that is simply supported on bothends.6. Repeat steps (a)-(c) of Exercise 4 for an eight foot beam, with the same parameters asin Exercise 3, that is embedded on the left end and simply supported on the right end.Then do the following:(d) It was intuitively clear where the maximum deflection occurred for the two previoussituations, but it is not so clear in this case. Take a guess as to about where youthink it should occur for this case. Then use the graph on your calculator, alongwith the trace function, to determine where the maximum deflection occurs, andhow much it is.7. The graphs below are those of some fourth degree polynomials. The points labeled A,D, E, H, I and L are maxima for their respective functions, and the points labeled B,C, F, G, J and K are inflection points. For each of the following boundary situations,give the endpoints of a section of graph that has the shape the deflection curve wouldtake. Assume that both ends of the beam are supported at the same level, and that thedashed lines are horizontal.(a) Simply supported at the left end, embedded at the right end.(b) Embedded at both ends.(c) Simply supported at both ends.ABCDEF GH IJKL138Section 7.3Give all answers in exact form.1. A 12 foot vertical column is pinned at both ends. For the material it is made of wehave E = 500 and, from its design, we have I = 200, both in the appropriate units.(a) Give the first four buckling modes, showing all steps of solving the IVP to get them.(b) Give the first four critical loads.(c) How many times larger is the third critical load than the first critical load?2. Now suppose we have a 6 foot vertical column with E = 500, I = 200, and both endspinned.(a) Give the first buckling mode, and find the corresponding critical loads.(b) Compare your critical loads with those from the 12 foot length (Exercise 1). Howdo the corresponding critical loads for the six foot column compare with those forthe 12 foot column? Does that make sense intuitively?(c) The first buckling mode for the six foot column is the same as which buckling modefor the 12 foot column? Draw a picture showing what is going on here.3. Consider now a 12 foot vertical column with E and I values of 500 and 200 again,but this time with both ends embedded. Suppose also that the ceiling is not allowedto drift, so the top and bottom of the column are vertically aligned.(a) Give the first three buckling modes.(b) Give the first three critical loads.(c) How does the third critical load compare to the first? (That is, how many timeslarger is it?)(d) How does the first critical load compare with the first critical load for the 12 footcolumn with pinned ends? (See Exercise 1(a).) How about the other critical loads?4. Repeat the previous exercise, but with the assumption that the ceiling is allowed todrift. (Hint: You should be able to use your computations from Exercise 3, ratherthan re-doing all of them.)5. Repeat parts (a) and (b) of Exercise 1 for a vertical column of length L that is pinnedat both ends, with modulus of elasticity E and moment of inertia I. This will givea general form of the buckling modes and critical loads for pinned ends.6. Repeat parts (a) and (b) of Exercise 4 for a vertical column of length L that isembedded at both ends, with the ceiling allowed to drift. Again use a modulus ofelasticity E and moment of inertia I. This will give a general form of the bucklingmodes and critical loads for embedded ends.Section 7.41. In this exercise you will be considering the first derivative operatorddx.139(a) The function y = e3x is an eigenfunction for the operator. What is the corre-sponding eigenvalue?(b) Give the eigenfunction of the operator that has eigenvalue 5.(c) Based on your answers to parts (a) and (b), what is the general form of any eigen-function of the first derivative operator and what is the corresponding generaleigenvalue?2. Now consider the second derivative operatord2dx2. There are three general forms ofthe eigenfunctions for this operator, depending on whether the eigenvalues are positive,negative or zero.(a) Give a specific function that has eigenvalue zero; that is, the second derivative ofthe function is zero.(b) Give the most general form of function whose eigenvalue is zero.(c) Weve already seen and worked with functions that have negative eigenvalues. Giveone general form of functions that have negative eigenvalues for this operator, andgive the corresponding eigenvalue for the general form. Then give the other generalform and its corresponding eigenvalue.(d) Give the eigenfunctions that will have eigenvalue nine; there are two of them!(e) Give the general form of the eigenfunctions of the operator that have positiveeigenvalues, and give the general eigenvalue.3. Let D be the operator D =d2dt2+ 5ddt+ 3, whose action on a function y = y(t) isdefined by Dy =d2ydt2+ 5dydt+ 3y.(a) Show that y = e2t is an eigenfunction for this operator, and determine thecorresponding eigenvalue.(b) In general, any function of the form ekt is an eigenfunction for D. Determine thegeneral eigenvalue.140A Review of Calculus and AlgebraA.1 Review of DifferentiationOutcomes: Apply the rules of differentiation, along with provided formulas, tofind the derivatives of functions.The purpose of this appendix is to remind you of how to find the derivatives of somefunctions of the sorts that we will encounter as we go on. For this we will usually takethe independent variable to be x and the dependent variable y; the statement y =f(x) indicates that y is a function of x; recall also the notations y = f (x) =dydxforthe derivative. Lets begin by giving the derivatives of some common functions, and somebasic rules of taking derivatives. The notation ( ) means the derivative of the quantityin the parentheses, and k represents an arbitrary constant.Derivatives of Some Functions(k) = 0 (x) = 1 (xn) = nxn1 (ex) = ex(sin x) = cosx (cos x) = sin x (lnx) = 1xFor the following, k again represents an arbitrary constant, and f(x), g(x), u =u(x) and v = v(x) are functions of the independent variable x.Derivative Rules[kf(x)] = kf (x) [u v] = u v(uv) = uv + vu(uv)=vu uvv2f [g(x)] = f [g(x)]g(x)Youll recall the third and fourth items above as the product rule and quotient rule,and the last item is the chain rule. The first two rules above tell us that the derivative issomething called a linear operator. This is the same idea as a linear transformation, forthose of you who have had linear algebra. Now well see how to use some of the rules andderivatives of common functions to find derivatives of some combinations of those functions.141Derivatives of PolynomialsYoull recall that to find the derivative of a polynomial like f(x) = 5x3 34x27x+2 wesimply use the various basic derivatives and rules as follows: Multiply the coefficient of eachterm by the corresponding power of x, and decrease the power by one, remember that thederivative of a constant is zero. Example B1: Find the derivative of f(x) = 5x3 34x2 7x+ 2.f(x) = 5x3 34x2 7x+ 2 = f (x) = 15x2 32x 7 1. Find the derivative of each polynomial:(a) y = 3x5 7x4 + x2 3x+ 2 (b) h(t) = 16t2 + 23.7t+ 3.5(c) f(x) = 23x4 + 5x3 18x2 + 3 (d) s(t) = t3 2t2 + 3t 5Second DerivativesFor certain applications we will need to take the derivative of the derivative of a function.Youll recall that such a derivative is called the second derivative (so the kind of derivativeyou did above is called a first derivative). The notation is this: If the first derivative isy or f (x), then the second derivative is y or f (x). If the first derivative isdydx,then the second derivative isd2ydx2. Note the different placement of the exponents in thenumerator and denominator of this expression - there is a reason for this, but its not tooexciting, so I wont go into that here.2. Find the second derivative of each of the functions in Exercise 1.Other Applications of the Power RuleSince we can write things like 1x2andx using negative and fractional exponents, wecan use the power rule to find their derivatives. Here are some examples: Example B3: Find the derivative of y = 5x3y =5x3= 5x3 dydx= 15x4 = 15x4 Example B4: Find the derivative of f(x) =3x4f(x) =3x4=14x13 f (x) = 112x23 =1123x2Note that the four in the denominator is a constant, but it is really 14, not 4. 1423. Find the derivative of each of the following. Give your final answers without negativeor fractional exponents.(a) f(x) =3x2(b) g(t) =t26 6t2(c) y = 15x (d) h(x) =43x4. Find the second derivatives of the functions from 3(a) and (b).The Chain RuleThis is perhaps the most confusing (but also most important!) of the derivative rules.Lets use an example to try to explain it; suppose that y = (x5 + 2x 4)8. Now if youknew x = 3 and wanted to compute y it would be sort of a two step process. First youwould compute 35 + 2(3) 4, then you would take the result to the eighth power. To dothe derivative, you do the derivative of the last part first[( stuff )8] = 8( stuff )7 ,and multiply by the derivative (x5 + 2x 4) = 5x4 + 2: Example B5: Find the derivative of y = (x5 + 2x 4)8.y = 8(x5 + 2x 4)7 (x5 + 2x 4) = 8(x5 + 2x 4)7(5x4 + 2) In the notation of the rule in the box on the previous page, g(x) = x5 + 2x 4 andf(u) = u8. Example B6: Find the derivative of f(t) = 5.6 sin(4.2t 1.3).f (t) = 5.6[cos(4.2t1.3)](4.2t1.3) = 5.6[cos(4.2t1.3)](4.2) = 23.6 cos(4.2t1.3) In this last example there is the additional constant 5.6 that is just multiplied by theresult of the derivative of the rest of the function. Note that the result of the derivative of4.2t 1.3 is brought to the front and multiplied by the original constant at the end of theprocess. More on this in a bit.5. Find the derivative of each of the following, utilizing the suggestions provided.(a) g(x) =16 x2 - If you write the square root as an exponent you will havesomething very much like the first example above.(b) y = 3 cos(pi2t)(c) A(t) = 500e0.3t - This is a two-step process (again, ignoring the constant of500), with the first step being 0.3t and the second step being the exponential.Remember that the derivative of ex is ex.6. Find the derivative of each of the following.(a) y = 5e2x (b) x = 4 sin(3t) (c) g(x) =2(x2 4x)7143(d) s(t) = cos(25t) (e) y = ex2In light of what you have seen, we can revise some of our basic derivatives some. Again,u is a function of x: u = u(x).Derivatives of Some Functionsddx(eu) = eududxddx(ln u) =1ududxddx(sin u) = cosududxddx(cosu) = sin ududxProper Manners(!) for Writing FunctionsYou will often have functions that are products of constants, powers of x and exponential,trig or log functions. In the case of such a function, the correct order to write the factors isconstant, power of x, exponential function, trigonometric functionOccasionally you will have a logarithmic function, but these will not generally occur withtrigonometric functions; a logarithmic function goes where the trigonometric function islisted above.7. For each of the following, put the product in the correct order.(a) e5t3 15t3 (b) 3[ sin(5x2)] 5 (c) 4 sin(5t+3) (e2t)8. For each of the following, multiply and put the final product in the correct order.(a) 4 sin(5t+ 3) (2e2t) (b) 7et cos(3t 1) 2t (c) (3 lnx)(4x3)The Product RuleSince (u + v) = u + v, one might hope that (uv) = uv. A simple example showsthat this is not true: Example B7: Let u(x) = x2 and v(x) = x3, and find [u(x)v(x)] and u(x)v(x).We see that[u(x)v(x)] = [x2x3] = (x5) = 5x4 and u(x)v(x) = (2x)(3x2) = 6x3.Therefore [u(x)v(x)] is NOT equal to u(x)v(x)! So how do we find the derivative of a product of two functions? Well, we use the productrule, which says that if u = u(x) and v = v(x) are two functions of x, then(uv) = uv + vu ord(uv)dx= udvdx+ vdudx144 Example B8: Find the derivative of y = 3x2 sin 5x.y = (3x2)(sin 5x) + (sin 5x)(3x2)= (3x2)(5 cos 5x) + (sin x)(6x)= 15x2 cos 5x+ 6x sin x 9. Use the product rule to find the derivative of each of the following.(a) f(t) = t2e7t (b) y = 3x cos 2x (c) h(t) = 2e3t sin pitThe Quotient RuleThe quotient rule is similar to the product rule. For two functions u and v of x,(uv)=v u u vv2 Example B9: Find the derivative of f(x) = e2xx5f (x) =(x5)(e2x) (e2x)(x5)(x5)2=(x5)(2e2x) (e2x)(5x4)x10=2x5e2x 5x4e2xx1010. Find the derivative of each of the following.(a) f(x) =3 sin 2te7t(b) y =4e5tt2(c) g(x) =cos 6x2x311. Use the quotient rule and the fact that tanx =sin xcosxto determine the derivative oftanx.145A.2 Review of Integration146A.3 Solving Systems of EquationsIn this course we will often need to solve systems of two equations in two unknowns. Thisis a simple process that you learned long ago, but if you make a mistake in carrying it out,you will have problems with the remainder of whatever else you have to compute afterward.Thus it is important to be able to do this quickly and correctly. There are two methods,the addition method and the substitution method. Each has its own advantages anddisadvantages; both methods will be demonstrated here.The Addition Method Example A1: Solve the system 3x y = 52x+ y = 15.The basic idea of the addition method is to add the two equations together so that one ofthe unknowns goes away. In this case, as shown below and to the left, nothing fancy needbe done. The remaining unknown is then solved for and placed back into either equationto find the other unknown as shown below and to the right.3x y = 5 3(4) y = 52x+ y = 15 12 y = 55x = 20 12 = y + 5x = 4 7 = yThe solution to the system is (4, 7). What made this work so smoothly is the = y in the first equation and the y in thesecond; when we add the two equations, the sum of these is zero and y has gone away.In the next example we see what to do in a slightly more difficult situation. Example A2: Solve the system 3x+ 4y = 13x+ 2y = 7.We can see that if we just add the two equations together we get 4x + 6y = 20, whichdoesnt help us find either of x or y. The trick here is to multiply the second equationby 3 so that the the first term of that equation becomes 3x, the opposite of the firstterm of the first equation. When we then add the two equations the x terms go away andwe can solve for y:3x+ 4y = 13 = 3x+ 4y = 13 x+ 2(4) = 7x+ 2y = 7 = 3x 6y = 12 x+ 8 = 7times 3 2y = 8 x = 1y = 4The solution to the system of equations is (1, 4).The steps shown above are not the only way to solve the system; we could have eliminatedy instead of x. To do so, wed multiply the second equation by 2 so that the the147y term of that equation becomes 4y, the opposite of the y term of the first equation.When we then add the two equations the y terms go away and we can solve for x:3x+ 4y = 13 = 3x+ 4y = 13 1 + 2y = 7x+ 2y = 7 = 2x 4y = 14 2y = 8times 2x = 1 y = 4The solution to the system of equations is of course the same, (1, 4). Example A3: Solve the system 2x 4y = 183x+ 5y = 5.In this case we cant multiply one equation or the other by a number to get it to match upwith the other equation; we must multiply each equation by something to do this. Wellchoose to eliminate the y terms, because they already have opposite signs. To do so wemultiply the first equation by five and the second by four:times 52x 4y = 18 = 10x 20y = 90 2(5) 4y = 183x+ 5y = 5 = 12x+ 20y = 20 10 4y = 18times 422x = 110 4y = 8x = 5 y = 2The solution to the system of equations is (5,2). We could have obtained the sameresult by multiplying the first equation by 3 and the second equation by 2, in order toeliminate the x terms. Lets summarize the steps for the addition method, which youve seen in the above ex-amples.The Addition MethodTo solve a system of two linear equations by the addition method,1) Multiply each equation by something as needed in order to make thecoefficients of either x or y the same but opposite in sign.2) Add the two equations and solve the resulting equation for whicheverunknown remains.3) Substitute that value into either equation and solve for the other un-known.Occasionally it is beneficial to do some slight variation on the above, as demonstrated inthe following example.148 Example A4: Solve the system 2x 3y = 44x+ 5y = 3.We multiply the first equation by 2 and add to the second equation:times 22x 3y = 4 = 4x+ 6y = 84x+ 5y = 3 = 4x+ 5y = 311y = 5y = 511Ordinarily we would substitute this value of y into one of the equations and solve for x,but that is a bit annoying in this case, since y is a fraction. It is easier to just repeat theaddition step, but in such a way as to eliminate y this time:times 52x 3y = 4 = 10x 15y = 204x+ 5y = 3 = 12x+ 15y = 9times 322x = 29x = 2922The solution to the system of equations is (2922, 511). The Substitution MethodThe substitution method can be done for any system of two equations in two unknowns,but it will generally be a bit messier than the addition method unless the coefficient of oneof the unknowns in one of the equations is one or negative one. In that case, the followingsteps will result in a solution to the system:The Substitution MethodTo solve a system of two linear equations by the substitution method,1) Pick one the equation in which the coefficient of one of the unknownsis either one or negative one. Solve that equation for that unknown.2) Substitute the expression for that unknown into the other equationand solve for the unknown.3) Substitute that value into the other equation and solve for the otherunknown.Youll find an example on the next page.149 Example A5: Solve the system of equations x 3y = 62x+ 5y = 5 using the substitutionmethod.Solving the first equation for x, we get x = 3y + 6. We now replace x in the secondequation with 3y + 6 and solve for y. Finally, that result for y can be substituted intox = 3y + 6 to find x:2(3y + 6) + 5y = 56y 12 + 5y = 5 x 3(7) = 6y 12 = 5 x+ 21 = 6y = 7 x = 15y = 7The solution to the system of equations is (15,7). 150A.4 Series and Eulers FormulaTry the following: Set your calculator for radians (as you always should in this course) andfind sin(0.05), sin(0.1) and sin(0.5). You should get (to five places past the decimal)0.04998, 0.09983 and 0.47943, respectively. These numbers are fairly close to the numbersthat you were finding the sine of, with the one closest to zero being the best approximation.This seems to indicate thatsin x xfor values of x near zero. Since of course sin(5) must be a number between 1 and1 this will certainly not hold in that case!Now consider the function f(x) = x 16x3+ 1120x5. Here we find that if we round to fiveplaces past the decimal we get f(0.5) = 0.47943, the value of sin(0.5) when rounded tothe same number of decimal places. Lets try something bigger:f(1.3) = 0.96477, sin(1.3) = 0.96356 f(2.0) = 0.93333, sin(2.0) = 0.90930It appears that this function f comes pretty close to approximating the sine function,especially for values of x nearer to zero.The fraction coefficients in the equation for the function f appear to be a bit mysterious,but it turns out that 6 = 3 2 1 and 120 = 5 4 3 2 1. We call the quantityn(n 1)(n 2) 3 2 1 n factorial, denoted by n! So 6 = 3!, 120 = 5! andsin x x x33!+x55!It turns out that (in a sense that you really must take Math 253 to understand)sin x = x x33!+x55! x77!+x99! This is called the (infinite) series representation of the sine function. As you have seen,for values of x near zero, very good approximations of sin x can be obtained by using justthe first term or few terms of this series. For values farther from zero, more terms must beused to get a good approximation.Cosine and the exponential function have series representations as well:cosx = 1 x22!+x44! x66!+x88! andex = 1 + x+x22!+x33!+x44!+x55!+x66!+ Recall now i, the imaginary unit (j for you electronics folks), for which i2 = 1. Wecan also compute things likei1 = i, i3 = i2 i = i, i4 = i3 i = i i = i2 = 1, i5 = i4 i = 1i = i151and so on. Because we can compute these, we can now find something like ei by using theseries representation of ex:ei = 1 + i +(i)22!+(i)33!+(i)44!+(i)55!+(i)66!+ = 1 + i +i222!+i333!+i444!+i555!+i666!+ = 1 + i 22! i33!+44!+ i55! 66!+ =(1 22!+44! 66!+ )+(i i33!+ i55!+ )=(1 22!+44! 66!+ )+ i( 33!+55!+ )= cos + i sin A similar computation can be done for ei. The result of the two of these isEulers Relations:ei = cos + i sin ei = cos i sin 152B Solutions to ExercisesB.1 Chapter 1 SolutionsSection 1.2 Solutions1. (a) The independent variable is time, and the dependentvariable is the amount of radioactive material.(b) See graph to the right.At2. (a) The independent variables are time t and the distance x out from the edge ofthe table, and the dependent variable is the deflection of the ruler at any point andtime.(b) 0 t, 0 x 6, where x is measured in inches. (Substitute 0.5 for 6 ifmeasuring in feet instead of inches.)3. 0 r 5 inches, 0 2pi, 0 t Note that, unless we are doing right triangletrigonometry, angles will be measured in radians.4. If measuring in feet, 0 x 1, 0 y 1, 0 z 1.5. (a) See graph to the right.(b) The shape is a parabola, opening downward.(c) We would model the height with a quadratic function.(d) In this case the domain is finite, of the form 0 t a forthe time a when the rock hits the ground, whereas inExercise 1(b) the domain was infinite, from time zero toinfinity.50 time (seconds)height(feet)Exercise 5(a)6.4080T (F)t(a)4030T (F)t(b)40T (F)t(c)7.0ty(a)0ty(b)153Section 1.3 Solutions1. (a)dydx= 6 cos 3x (b) y = 2e0.5t (c) x = 2t+ 5(d) y = 4.42 sin(1.3t 0.9) (e) dydt= 3te3t + e3t(f) x = 12e2t cos(3t + 5) 8e2t sin(3t+ 5)2. (a)d2ydx2= 18 sin 3x (b) y = e0.5t (c) x = 23. At seven minutes, the temperature is increasing at 2.7F per minute.4. At 12.5 minutes, the amount of salt in the tank is decreasing at 1.3 pounds per minute.5. (a) At 2 seconds the mass is moving downward at 5 inches per second.(b) At 2 seconds the mass is accelerating at 3 inches per second per second (in/sec2).(c) At 2 seconds the mass is speeding up because the acceleration is in the same direc-tion as the velocity.6. (a) y = e3x, y = Ce3x for any constant C.(b) y = sin 3t or y = cos 3t. Any function of the form y = C1 sin 3t + C2 cos 3t willdo it, for any constants C1 and C2.(c) y = e3t, y = Ce3t for any constant C.(d) y = sin5 x or y = cos5 x. Any function y = C1 sin5 x + C2 cos5 x willdo it, for any constants C1 and C2.Section 1.4 Solutions1. (a) The parameters are the the initial temperature T0, the temperature Tm of themedium, and the constant k.(b) The independent variable is time t.(c) The dependent variable is temperature T .2. The parameters are the amount of the mass, the stiffness of the spring (the springconstant), the amount the mass is pulled downward before letting it go, and the viscosityof the oil in the oil bath.3. The independent variable is time t and the dependent variable is current i. Theparameters are the inductance L, the resistance R, the voltage E and the initialcurrent i0.154B.2 Chapter 2 SolutionsSection 2.2 Solutions1. (a) Independent variable: x Dependent variable: y(b) Independent variable: cant determine Dependent variable: y(c) Independent variables: x1, x2, x3, t Dependent variable: u(d) Independent variable: t Dependent variable: y(e) Independent variable: x, t Dependent variable: u(f) Independent variable: x Dependent variable: u(g) Independent variable: x Dependent variable: y(h) Independent variable: x Dependent variable: y(i) Independent variable: r, t Dependent variable: u2. (a) Independent variable: x Dependent variable: y(b) Independent variable: t Dependent variable: y3. no 4. yes9. (a) no (b) y = ce3x is only a solution if c = 2(c) y = 2e3x (d) y = cex is a solution for any value of cSection 2.3 Solutions1. (a), (b), (d), (f), (g), (h)2. (a), (c), (e) and (i) are first order. (b), (d), (f) and (g) are second order, and (h) isfourth order.3. (a) F (x, y) = 2y (c) F (x, y) = yy2 (e) F (x, y) = xxy (i) F (x, y) = 1xy4. (a) f(x) = 0, a0(x) = 2, a1(x) = 1(b) f(x) = 0, a0(x) = 1, a1(x) = 0, a2(x) = 1(c) not linear(d) f(t) = 26e2t, a0(t) = 9, a1(t) = 0, a2(t) = 1(e) f(x) = 1, a0(x) = 1, a1(x) =1x(f) f(t) = 10 sin t, a0(t) = 6, a1(t) = 5, a2(t) = 1155(g) not linear(h) f(x) = w, a0(x) = a1(x) = a2(x) = a3(x) = 0, a4(x) = 1(i) f(x) = 1, a0(x) = x, a1(x) = 15. (a) g(x) = 1, h(y) = 2y (c) g(x) = 1, h(y) = y y2(e) g(x) = x, h(y) = 1 y (i) not separable6. (a) and (c)Section 2.5 Solutions2. Yes, it is a solution.3. (a) Not a solution, y(0) 6= 1 and y is not a solution to the ODE.(b) Not a solution, y is not a solution to the ODE.(c) Solution. (d) Solution.4. A = 235, B = 3B.3 Chapter 3 SolutionsSection 3.2 Solutions1. (a) sin y = 12x2 + C (b)12y2 =12x2+ C ory2 =1x2+ C(c) 15y5 = 13x3 + C or (d) ln |y| = 12ln |2x+ 3|+ C3y5 = 5x3 + C(e) ey = 1x+ C or (f) 12y2 = 52x2 + 3x+ C orey = 1x+ C y2 = 5x2 + 6x+ C2. (a) ln |y| = 12x2 + ln 3 12(b) 12x2 = 52t2 + 3t+ 12 orx2 = 5t2 + 6t+ 24(c) ey = 32x2 + e2 (d) 13y3= sin t 1243. (a) y = 4e3x (b) not separable(c) ln |y| = 2x2 + ln 2 (d) y = 7e2e12x2 2(e) not separable (f) y = 12x+ 521564. (a) y = ln(x2 + x+ 1) (b) y = ln(x2 + x+ e3 2)5. (b) y = Cet2(c) y = 7et2157Section 3.3 Solutions1. y = e4x 2e3x2. y = 0.4te2t + Ce2t 4. y = 0.4te2t + 3e2t 5. y = 32e12x6. (a) y = 629sin 2t 1529cos 2t+ Ce5t (b) y = 629sin 2t 1529cos 2t 10129e5t7. (a) y = Ce3t + 13t2 + 139t 2227(b) y = 7627e3t + 13t2 + 139t 22278. (a) m = 5 (b) y = 629sin 2t 1529cos 2t (c) y = 629sin 2t 1529cos 2t + Ce5t9. (a) y = 12e3x + 72ex (b) y = x ln x+ 2xSection 3.4 Solutions1. (a) A(0) = 80 (b) A = 80ekt (c) k = 0.104, A = 80e0.104t2. (a)dAdt= 310A, A(0) = 87 (b) A = 87e310t (c) t = 9.52 hours3. (a) T = 70 + Cekt (b) C = 38, so T = 70 38ekt(c) k = ln 196 1.15 (d) Steady-state: 70 Transient: 38e ln 196 t 38e1.15t4. (a) i = 65 65e20t or i = 65(1 e20t)(b) i = 5101(10 cos 2t+ sin 2t) 50101e20t or i = 50101cos 2t + 5101sin 2t 50101e20t(c) The transient part is 50101e20t and the steady-state part is 5101(10 cos 2t+sin 2t).5. (a) If T0 = Tm the temperature will not change because the initial temperature ofthe object will be the same as the temperature of the medium; the solution will becompletely steady-state.(b) The steady-state solution is Tm.(c) The transient solution is (T0 Tm)ekt158B.4 Chapter 4 SolutionsSection 4.4 Solutions1. The ODEs in parts (a), (c), (d) and (f) are autonomous.2.0y0 ty(a)y = 0 is an unstable equilibrium30y03tyy = 0 is an unstable equilibriumy = 3 is a stable equilibrium(c)52y25yy = 2 is a stable equilibriumy = 5 is an unstable equilibrium(d)30y03yy = 0 is an unstable equilibriumy = 3 is a semi-stable equilibrium(f)3. (a) y = 0 is a stable equilibrium, y = 2 is an unstable equilibrium, y = 6 is a stableequilibrium(b) y = 1 is a semi-stable equilibrium, y = 4 is a stable equilibrium(c) y = 0 is an unstable equilibrium, y = 2 is a stable equilibrium, y = 5 is anunstable equilibrium4.620y(a)14y(b)025y(c)1595. (a)dydt= y(y2)(y6) (b) dydt= (y4)(y1)2 (c) dydt= y(y2)(y5)Section 4.5 Solutions1. 2. (a) yn+1 = 1.05yn 0.05tnn tn yn y(tn) % error0 0.0 2 2 01 0.2 2.4 2.4214 0.882 0.4 2.84 2.8918 1.793 0.6 3.328 3.4221 2.754 0.8 3.8736 4.0255 3.775 1.0 4.48832 4.7183 4.87n tn yn y(tn) % error0 0.00 1.4 1.4 0.001 0.05 1.47 1.4705 0.032 0.10 1.541 1.5421 0.073 0.15 1.6131 1.6147 0.104 0.20 1.6863 1.6886 0.14(b)3. (a) n tn yn y(tn) % error0 0.0 2 2 0.001 0.1 2 2.01 0.502 0.2 2.02 2.0404 1.003 0.3 2.0604 2.0921 1.52(b) y = 2e12t24. (a) 1.1yn 0.1tn (b) yn+1 = 1.22yn 0.22tn 0.02(c) n tn yn y(tn) % error0 0.0 2 2 01 0.2 2.42 2.4214 0.062 0.4 2.8884 2.8918 0.123 0.6 3.4158 3.4221 0.184 0.8 4.0153 4.0255 0.255 1.0 4.7027 4.7183 0.33(d) The error using the midpoint method are much smaller than those obtained usingEulers method with the same step size.5. (a) k1 = .2yn .2tn, k2 = 0.22yn0.22tn0.02, k3 = 0.222yn0.222tn0.022,k4 = 0.2444yn 0.0444tn 0.044 0.2tn+1Note the appearance of both tn and tn+1 in k4.(b) n tn yn k1 k2 k3 k40 0.0 2 0.4 0.42 0.422 0.44441 0.2 2.4214 0.4443 0.4687 0.4712 0.49852 0.4 2.8918 0.4984 0.5282 0.5312 0.56463 0.6 3.4221 0.5644 0.6009 0.6045 0.64534 0.8 4.0255 0.6451 0.6896 0.6941 0.74395 1.0 4.7182(c) The approximations obtained using this method are very close to the exact values.160B.5 Chapter 5 SolutionsSection 5.2 Solutions1. (a) y = C1e4t + C2et (b) y = C1e12t + C2e5t2. (a) y = C1e3t + C2te3t (b) y = C1e2t + C2te2t3. (a) y = C1et + C2e3t (b) y = et(C1 sin 3t+ C2 cos 3t)(c) y = C1e5t + C2te5t (d) y = e3t(C1 sin2 t + C2 cos2 t)(e) y = C1et + C2e2t (f) y = C1 sin2 t+ C2 cos2 t(g) y = C1et + C2tet (h) y = e2t(C1 sin 5t+ C2 cos 5t)(i) y = e1.55t(C1 sin 1.45t+ C2 cos 1.45t)Section 5.3 Solutions1. (a) yp = 56 t 1336 (b) yp = 35986 sin 5t 217986 cos 5t (c) yp = 45te2t3. (a) y = 76sin 3t+ 2 cos 3t+ 12sin t (b) y = 65e3t + 45e2t + t2 t(c) y = 15e2t te2t + 15e3t (d) y = e2t + 3e5t + 2te2t(e) y = 75e5t 15te5t + 65t + 35(f) y = 138sin 2t+ 12cos 2t 34t cos 2t4. (a) Steady-state: 23sin 3t+ 53cos 3t Transient: none(b) Steady-state: 34cos 7t Transient: e3t(4 sin t+ 7 cos t)(c) Steady-state: 35sin 5t 65cos 5t Transient: 72e2t(d) Steady-state: none Transient: 3te5t 7e5t + etSection 5.6 Solutions1. (a) linearly independent (b) c1 = 1, c2 = 2 (c) c1 = 2, c3 = 3 (d)linearly independent2. (b) c1 = 1, c2 = 1 (c) c1 = 1, c3 = 13. (a) W (x) = 6x2 6x 10 Any value of x will give W (x) 6= 0.(b) W (x) = 9e7x Any value of x will give W (x) 6= 0.4. W (t) = e2kt Any value of t will give W (t) 6= 0.5. W (t) = 2 Any value of t will give W (t) = 2 6= 0.8. y2 = x329. y2 =1x2161B.6 Chapter 6 SolutionsSection 6.2 Solutions1. 2. 22.214.171.124ytyt22yty(0) = 2.5, y(0) = 0 y(0) = 0, y(0) = 3 y(0) = 2, y(0) = 84. (a) 34y + 15y = 0, y(0) = 2.5, y(0) = 0(b) y = 2.5 cos 25 t if working in centimeters, y = 0.025 cos 25 t if working inmeters5. (a) y = 1.8 sin(25t)+2.0 cos(25t) if working in centimeters, y = 0.018 sin(25t)+0.02 cos(25 t) if working in meters(c) y = 0.031 sin(25 t+ 4.1) or y = 3.1 sin(25 t+ 4.1)Section 6.3 Solutions1. (a) x(t) = 0.4 cos(2.2t) 0.4 cos(5t)(c) We should expect no transient solution because there is no damping. Note thatwhenever there is damping the homogeneous solution is the transient part of thesolution. When there is no damping, the homogenous solution will not be transient.2. (a) x(t) = 1.8t sin(2.2t)(c) What causes the resonance is that the frequency of the forcing function f(t) =8 cos(2.2t) is the same as the natural frequency of the system, which is seen in thehomogeneous solution.3. (a) x(t) = 9.5 cos(2t) 9.5 cos(2.2t)(b) x(t) = 19 sin(.2t) sin(4.2t) The first sine function (and the factor of 19) acts asa sort of variable amplitude for the higher frequency second sine function.(c) The beats are being caused by the fact that the frequency of the forcing functionis close to, but not the same as, the natural frequency of the system.Section 6.4 Solutions1. (a) y(t) = e0.6t(1.2 sin 4.0t+ 2.0 cos 4.0t) y(t) = 2.3e0.6t sin(4.0t+ 4.2)2. (a) y(t) = 113e2t 53e8t3. (a) = 40 (b) y(t) = 2e4t + 14te4t5.yt(a)yt(b)yt(c)yt(d)162Section 6.5 Solutions1. (a) y = et(1.4 sin 3t+ 5.2 cos 3t) + sin t 0.2 cos t(c) Transient part: et(1.4 sin 3t + 5.2 cos 3t) Steady-state part: sin t 0.2 cos tSection 6.7 Solutions1. (a) Undamped: ii, iv, v, viii Under-damped: vi, viiCritically damped: iii Over-damped: i(b) Transient: i, iii, vi, vii Steady-state: ii, iv, v, vi, viii(c) Resonance: iv Beats: viii2. (a) (i) entire solution is transient (ii) entire solution is transient(iii) entire solution is steady-state (iv) entire solution is transient(v) Transient: e0.4t[C1cos2t+ C2 cos 2t] Steady-state: 1.3 cos 7t(vi) entire solution is steady-state (vii) Steady-state: C1 cos 3t+ c2 sin 3t(viii) entire solution is steady-state(b) i, ii, iii, iv(c) Undamped: iii, vi, vii, viii Under-damped: iv, vCritically damped: ii Over-damped: i3. (a) Undamped: ii, iii, v, vii Under-damped: iv, vi Critically or over-damped:i, vi(b) i, vi(c) Equation Solution Graphi i iii vi iiiii ii iiv vii viiv iii iiivi v ivvii iv viviii viii v163B.7 Chapter 7 SolutionsSection 7.2 Solutions1. (a) y(0) = 0, y(0) = 0, y(12) = 0, y(12) = 0(b) y(0) = 0, y(0) = 0, y(8) = 0, y(8) = 0(c) y(0) = 0, y(0) = 0, y(20) = 0, y(20) = 0, y(6) = 0(d) y(0) = 0, y(0) = 0, y(15) = 0, y(15) = 02. Only (a) is possible, y(0) = 0, y(0) = 0, y(8) = 0, y(8) = 03. (a) (30)(80)d4ydx4= 150, y(0) = y(0) = y(8) = y(8) = 0(b) y = 1384x4 124x3 + 16x2(d) The maximum deflection should appear at the middle of the beam (x = 4). Thedeflection there is 23.4. (b) y = 112x4 103x3 + 50x2(d) The maximum deflection is 2500 at x = 10, the right hand end of the beam.5. (b) y = 112x4 53x3 + 2503x(d) The maximum deflection is 312512at x = 56. (a) (30)(80)d4ydx4= 150, y(0) = y(0) = y(8) = y(8) = 0(b) y = 1384x4 596x3 + 14x2(d) The maximum deflection is about 1.39 at x = 4.67. (a) J and L (b) E and H (c) F and GSection 7.3 Solutions1. (a) y = C sin pi12x, y = C sin 2pi12x, y = C sin 3pi12x, y = C sin 4pi12x, . . .(b) P = 62509pi2, 4(62509pi2), 9(62509pi2), 16(62509pi2), . . .(c) The third critical load is nine times the first critical load.2. (a) y = C sin pi6x, y = C sin 2pi6x, y = C sin 3pi6x, y = C sin 4pi6x, . . .P = 250009pi2, 4(250009pi2), 9(250009pi2), 16(250009pi2), . . .(b) Each critical load is four times the corresponding critical load for the twelve footcolumn.(c) The first buckling mode for the six foot column is the same as the second bucklingmode for the twelve foot column, but is half as high, so it only includes half a periodof the sine function, whereas the twelve foot columns second buckling mode has afull period of the sine function.3. (a) y = C cos 2pi12x, y = C cos 4pi12x, y = C cos 6pi12x, . . .(b) P = 4(62509pi2), 16(62509pi2), 36(62509pi2), . . .(c) The third critical load is nine times the first critical load.164(d) Each critical load is four times the corresponding critical load for the column withpinned ends.4. (a) y = C cos pi12x, y = C cos 2pi12x, y = C cos 3pi12x, y = C cos 4pi12x, . . .(b) P = 62509pi2, 4(62509pi2), 9(62509pi2), 16(62509pi2), . . .(c) The third critical load is (again!) nine times the first critical load.(d) The critical loads are the same as those for the pinned ends.5. (a) y = C sin piLx, y = C sin 2piLx, y = C sin 3piLx, y = C sin 4piLx, . . .(b) P = EIL2pi2, 4(EIL2pi2), 9(EIL2pi2), 16(EIL2pi2), . . .5. (a) y = C cos piLx, y = C cos 2piLx, y = C cos 3piLx, y = C cos 4piLx, . . .(b) P = EIL2pi2, 4(EIL2pi2), 9(EIL2pi2), 16(EIL2pi2), . . .Section 7.4 Solutions1. (a) The eigenvalue is three. (b) y = e5x(c) Eigenfunction: y = ekx Eigenvalue: k2. (a) y = 3x, y = 5, y = 2x 1, etc. (b) y = Ax+B(c) Eigenfunction: y = sin kx Eigenvalue: k2Eigenfunction: y = cos kx Eigenvalue: k2(d) y = e3x, y = e3x (e) Eigenfunction: y = ekx Eigenvalue: k23. (a) D(e2t) = 4e2t 10e2t + 3e2t = 3e2t, the eigenvalue is 3(b) D(ekt) = k2ekt + 5kekt + 3ekt = (k2 + 5k + 3)ekt, the eigenvalue is k2 + 5k + 3B.8 Solutions for AppendicesAppendix B Solutions1. (a) y = 15x4 28x3 + 2x 3 (b) h(t) = 32t+ 23.7(c) f (x) = 83x3 + 15x2 14x (d) s(t) = 3t2 4t+ 32. (a) y = 60x3 84x2 + 2 (b) h(t) = 32(c) f (x) = 8x2 + 30x 14(d) s(t) = 6t 43. (a) f (x) = 6x3(b) g(t) =13t +12t3(c) y =110x(d) h(t) = 433x44. (a) f (x) =18x4(b) g(t) =13 36t45. (a) g(x) =x16 x2 (b) y = 3pi2sin(pi2t)(c) A(t) = 150e0.3t6. (a) y = 10e2x (b) x = 12 cos 3t (c) g(x) = 14(2x4)(x24x)7(d) s(t) = 25sin(25t)(e) y = 2xex21657. (a) 15t3e5t3(b) 15 sin(5x 2) (c) 4e2t sin(5t+ 3)8. (a) 8e2t sin(5t+ 3) (b) 14tet cos(3t 1) (c) 12x3 lnx9. (a) f (t) = 7t2e7t + 2te7t (b) y = 6x sin 2x+ 3 cos 2x(c) h(t) = 2pie3t cospit 6e3t sin pit10. (a) f (x) =6e7t cos 2t 21e7t sin 2te14t(b) y =20t2e5t 8te5tt4(c) g(x) =12x3 sin 6x 6x2 cos 6x4x6166Indexamplitude, 105analytic solution, 53angular frequency, 105autonomous equation, 23, 59auxiliary equation, 81bending moment, 124Bessel equation, 22boundary conditions, 11, 24, 25, 124boundary value problem, 124buckling mode, 128critical load, 128critical point, 60damped vibration, 107damping, 107deflection of a horizontal beam, 123dependent variable, 3differential equation, 7, 8differential operator, 78direction field, 58domain, 4eigenfunction, 132eigenvalue, 132electric circuit, 45embedded, 123equilibrium solution, 59Eulers formula, 83eulers method, 63family of solutions, 19, 24forced, damped vibration, 109forced, undamped vibration, 106forcing function, 86, 109Fourier series, 135free boundary condition, 125free, undamped vibration, 103frequency, 105general solution, 76, 80heat equation, 133homogeneous equation, 19, 22, 75homogeneous solution, 76, 80independent variable, 3initial conditions, 10, 24initial value problem, 27, 76, 89integrating factor, 34, 39iterative method, 64linear combination, 78, 94linear differential equation, 21linear operator, 78, 131linearly independent functions, 94linearly independent solutions, 94midpoint method, 73mixing problem, 42modulus of elasticity, 11, 124Newtons Law of Cooling, 44non-homogeneous equation, 20, 22non-linear differential equation, 22numerical analysis, 65numerical method, 65Numerical methods, 63numerical methods, 54one-parameter family of solutions, 20operator, 130order of a differential equation, 17, 21, 75ordinary differential equation, 4, 17parameter, 10, 105, 108partial differential equation, 4, 17particular equation, 20particular solution, 76, 80, 86period, 105phase portrait, 60phase shift, 104, 105pinned boundary condition, 126population growth, 41qualitative methods, 53radioactive decay, 41reduction of order, 96relation, 35Runge-Kutta method, 73second order equation, 75semi-stable equilibrium, 62167separable equation, 23separation of variables, 34, 35, 135simple harmonic motion, 9simply supported, 125slope field, 58solution curve, 56solution solution, 20solution to a differential equation, 18spring constant, 9stable equilibrium, 60steady-state solution, 46step size, 63Taylor polynomial, 65transient solution, 46two-parameter family of solutions, 20undetermined coefficients, 76unstable equilibrium, 61variable, 10vertical column, 126wave equation, 22Wronskian, 951681 Functions and Derivatives, Variables and Parameters1.1 Introduction1.2 Functions and Variables1.3 Derivatives and Differential Equations1.4 Parameters and Variables1.5 Chapter 1 Summary1.6 Chapter 1 Practice Exercises2 Introduction to Differential Equations2.1 Introduction2.2 Differential Equations and Their Solutions2.3 Classification of Differential Equations2.4 Initial Conditions and Boundary Values2.5 Initial Value Problems2.6 Chapter 2 Summary2.7 Chapter 2 Practice Exercises3 Solving First Order Equations3.1 Introduction3.2 Solving By Separation of Variables3.3 Solving With Integrating Factors3.4 Applications of First Order ODEs3.5 Some Theoretical Concerns3.6 Chapter 3 Summary3.7 Chapter 3 Practice Exercises4 Qualitative and Numerical Methods: First Order4.1 Introduction4.2 Solution Curves4.3 Direction Fields4.4 Phase Portraits and Stability4.5 Numerical Solutions4.6 Chapter 4 Summary4.7 Chapter 4 Practice Exercises5 Second Order Linear Constant Coefficient ODEs5.1 Introduction5.2 An Initial Exploration5.3 Homogeneous Equations With Constant Coefficients5.4 Particular Solutions, Part One5.5 Initial Value Problems5.6 Particular Solutions, Part Two5.7 Linear Independence of Solutions, Reduction of Order5.8 Chapter 5 Practice Exercises6 Applications of Second Order Differential Equations6.1 Introduction6.2 Free, Undamped Vibration6.3 Forced, Undamped Vibration6.4 Damped Vibration6.5 Forced, Damped Vibration6.6 RLC Electrical Circuits6.7 Chapter 6 Summary6.8 Chapter 6 Practice Exercises7 Boundary Value Problems7.1 Introduction7.2 Deflection of Horizontal Beams7.3 Deflection of Vertical Columns7.4 Eigenfunctions and Eigenvalues7.5 The Heat Equation in One Dimension7.6 Chapter 7 Summary7.7 Chapter 7 Practice ExercisesA Review of Calculus and AlgebraA.1 Review of DifferentiationA.2 Review of IntegrationA.3 Solving Systems of EquationsA.4 Series and Euler's FormulaB Solutions to ExercisesB.1 Chapter 1 SolutionsB.2 Chapter 2 SolutionsB.3 Chapter 3 SolutionsB.4 Chapter 4 SolutionsB.5 Chapter 5 SolutionsB.6 Chapter 6 SolutionsB.7 Chapter 7 SolutionsB.8 Solutions for AppendicesIndex
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