oregon state university mathematics department...

Oregon State UniversityMathematics Department

Study Guide for

Mth 256 - Applied Differential Equations

to accompany the text

Elementary Differential Equationsby William F. Trench

0 10 20 30 40 50

0

50

100

150

t

y

written by

Dr. Filix Maisch

and

Dr. Stephen Scarborough

Summer/Fall 2015

1

Contents

0 Preliminaries 4

0.1 Schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

0.2 Introduction and Notes to Students . . . . . . . . . . . . . . . . . . . . . . . . 5

0.3 Basic Integral Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.4 Table of Elementary Laplace Transforms . . . . . . . . . . . . . . . . . . . . . 7

Lesson 1: Introduction to Differential Equations 9

Lesson 2: Direction Fields 13

Lesson 3: Linear First Order Equations, Use of Integrating Factors 15

Lesson 4: Separable Equations 20

Lesson 5: Existence and Uniqueness Theorems 22

Lesson 6: Transformations 23

Lesson 7: Exact Equations 25

Lesson 8: Autonomous First Order Differential Equations 27

Lesson 9: Applications 35

Lesson 10: Homogeneous Linear Equations 43

Lesson 11: Constant Coefficient Homogeneous Equations 44

Lesson 12: Euler Equations 47

Lesson 13: Nonhomogeneous Linear Equations 50

Lesson 14: The Method of Undetermined Coefficients 52

Lesson 15: Reduction of Order 56

Lesson 16: Variation of Parameters 59

Lesson 17: Spring Problems I 62

Lesson 18: Spring Problems II and the RLC Circuit 64

Lesson 19: Introduction to the Laplace Transform 702

Lesson 20: The Inverse Laplace Transform 73

Lesson 21: Solution of Initial Value Problems 75

Lesson 22: The Unit Step Function 77

Lesson 23: Const. Coeff. Equations with Piecewise Cont. Forcing Functions 79

Lesson 24: Convolution 81

Lesson 25: Constant Coefficient Equations with Impulses 83

Lesson 26: Euler’s Method and Runge-Kutta Method 86

3

0 Preliminaries

0.1 Schedule

Study Guide Textbook Suggested NumberLesson Section(s) Topic of Lectures

1 1.1,1.2 Introduction to 1Differential Equations

2 1.3 Direction Fields 13 2.1 Linear First Order 1

(use of integrating factors)4 2.2 Separable Equations 15 2.3 Existence and Uniqueness 1

Theorems6 2.4 Transformations 17 2.5,2.6 Exact Equations 18 Autonomous Equations 1

(material not in text)9 4.1,4.2,4.3 Applications 110 5.1 Homogeneous Linear Equations 111 5.2 Constant Coefficient 1

Homogeneous Equations12 7.4 Euler Equations 113 5.3 Nonhomogeneous Linear Equations 114 5.4,5.5 Undetermined Coefficients 115 5.6 Reduction of Order 1

(homogeneous case only)16 5.7 Variation of Parameters 117 6.1,6.2 Spring Problems I 118 6.3 Spring Problems II and 1

the RLC Circuit19 8.1 Introduction to 1

Laplace Transforms20 8.2 The Inverse Laplace Transform 121 8.3 Solution of IVPs 122 8.4 The Unit Step Function 123 8.5 Constant Coefficient Equations with 1

Piecewise Continuous Forcing Functions24 8.6 Convolution 125 8.7 Constant Coefficient Equations 1

with Impulses26 3.1 Euler’s Method and the 1

Range-Kutta MethodCatch-up and Review 1

Exams 2Number of Lectures 29

4

0.2 Introduction and Notes to Students

Introduction:

This study guide is set up for a ten week term with 29 lectures and 9 to 10 recitations. Pleasecheck with your instructor for a detailed syllabus for your section of the course.

Prerequisites:

It is expected that you are comfortable with calculus. This includes familiarity with materialfrom previous courses, such as basic algebraic manipulations, including those involving fractionsand exponents, as well as trigonometry, differentiation, integration, partial fractions, multivariablechain rule. The most common difficulty for students in this course is with “old concepts such asalgebra and basic calculus; if your skills are rusty, practice now, and get help early.

Comments and suggestions on study habits:

There is much new material presented in this course. It is very important to develop good studyhabits and to keep up with the course material right from the beginning of the course. Listed beloware a number of things to watch out for.

Time: It is very important to devote enough time on a daily basis for the course. You should planon spending two to three hours going over the material and assignments for each hour of lecture. Ifyou have seen some of this material before, it is easy to fall into the pattern of not devoting enoughtime to the material at the beginning of the course. When more difficult material is presented in afew weeks you then will find yourself too far behind.

Algebra skills: The most common difficulty for students in this course is remembering the correctalgebraic manipulations from pre-Calculus courses. If you find yourself having difficulty, get helpearly in the course. Don’t wait until it is too late. The material in the course is based on theassumption that you remember the algebra and trigonometry that you learned in your previouscourses.

Attendance: Many students make the mistake of thinking that they can learn all the requiredmaterial by just working the homework problems without attending the lectures. Often, yourlecturer will present an additional viewpoint on the material that is not presented in the text. Theexams for the course are based on the text material and on the presentations in lecture. Make surethat you attend all the lectures. If for some reason you miss a lecture, get notes for the missedclass from one of the other students in the course.

Homework: Often, students think that they have mastered the material if they can get the answersin the back of the book. You should look at the answers in the back of the book only after you havecompleted the assignment. Relying on the solutions can give you a false sense of security. Yourhomework assignments will include problems that are not available through the online homework.Do not make the mistake of not doing these problems. They often cover important concepts thatare difficult to practice in an online environment.

Other Resources:

The Mathematics Learning Center (MLC) provides drop-in help for all lower division mathematicscourses. The MLC is located on the ground floor of Kidder Hall in room 108, and is normally openMonday through Thursday from 9 a.m. to 6 p.m. and on Fridays from 9 a.m. to 5 p.m., from thesecond week of the term through the Dead Week. The MLC also provides evening tutoring in theValley Library, in general Sunday through Thursday from 6 p.m. till 9 p.m. Current hours can befound at the MLC homepage, http://www.math.oregonstate.edu/mlc

5

0.3 Basic Integral Table

k is any nonzero constant

1.∫

kf(x)dx = k

∫f(x)dx

2.∫

(f(x) + g(x)) dx =∫

f(x)dx +∫

g(x)dx

3.∫

(f(x)− g(x)) dx =∫

f(x)dx−∫

g(x)dx

4.∫

xpdx =xp+1

p + 1+ C, p 6= −1

5.∫

1x

dx = ln |x|+ C

6.∫

ekxdx =1kekx + C

7.∫

ln(x)dx = x ln(x)− x + C

8.∫

sin(kx)dx = −1k

cos(kx) + C

9.∫

cos(kx)dx =1k

sin(kx) + C

10.∫

tan(kx)dx =1k

ln |sec(kx)|+ C

11.∫

cot(kx)dx = −1k

ln |csc(kx)|+ C

12.∫

sec(kx)dx =1k

ln |sec(kx) + tan(kx)|+ C

13.∫

csc(kx)dx = −1k

ln |csc(kx) + cot(kx)|+ C

14.∫

sec2 (kx) dx =1k

tan(kx) + C

15.∫

csc2(kx)dx = −1k

cot(kx) + C

16.∫

sec(kx) tan(kx)dx =1k

sec(kx) + C

17.∫

csc(kx) cot(kx)dx = −1k

csc(kx) + C

18.∫

dx

x2 + k2=

1ktan−1

(x

k

)+ C, k > 0

19.∫

dx√k2 − x2

dx = sin−1(x

k

)+ C, k > 0, |x| < k

20.∫

dx

x√

x2 − k2dx =

1ksec−1

∣∣∣xk

∣∣∣ + C, k > 0, |x| > k

6

0.4 Table of Elementary Laplace Transforms

f(t) = L−1{F (s)} F (s) = L{f(t)}

11

(n = 0, 1, 2, ...)1s

s > 0

2tn

(n = 0, 1, 2, ...)n!

sn+1s > 0

3 tp (p > −1)Γ(p + 1)

sp+1s > 0

4 eat 1s− a

s > a

5tneat

(n = 1, 2, 3, ...)n!

(s− a)n+1s > a

6 cos (ωt)s

s2 + ω2s > 0

7 sin (ωt)ω

s2 + ω2s > 0

8 eλt cos (ωt)s− λ

(s− λ)2 + ω2s > λ

9 eλt sin (ωt)ω

(s− λ)2 + ω2s > λ

10 cosh (bt)s

s2 − b2s > |b|

11 sinh (bt)b

s2 − b2s > |b|

12 t cos (ωt)s2 − ω2

(s2 + ω2)2s > 0

7

13 t sin (ωt)2ωs

(s2 + ω2)2s > 0

14 sin (ωt)− ωt cos (ωt)2ω3

(s2 + ω2)2s > 0

15 ωt− sin (ωt)ω3

s2 (s2 + ω2)2s > 0

161t

sin (ωt) arctan(ω

s

)s > 0

17 eatf(t) F (s− a)

18 tkf(t) (−1)kF (k)(s)

19 f(ωt)1ω

F( s

ω

)ω > 0

20 u(t− τ)e−τs

ss > 0

21 u(t− τ)f(t− τ) (τ > 0) e−τsF (s)

22∫ t

0f(τ)g(t− τ) dτ F (s)G(s)

23 δ(t− a) e−as s > 0

24 f ′(t) sF (s)− f(0)

25 f ′′(t) s2F (s)− sf(0)− f ′(0)

26 f (n)(t) snF (s)− sn−1f(0)− sn−2f ′(0)− · · · − f (n−1)(0)

8

Lesson 1

Introduction to Differential Equations

Sections 1.1 and 1.2

A common model assumes that the rate of growth of some quantity N is proportional to itself.

dN

dt= kN , where k is the proportional constant.

To solve this we separate the variables:

(dN

dt

)

N= k

∫(

dN

dt

)

Ndt =

∫k dt

ln |N | = kt + C1

|N | = ekt+C1

|N | = eC1ekt

|N | = C2ekt where C2 = eC1

Notice that with this model if N is ever 0 then it is always 0. So N is either positive or negative.In most models N will always be positive.

Let C =

C2 if 0 ≤ N

−C2 if N < 0

N = Cekt

If k > 0 this model is called exponential growth.If k < 0 this model is called exponential decay.

9

A differential equation is an equation that contains derivatives.

Examples:

1.dy

dt= t2. In this example y is the dependent variable and t is the independent variable.

2.dy

dx= xy.

3. y′′ + y = cos (t).

4.dv

dt= g − βv.

Ordinary vs Partial

A partial differential equation (p.d.e.) is a differential equation that contains partial derivatives.

Examples:

1.∂u

∂t= a

∂2u

∂x2is the heat equation.

2. utt = auxx is the wave equation. Both of these equations have a wide range of applications.

An ordinary differential equation (o.d.e.) is a differential equation that only contains ordinaryderivatives. In this course we will restrict our studies to ode’s. One of the major reasons to studyode’s is to gain the background needed to understand and solve pde’s.

Systems

Often differential equations make their appearance as systems. One famous example are the Lotka-Volterra equations that are used to model predator prey interactions in biology.

x = number of hares (prey)y = number of lynx (predators)

dx

dt= ax− αxy

dy

dt= −cy + γxy

We will not be studying systems of differential equations in this course.

10

Order

The order of a differential equation is the order of the highest derivative that appears.

Examples:

1.dy

dt= t2 first order

2.dy

dx= xy first order

3. y′′ + y = cos (t) second order

4.dv

dt= g − βv first order

Linear vs Nonlinear

An nth order ordinary differential equation is linear if it can be written in the form:

an(t)y(n) + an−1(t)y(n−1) + · · ·+ a1(t)y′ + a0(t)y = g(t)

otherwise it is nonlinear.

Examples:

These are all linear

1.dy

dt= t2

2.dy

dx= xy

3. y′′ + y = cos (t)

4. x2y′′ = 5xy′ − 6y + sin (3x)

5.y′ + t2

y′ cos (t) + y= sin (t)

These are all nonlinear

6. y′y = x

7.y

y′′′+ ex = y′

8. 5dy

dt+ y = t + y2

A linear differential equation is homogeneous if g(t) = 0; otherwise it is nonhomogeneous. Inthe examples above 1., 3., 4., and 5. are nonhomogeneous, while 2. is homogeneous.

11

A solution of a differential equation is a function that solves the differential equation on some open

interval. For example, the function y = ex2

2 is a solution to the differential equationdy

dx= xy. To

verify this just notice thatdy

dx= xe

x2

2 = xy. The function y = 5ex2

2 is also a solution. The function

y = Cex2

2 is a solution for any constant C. It can be shown that any solution to this differential

equation is of the form y = Cex2

2 . We call this a general solution to the differential equation. Todetermine C we need a data point, called an initial condition.

Example: Find a solution tody

dx= xy, y(2) = 7.

y = Cex2

2

y(2) = Ce2 = 7C = 7e−2

y = 7e−2ex2

2

y = 7ex2

2−2

We call this type of problem an initial value problem, often referred to as an IVP.

Suggested Homework:Note: The Assignment given by your instructor may be different.

Read Sections 1.1 and 1.2, Pages 1 - 14Section 1.2 - { 1a, 2a b c d e, 3a b c e, 4a }

12

Lesson 2

Direction Fields

Section 1.3

Consider a differential equation of the formdy

dx= f(x, y). At each point (x, y) we are given the

value of the rate of change of y. We call f the rate function.

A plot in which short line segments are drawn giving the slope at each point (x, y) is called adirection field or a slope field. As a practical matter, we draw these line segments at enoughpoints to get a representation of the direction field.

Here’s a direction field plotted on R = {(x, y)| − 1 ≤ x ≤ 4,−1.5 ≤ y ≤ 4.5}:

-1 1 2 3 4

-1

1

2

3

4

slope field for y¢ = x2 y

The graph of a solution to a differential equation is called an integral curve. The slope of an

integral curve fordy

dx= f(x, y) at (x0, y0) is f(x0, y0).

13

Here’s a direction field fordy

dx= x2 − y with several integral curves plotted:

-4 -2 0 2 4

0

2

4

6

x

y

In general plotting direction fields by hand is tedious at best. There are many applications andJava apps that can plot direction fields.


Read Section 1.3, Pages 16 - 21Section 1.3 - { 1 - plot the integral curve where y(0) = 0 , 2 - plot the integral curve where y(0) = 0.5,3 - plot the integral curve where y(0) = 0.6, 4 - plot the integral curve where y(0) = 0, 7 - plot the in-tegral curve where y(0) = 1, use an online tool ( such as http://www.mathscoop.com/calculus/differential-equations/slope-field-generator.php ) to create direction fields for 12, 13, 13 }

14

Lesson 3

Linear First Order Equations, Use of Integrating Factors

Section 1.3

Trench uses the method of variation of parameters to solve first-order linear differential equa-tions. While this non-traditional method works, the traditional method of integrating factorsis, in our opinion, simpler to learn and use.

The following can be used to replace pages 31 through 35 in the textbook.

Consider a first-order linear differential equation

a1(x)dy

dx+ a0(x)y = g(x).

By dividing through by a1(x) we get the standard form:

dy

dx+ p(x)y = f(x). (∗)

The left hand side almost looks like the product rule applied to y(x) and some function µ(x):

d

dx[µy] = µy′ + µ′y.

Multiplying both sides of (∗) by µ yields

µy′ + µpy = µf.

Matching the left hand side to the form of the product rule for µy, we want a function µ such that

µ′ = µp.

From here we getµ′

µ= p, and by integrating both sides

ln |µ| =∫

p dx.

So the possibilities for µ areµ = ±e

∫p dx.

These are called integrating factors.15

Since we only need one we will always choose µ = e∫

p dx and set the constant of integration to 0.

With this choice µy′ + µpy = µf becomes [ye∫

p dx]′ = fe∫

p dx. From here we can integrate bothsides and solve for y.

Examples

1. y′ +3x

y =4x5

.

We can use an integrating factor of µ(x) = e∫

3x

dx = e3 ln |x| = ±x3.

By multiplying both sides of y′ +3x

y =4x5

by µ = x3 we get

x3y′ + 3x2y =4x2

.

This equation can be rewritten as

(x3y)′ =4x2

.

By integrating both sides,

x3y = −4x

+ C.

So y =C

x3− 4

x4is the general solution.

2. xy′ − 3y = x4ex.

First we rewrite this in the standard form to get y′ − 3x

y = x3ex.

So an integrating factor is µ(x) = e∫ − 3

xdx = e−3 ln |x| = ±x−3. Notice the integrand’s sign

here. A common mistake is to leave it behind.

By multiplying both sides of y′ − 3x

y = x3ex by µ = x−3 we get

x−3y′ − 3x−4y = ex.

This equation can be rewritten as(x−3y)′ = ex.

By integrating both sides,x−3y = ex + C.

So y = x3ex + Cx3 is the general solution.16

3. y′ + 2xy = x3.

An integrating factor is µ = e∫

2x dx = ex2.

By multiplying both sides of y′ + 2xy = x3 by µ we get

ex2y′ + 2xex2

y = x3ex2.


(ex2y)′ = x3ex2

.

By integration by parts,

ex2y =

12ex2

(x2 − 1) + C.

Hence y = 12(x2 − 1) + Ce−x2

is the general solution.

4. y′ cos (t) + y sin (t) = 1, y(0) = 2.

First we rewrite this in the standard form to get y′ + y tan (t) = sec (t).

So an integrating factor is e∫

tan (t) dt = eln | sec (t)| = ± sec (t).

By multiplying both sides of y′ + y tan (t) = sec (t) by µ = sec (t) we get

sec (t)y′ + sec (t) tan (t)y = sec2 (t).


(sec (t)y)′ = sec2 (t).

By integrating both sides,sec (t)y = tan (t) + C.

Then y = sin (t) + C cos (t) is the general solution. Imposing the initial condition of y(0) = 2yields C = 2 and a solution of

y = sin (t) + 2 cos (t).

17

Sometimes you have to leave your answer in an integral form and use numerical integration.

5. y′ − 2xy = 1, y(1) = 2.

An integrating factor is µ = e∫ −2x dx = e−x2

.

By multiplying both sides by µ we get

e−x2y′ − 2xe−x2

y = e−x2


(e−x2y)′ = e−x2

.

But the integral of e−x2does not have an expression in terms of elementary functions. Looking

at the initial condition of y(1) = 2 we will use the integral function F (x) =∫ x

1e−t2 dt as an

antiderivative of f(x) = e−x2.

So we may write that e−x2y =

∫ x

1e−t2 dt + C. Then the general solution is

y = ex2

∫ x

1e−t2 dt + Cex2

.

Imposing the initial condition of y(1) = 2 yields C = 2/e. So the solution is

y = ex2

∫ x

1e−t2 dt + 2ex2−1.

Try doing the homework for this section using the method of integrating factor instead of themethod of variation of parameters shown in the text.

18

Existence and Uniqueness Theorem for Linear First Order Differential Equations:

Suppose that p and f are continuous on an open interval (a, b) that contains the point x0, then theinitial value problem:

y′ + p (x) y = f (x) , y(x0) = y0,

has a unique solution that is valid on the interval (a, b).

The proof for this result can be found in the textbook at the end of section 2.1.

Example: What is the largest interval on which the initial value problem

(x + 1)dy

dx+ xy = tanx, y (0) = 7,

is guaranteed a unique solution?

The first step is to put the differential equation into standard form so that p and f can be identified.

dy

dx+

x

x + 1y =

tanx

x + 1

Then p (x) =x

x + 1and f (x) =

tanx

x + 1.

We see that p has continuity problems at x = −1 and f is not continuous at x = −1 andx = π

2 + nπ, n = 0,±1,±2,±3, · · · .

The largest interval on which both p and f are continuous that contains 0 is(−1, π

2

). This is the

largest interval on which we are guaranteed a unique solution.


Section 2.1 { 1, 2, 5, 6, 10, 16, 18, 19, 20, 21, 31, 33, 38 }

19

Lesson 4

Separable Equations

Section 2.2

A first order differential equation is separable if the rate function can be factored into pure single-variable functions of the independent variable and the dependent variable as shown below.

dy

dx= g (x) h (y)

To solve a separable differentiable equation, separate and integrate.∫

1h (y)

dy

dxdx =

∫g (x) dx

∫1

h (y)dy =

∫g (x) dx

Examples of separable differential equations include:

y′ = y

dy

dx=

x

y

dy

dt= t− 1 + ty − y = (t− 1) + y (t− 1) = (t− 1) (1 + y)

Example: Consider the non-linear first order homogeneous (since every term involves y or a deriva-tive of y) separable o.d.e. (x2 + 1)y′ −√y = 0. Find the general solution, and then solve the IVPwith the initial condition y(π/4) = 0.

(x2 + 1)dy

dx=√

y

1√y

dy

dx=

11 + x2

∫1√y

dy

dxdx =

∫1

1 + x2dx

∫dy√

y= tan−1 (x) + C

2√

y = tan−1 (x) + C

20

y =(0.5 tan−1 (x) + c

)2

Let’s solve the IVP with y(π/4) = 0. 0 = (0.5 + c)2 implies c = −0.5 so

y =(0.5 tan−1 (x)− 0.5

)2.

When you read the textbook, please play special attention to the section on page 51, DifferencesBetween Linear and Nonlinear Equations.


Read Section 2.2, Pages 40 - 47Section 2.2 { 1, 2, 3, 11, 17, 18, 19, 21, 26 }

21

Lesson 5

Existence and Uniqueness Theorems

Section 2.3

Consider the following initial value problem:

y′ =√

y, y (0) = 0.

It is easy to see that y = 0 is a solution. However, if we did not notice this and proceeded to solvethis separable differential equation we see that:

y−12 y′ = 1

2√

y = t + C

Apply the initial condition to see that C = 0. So 2√

y = t or y =t2

4is also a solution.

We have found two different solutions to this initial value problem. The first thing to notice isthat this differential equation is not linear. The existence and uniqueness theorem for nonlinearequations is not as strong or useful as the one for linear equations. The proof for this theorem isbeyond the scope of this course. Please see the textbook (section 2.3) for a statement of this resultand some examples on applying this theorem.


Read Section 2.3, Pages 50 - 55Section 2.3 { 1, 2, 3, 17, 19 }

22

Lesson 6

Transformations

Section 2.4

A first order differential equation of the form y′ + p(x)y = f(x)yr is called a Bernoulli equation.For r = 0 or r = 1 the differential equation is linear, so we will restrict our attention to the casewhere r 6= 0 or 1 . These can be reduced to a linear equation by the substitution u = y1−r.

Example: xy′ − 5y = x3y3.

We can reduce this Bernoulli equation to a linear equation by the substitution u = y1−3 = y−2:

Then u′ = −2y−3y′ which gives y′ = −12y3u′. Substituting into the differential equation gives

−12xy3u′ − 5y = x3y3.

This simplifies to

xu′ + 10y−2 = −2x3.

Then recalling that u = y−2 we see that we have a standard first order linear differential equation:

xu′ + 10u = −2x3.

This can be solved using integrating factors.

Please verify that the solution is

u = − 213

x3 + C1x−10.

Theny−2 = C1x

−10 − 213

x3.

Which simplifies to

y2 =13x10

C − 2x13where C = 13C1.

23

Another type of first order differential equation that can be solved by use of a transformation is aso-called nonlinear homogeneous equation.

A nonlinear first order differential equationdy

dx= f (x, y) is homogeneous if f(x, y) only depends

on the ratio yx , which means

dy

dx= f

( yx

).

The substitution u = yx will turn such homogeneous differential equations into separable differential

equations. We can see this since y = ux and by the product rule

dy

dx= u + x

du

dx.

The differential equation them becomes

u + xdu

dx= f (u) .

which is separable.


Read Section 2.3, Pages 50 - 55 and Section 2.4, Pages 57 - 63Section 2.4 { 1, 2, 7, 9, 15, 16, 22, 23 }

24

Lesson 7

Exact Equations

Section 2.5

A convenient way to write a first order o.d.e. is with differentials:

M(x, y) dx + N(x, y) dy = 0.

This is equivalent to M(x, y) + N(x, y)y′ = 0 for y is a function of x.

We will say that F (x, y) = c is an implicit solution of M(x, y) dx + N(x, y) dy = 0 if everydifferentiable function y(x) such that F (x, y(x)) = c is a solution of M(x, y) + N(x, y)y′ = 0.

By the Chain Rule F (x, y) = c is an implicit solution of Fx dx + Fy dy = 0 where Fx, Fy denote thex, y partials respectively.

This motivates the following important definition:

M(x, y) dx + N(x, y) dy = 0 is exact on an open rectangle R if there is a function F (x, y) suchthat Fx, Fy are continuous on R, and

Fx(x, y) = M(x, y) and Fy(x, y) = N(x, y)

for all (x, y) in R.

It turns out that M(x, y) dx+N(x, y) dy = 0 is exact on an open rectangle R if and only if My = Nx

(assuming My, Nx are continuous on R).

Finding an implicit solution for an exact equation M(x, y) dx + N(x, y) dy = 0 is straightforward:

Integrate M with respect to x, producing a constant of integration that can be a function of y.Then take the y-partial of the result to determine the constant of integration.

Note: One can do the above with the roles of M and N switched and the roles of x and y switched.

This gives F (x, y) and an implicit solution of F (x, y) = c. If possible, find a solution y(x) fromF (x, y) = c.

25

Examples:

1. (y3 − 1)ex dx + 3y2(ex + 1) dy = 0.

∂

∂y

((y3 − 1)ex

)= 3y2ex and

∂

∂x

(3y2(ex + 1)

)= 3y2ex.

So this equation is exact.

Integrating (y3 − 1)ex with respect to x yields

F (x, y) = (y3 − 1)ex + g(y).

Then Fy = 3y2ex + g′(y). We want g′(y) = 3y2. So set g(y) = y3. Thus, (y3 − 1)ex + y3 = cis an implicit solution. We can take this one further!

This implicit solution is equivalent to y3(ex + 1) = ex + c and therefore, y = 3

√ex + c

ex + 1is the

explicit general solution.

2. 3x2y2 dx + 4x3y dy = 0.

The y-partial of 3x2y2 is 6x2y and the x-partial of 4x3y is 12x2y. Hence there is no openrectangle on which this equation is exact.

Now suppose M(x, y) dx + N(x, y) dy = 0 is not exact, but can be made exact by multiplying bothsides by some function µ(x, y). Such a function is called an integrating factor.

Example: Consider the o.d.e. (3xy + 2y + 2) dx + (x2 + x) dy = 0.

The reader is left to check that this equation isn’t exact.

Multiplying both sides by µ(x, y) = x we get (3x2y + 2xy + 2x) dx + (x3 + x2) dy = 0.

The reader is left to check that this equation is exact.

Now we can solve as before:

Taking the integral of x3 + x2 with respect to y yields x3y + x2y + g(x). Taking the x-partial andequating with 3x2y + 2xy + 2x implies that we may set g(x) = x2.

So an implicit solution is x3y + x2y + x2 = c which we can solve for y to get the explicit solution:

y =c− x2

x3 + x2.


Read Section 2.5, Pages 66 73Section 2.5 { 1 9, 18, 19 }

26

Lesson 8

Autonomous First Order Differential Equations

Not in textbook

A first order differential equation of the formdy

dt= f(y) is said to be autonomous.

The rate function f does not depend on the independent variable t.

Here are a few useful properties of first order autonomous differential equations

1. They are separable.

2. The slopes in the direction field only depend on y.

3. The general solution is invariant under horizontal translations.

Example: Exponential growth or decay.

Let q(t) be the quantity of something. This common model assumes that the rate of growth in q isproportional to itself:

dq

dt= rq.

Examples include population growth (r > 0) and radioactive decay (r < 0). As we have alreadyseen, with an initial condition q(t0) = q0, we have q(t) = q0e

r(t−t0).

27

Example: Population dynamics.

Let p(t) represent the size of a population at time t. When resources are restricted we can employa competition factor, −bp2, so that

dp

dt= rp− bp2 = rp

(1− p

K

),

where K =r

b.

This equation is called the logistic equation. This differential equation has equilibrium solutionsat y = 0 and y = K. We call K the carrying capacity and r the intrinsic growth rate.

Let’s determine the “stability” of these equilibrium values. Assume r > 0 and K > 0 (negativevalues would make no sense for this model).

To do this we plotdp

dtversus p for

dp

dt= rp

(1− p

K

):

KpHtL

p¢HtL

The graph is a parabola with x-intercepts at x = 0 and x = K.

28

Note:If 0 < p < K then

dp

dt> 0 and p is increasing. If p > K then

dp

dt< 0 and p is decreasing. So if p is

close to K then p is “pushed” towards K. A similar analysis shows that if p is close to 0 it will be“pushed” away from 0.

(While we getdp

dt< 0 when p < 0, it’s not relevant to the model as p cannot be negative.)

We can use arrows to indicate how p is “pushed.” Here it is with the logistic equation:

K0pHtL

p¢HtL

The solution p = K is an asymptotically stable equilibrium solution, since small perturbations(changes) are “pushed” back towards K. The solution p = 0 is an unstable solution, since anyperturbations will “push” p away from 0.

With the arrows, the p-axis is called the phase line.

29

More Examples:

Find and classify the equilibrium solutions ofdy

dt= (1− y)(2− y).

1 2yHtL

y¢HtL

The equilibrium solutions are y = 1 and y = 2. Since y′ is positive outside of the interval [1, 2] andnegative inside (1, 2), y = 1 is an asymptotically stable solution and y = 2 is an unstable solution


dt= y(2− y)2.

0 2yHtL

y¢HtL

The equilibrium solutions are y = 0 and y = 2. y = 0 is an unstable solution and y = 2 is asemistable equilibrium solution since it is stable from one side (in this case from below) whileunstable from the other side (in this case from above).

30


dt= sin (y).

-Π 0 Π 2Π 3ΠyHtL

y¢HtL

Unstable equilibria exist at t = 0,±2π,±4π, ...

Stable equilibria exist at t = ±π,±3π, ...

Let’s return to and solve the first order logistic equationdy

dt= ry

(1− y

K

)for y, t > 0 where K, r

are positive constants.

y′ = ry(1− y

K

)

y′

y(1− y

K

) = r

∫1

y(1− y

K

) y′ dt =∫

r dt

∫ (1K

1− yK

+1y

)dy = rt + C1

− ln∣∣∣1− y

K

∣∣∣ + ln |y| = rt + C1

ln

∣∣∣∣∣∣y

1− y

K

∣∣∣∣∣∣= rt + C1

y

1− y

K

= C2ert (where C2 = ±eC1)

31

1− y

Ky

= C3e−rt (where C3 = 1/C2)

1y− 1

K= C3e

−rt

1y

=Ce−rt + 1

K(where C = KC3)

y =K

1 + Ce−rt.

Next we use an initial condition y(0) = y0 6= 0 to find C. If y(0) = 0 then y = 0.

By setting y0 =K

1 + Cwe have C =

K − y0

y0.

So the solution to the IVPdy

dt= ry

(1− y

K

), y(0) = y0 is

y =y0K

y0 + (K − y0)e−rt.

Let’s compare solutions when K = 100 and r = 0.001 for initial values y(0) = 25 and y(0) = 125.Here are the graphs (shown within the slope field):

0 10 20 30 40 50

0

50

100

150

t

y

32

References:

Elementary Differential Equations, 9th ed., by Boyce, DiPrima (Pages 88-100)


Find and classify the equilibrium solutions for the differential equations (1) through (6):

1. y′ = 3− y

2. y′ = 5 + y

3. y′ = y2(3− y)

4. y′ = y2 − 5y + 6

5. y′ = (1− y)(y − 2)(y − 3)

6. y′ = y − y3

7. Show that for the logistic equation,dy

dt= ry

(1− y

K

), that the steepest ascent occurs at

y =K

2.

8. Let r,K be positive constants. Consider the IVPdy

dt= ry ln

(K

y

), y(0) = y0.

(a) Find and classify the equilibrium solutions.

(b) Solve for y.

33

Answers to selected exercises:

1. y = 3; stable equilibrium.

3. y = 0; semistable equilibrium. y = 3; stable equilibrium.

5. y = 1; stable equilibrium. y = 2; unstable equilibrium. y = 3 stable equilibrium.

34

Lesson 9

Applications

Section 4.1, 4.2, 4.3

We start with the classic exponential model : y′(t) = ky, y(t0) = y0.

The general solution (without an initial value) is y(t) = cekt. If k > 0 the model is called expo-nential growth and if k < 0 the model is called exponential decay. The constant k is called agrowth constant and decay constant respectively.

Applying the initial value, y0 = cekt0 , so c = y0e−kt0 . The initial value problem has the solution

y = y0ek(t−t0).

We assume that you have seen examples (in algebra and calculus) of exponential growth and decay,such as least population growth, compound interest, and radioactive decay.

Let’s consider a variation on the exponential model:

Suppose that a radioactive substance has decay constant k > 0 (reported as a positive number, souse y′ = −ky). At the same time the substance is being produced at a constant rate of a units ofmass per unit time.

Let y(t) be the mass at time t and y(0) = y0 be the initial mass. Let’s derive an IVP and solve it.Then we can take the limit of y(t) as t →∞ to find the equilibrium solution.

First the IVP: y′(t) = rate of increase in y(t) − rate of decrease in y(t) = a− ky(t), y(0) = y0.

Since y′(t) = a− ky is separable let’s solve it by separating and integrating:

1a− ky

y′ = 1 → −1k

ln |a− ky| = t + c1 → a− ky = ce−kt → y =a

k+ ce−kt.

Imposing the initial condition we get: y0 =a

k+ c → c = y0 − a

k.

So the solution to the IVP isy =

a

k+

(y0 − a

k

)e−kt.

The equilibrium solution is y =a

k. In particular, if y0 >

a

kthen y decreases toward this solution,

and if y0 <a

kthen y increases towards this solution.

35

Example: An immortal alchemist creates bitcoin at the rate of 1 bitcoin per day. Going unnoticed,a cyber thief has setup a script that is continuously stealing the aclchemist’s bitcoin at a rate of2.5 percent of whatever is there. What is the number of bitcoins b(t) the alchemist has at time t?Assume b(0) = 1. How many bitcoins will the alchemist have in the long run (as t →∞)?

b′(t) = rate of increase in bitcoin − rate of decrease in bitcoin.

So b′ = 1− 0.025b. Using a = 1, k = 0.025, and b0 = 1 we get b(t) = 40− 39e−0.03t.

In the long run the alchemist will have 40 bitcoins.

Newton’s Law of Cooling:

If an object of temperature T (t) at time t is in a medium of temperature Tm(t) at time t then therate of change in T (t) is proportional to the ∆T = T (t) − Tm(t). When T (t) > Tm(t) we haveT ′(t) < 0 and when T (t) < Tm(t) we have T ′(t) > 0.

So T (t) satisfies T ′ = −k(T − Tm) where k > 0.

Note: You don’t need to convert temperatures to Kelvin units as the model is about relativetemperatures.

Example: Suppose a cup of coffee brewed at 150 degrees Fahrenheit is cooling in a room of constanttemperature 70 degrees Fahrenheit. After 10 minutes the temperature of the coffee is 100 degreesFahrenheit. How hot is the temperature after 20 minutes?

First we setup the IVP: T ′ = −k(T − 70), T (0) = 150.

Then ln |T − 70| = −kt + c1 implies T − 70 = ce−kt and hence T = 70 + ce−kt. By imposing theinitial condition, T (t) = 70 + 80e−kt.

Using T (10) = 100 we get that 100 = 70 + 80e−10k → e−k = −0.1 ln (3/8) = ln (8/3)/10.

Then after 20 minutes the cup of coffee is at T (20) = 70 + 80e−2 ln (8/3) = 70 + 80(9/64) = 81.25degrees Fahrenheit.

36

The following type of problem is usually referred to as a mixing problem (technically we havealready seen an example)

Example: A tank initially contains 10 kg of salt dissolved in 200 L of water. Brine that contains0.5 kg of salt per liter is pumped into the tank at 5 liters per minute and at the same time wateris drained from the tank at 5 liters per minute. Assume the tank is always kept uniformly mixed.Calculate the amount of salt in the tank in the long run (the equilibrium solution as t →∞).

Let’s pose this as an IVP:

Let y(t) be the amount of salt in kg in the tank (which always contains 200 L of water).

Then y′ = rate of salt in − rate of salt out.

So

y′ =(

0.5 kgL

)(5 Lmin

)−

(y(t) kg200 L

)(5 Lmin

), y(0) = 10.

Again, we could choose to apply the formula derived in general on the first page of this lesson,however it’s good practice to solve y′ = 2.5− y/40 as separable equation:

12.5− y/40

y′ = 1 → −40 ln |2.5− y/40| = t + c1 → 2.5− y/40 = ce−t/40 → y = 100 + ce−t/40.

Imposing the initial condition yields c = −90, so y = 100− 90e−t/40. Thus as t →∞, y → 100. Soin the long run the amount of salt in the tank is 100 kg.

37

Here another mixing problem example with a wrinkle that was not present in the last one:

A tank of maximum capacity 1, 000 gallons initially contains 10 lb of salt dissolved in 500 gallonsof water. Brine that contains 0.5 lb of salt per gallon is pumped into the tank at 10 gallons perminute and at the same time water is drained from the tank at 5 gallons per minute. Assumethe tank is always kept uniformly mixed. Calculate the amount of salt in the tank the moment itbegins to overflow.

The change here is the amount of water in the tank at time t isn’t constant: V (t) = 500 + 5t is thevolume of the tank in gallons after t minutes where V (0) = 500.

We still set it up like before: Let y(t) be the amount of salt in the tank in lb at time t, wherey(0) = 10.

Then y′ = rate of salt in − rate of salt out.

So

y′ =(

0.5 lbgal

)(10 galmin

)−

(y(t) lb

500 + 5t gal

)(5 galmin

), y(0) = 10.

So we have to solve the following IVP: y′ = 5− 1100 + t

y, y(0) = 10.

Let’s rewrite the o.d.e. as y′ +1

100 + ty = 5. An integrating factor is found by computing

µ = e∫

1100+t

dt = eln |100+t| = ±(100 + t).

Let’s multiply both sides of the o.d.e. by 100 + t:

(100 + t)y′ + y = 5(100 + t).

This becomes((100 + t)y)′ = 5(100 + t).

Integrating both sides yields

(100 + t)y = 5(100t + 0.5t2) + c → y =500t + 2.5t2 + c

100 + t.

By imposing the initial condition y(0) = 10 we get 10 =c

100, so c = 1, 000. So

y =1, 000 + 500t + 2.5t2

100 + t.

Now to find the amount of salt at the moment the tank overflows: Set V (t) = 500 + 5t = 1, 000 toget t = 100 minutes.

Then y(100) = 380 lbs of salt when the tank overflows.38

Let’s look at motion with resistance.

We assume an object is traveling vertically through a medium (air or water for example) and thatthe only forces acting on the object are a constant gravitational force pulling the object down andthe resistance of the medium (assumed to be directly proportional to the speed of the object).

Suppose an object of mass m > 0 moves through vertically a medium under a constant downwardgravitational force Fg = mg (where g is the constant acceleration due to gravity) and the mediumexerts a force of Fmed = k|v|, k > 0, in the opposite direction of the motion, where v is the velocityof the object.

Let’s derive a model for this scenario:

Let v be the vertical velocity of an object (where moving up corresponds to a positive velocity).

Newton’s second law of motion asserts that Fnet = ma where a = v′.

Suppose the object is moving up. Then the resistance acts downward: Fmed = −k|v| = −kv.

Now suppose the object is moving down. Then the resistance acts upward: Fmed = k|v| = k(−v) =−kv.

Either way, the net force on the object is and Fnet = −mg − kv.

So we get the following first order linear o.d.e. mv′ = −mg − kv.

This can be expressed as v′ +k

mv = −g.

While this is separable, we can solve it quickly by finding an integrating factor: µ = e(k/m)t.

39

e(k/m)tv′ +k

m= −ge(k/m)t → (e(k/m)tv)′ = −ge(k/m)t → e(k/m)tv = −mg

ke(k/m)t + c.

Hencev(t) = −mg

k+ ce−(k/m)t.

(Exercise left for the reader: Solve it as a separable differential equation.)

Finally, we see that limt→∞ v(t) = −mg

kis the equilibrium solution, which is called the terminal

velocity of the object in the medium.

Given an initial value v0 for the velocity, we get c = v0 + mgk .

So the solution to the IVP v′ +k

mv = −g, v(0) = v0 is

v(t) = −mg

k+

(v0 +

mg

k

)e−(k/m)t.

Example:

A 4 kg object is launched vertically into the air at 40 meters per second “near” the surface of theEarth (though we assume it’s far enough up that the object will get close to its terminal velocityduring the fall).

Suppose the air resists motion with a force of 0.49 Newtons for each m/s of speed. Using g = 9.8meters per square second, find a model for the objects vertical velocity and find the terminalvelocity.

We can just plug-in to the IVP solution we just derived: v(t) = −4(9.8)0.49 +

(40 + 4(9.8)

0.49

)e−(0.49/4)t.

So v(t) = −80 + 120e−49t/400.

The terminal velocity is −80. Meaning that in the “end” the object falls at a constant rate of 80meters per second.

40

(Note to the instructor : Coverage of motion with resistance (previous pages) is typically included inthe course, but coverage of escape velocity (pages that follow) is often skipped, and it is suggestedthat you do so if pressed for time.)

More realistic models don’t assume that the force due to gravity is a constant.

Rather they assume that the force due to gravity that a “massive object” exerts on another object isinversely proportional to the square of the distance from the center of the massive object (assumedto be a sphere) to the other object. This is an example of an inverse square law.

Assume a spacecraft launches from the surface of the Earth and exhausts its fuel as it reaches aheight h that is sufficiently large that atmospheric resistance is negligible and can be assumed tobe effectively 0. Let t = 0 be this time (known as “burnout”) and consider that the force of gravityon the spacecraft at altitude y ≥ h is

Fg = − c

(R + y)2where c is a constant R is the radius of the Earth.

Using Fg = −mg when y = 0 where g is the acceleration due to gravity on the surface of the Earth,we get −mg = − c

R2 , so c = mgR2.

Thus

Fg = − mgR2

(R + y)2.

Since Fg is assumed to be the only force acting on the spacecraft for t ≥ 0,

Fg = ma = my′′ = − mgR2

(R + y)2→

y′′ = − gR2

(R + y)2.

This equation is second order, but we can transform it into a first order o.d.e. involving velocities:Let v(y) be the vertical velocity at altitude y.

By the chain rule,d2y

dt2=

dv

dt=

dv

dy

dy

dt= v

dv

dy.

Now we solve the o.d.e. vdv

dy= − gR2

(R + y)2.

This equation is separable, and integrating both sides yields

0.5v2 =gR2

R + y+ c → v =

√2gR2

R + y+ c.

41

Now suppose we have an initial velocity of v(h) = v0.

Then v20 =

2gR2

R + h+ c.

So c = v20 −

2gR2

R + hand hence

v =

√2gR2

R + y+ v2

0 −2gR2

R + h.

If v0 ≥√

2gR2

R + hthen v ≥ 0 for all y ≥ h and the spacecraft continues on up into space!

The expression ve =

√2gR2

R + his called escape velocity.

We will show that if v0 < ve the spacecraft falls back to Earth:

If v0 < ve then v =

√2gR2

R + y+ v2

0 −2gR2

R + h= 0 for y such that

2gR2

R + y+ v2

0 −2gR2

R + h= 0,

which occurs at y =2gR2h + R(R + h)v2

0

2gR2 − (R + h)v20

≥ h since

2gR2

R + y=

2gR2

R + h− v2

0 ≤2gR2

R + h.


Read Sections 4.1, Pages 123 130, 4.2, Pages 133 139, 4.3, Pages 142 - 151Section 4.1 { 1, 2, 6, 7, 11, 14, 16, 17, 18 }Section 4.2 { 1, 2, 4, 6, 7, 8, 9, 11, 14 }Section 4.3 - { 1, 2, 10 }

42

Lesson 10

Homogeneous Linear Equations

Section 5.1

In this lesson be begin the study of linear second order differential equations. A second orderdifferential equation

y′′ = F(y′, y, t

)

is linear if it can be written in the form:

a2 (t) y′′ + a1 (t) y′ + a0 (t) y = g (t) .

We can divide through by a2 (t) to put the differential equation into standard form

y′′ + p (t) y′ + q (t) y = f (t) .

If g (t) = 0 the differential equation is homogeneous, otherwise it is nonhomogeneous. We call thefunction f a forcing function.

In this lesson we will cover the existence and uniqueness theorem, the concepts of linear combi-nations, linear independence, linear dependence, trivial solution, fundamental set of solutions, andthe general solution. Make sure you thoroughly understand these concepts from the reading.

We will also learn about the Wronskian, a tool that is useful for testing for linear independenceand has many other applications. A useful theorem related to the Wronskian is Abel’s Theorem(Theorem 5.1.4 in the textbook.)

Suggested Homework:Read Section 5.1, Pages 184 193Section 5.1 { 1, 2, 3, 5, 7, 24, 37, 38, 39 }

43

Lesson 11

Constant Coefficient Homogeneous Equations and Euler Differential Equations

Section 5.2

Let’s consider some very simple, but important linear second order differential equations.

If a, b, c are real numbers and a 6= 0 then the linear second order o.d.e.

ay′′ + by′ + cy = f(x)

is said to be a constant coefficient equation. If f(x) = 0 it is a homogeneous constant coefficientequation. Otherwise it is non-homogeneous.

Suppose we are trying to find the general solution to the linear homogeneous constant coefficientsecond order o.d.e. ay′′ + by′ + cy = 0.

The quadratic p(r) = ar2 + br + c is called the characteristic polynomial of ay′′ + by′ + c = 0. Theequation p(r) = 0 is the characteristic equation.

By the quadratic formula we have the the roots of the characteristic polynomial are

r =−b±√b2 − 4ac

2a.

So we consider three cases:

Case 1: b2 − 4ac > 0. In this case the characteristic polynomial has two distinct real roots.

Let r1 6= r2 be the real roots. Then y1 = er1x and y2 = er2x are both solutions ofay′′ + by′ + cy = 0. The reader is left to verify this. Since y2/y1 = e(r2−r1)x is non-constant asr2−r1 6= 0 we have that y1, y2 are linearly independent. Thus the general solution to ay′′+by′+c = 0is y = c1e

r1x + c2er2x.

Case 2: b2 − 4ac = 0. In this case the characteristic polynomial has one repeated real root.

Let r = −b/(2a) be the repeated root. Then y1 = erx and y2 = xerx are solutions of ay′′+by′+cy =0. The reader is left to verify this.

Moreover, y2/y1 = x is non-constant, so y1, y2 are linearly independent (on any interval). Hencethe general solution of ay′′ + by′ + cy = 0 is

y = c1er1x + c2xer1x = (c1 + c2x)er1x.

44

Case 3: b2 − 4ac < 0. Then the roots of the characteristic polynomial are complex conjugatesλ± ωi, where λ, ω are real numbers and ω 6= 0.

Then y1 = eλx cos (ωx) and y2 = eλx sin (ωx) are solutions of ay′′ + by′ + cy = 0. The reader is leftto verify this.

Since y2/y1 = tan (ωx) is non-constant y1, y2 are linearly independent, and hence the generalsolution to ay′′ + by′ + cy = 0 is

y(t) = c1eλx cos (ωx) + c2e

λx sin (ωx) = (c1 cos (ωx) + c2 sin (ωx))eλx.

Examples:

1. Find the general solution of y′′ + 5y′ − 14y = 0.

The characteristic polynomial is r2 + 5r − 14 = (r + 7)(r − 2). The roots are r = −7, 2. Sothe general solution is

y = c1e−7x + c2e

2x.

2. Solve the IVP y′′ − 6y′ + 9y = 0, y(0) =1, y′(0) = 5.

The characteristic polynomial is r2 − 6r + 9 = (r − 3)2. The repeated root is r = 3. So thegeneral solution is

y = c1e3x + c2xe3x.

Then y′ = 3c1e3x + c2e

3x + 3c2xe3x.

So imposing the initial conditions yields

c1 = 1 and 3c1 + c2 = 5 → c1 = 1 and c2 = 2.

So the solution isy = e3x + 2xe3x.

3. Solve the IVP y′′ + 4y = 0, y(0) = −1, y′(0) = 2.

The characteristic polynomial is r2 +4, which has complex roots r = ±2i. Since ±2i = 0±2i,the general solution is

y = c1 cos (2x) + c2 sin (2x).

Then y′ = −2c1 sin (2x) + 2c2 cos (2x).

By imposing the initial conditions we get c1 = −1 and 2c2 = 2, so c1 = −1 and c2 = 1. Sothe solution is y = − cos (2x) + sin (2x).

45


Read Section 5.2, Pages 200 206Section 5.2 - { 1 - 12, 13, 14, 17, 22, 23, 24, 25 }

46

Lesson 12

Euler Differential Equations

Section 7.4

A second order differential equation that can be written in the form

ax2 d2y

dx2+ bx

dy

dx+ cy = g (x)

where a, b, and c are real constants with a 6= 0 is called an Euler equation.

By the existence and uniqueness theorem we see that if g is continuous everywhere, except possiblyat 0 that an Euler equation has solutions defined on (−∞, 0) and (0,∞). To see this put the dif-ferential equation into standard form and observe that we have a discontinuity at 0 for both p (x)and q (x).

We will restrict our attention to the interval (0,∞).

To find the general solution of the homogeneous Euler equation

ax2 d2y

dx2+ bx

dy

dx+ cy = 0

we will assume that there is a solution of the form y = xm . Then y′ = mxm−1 and y′′ =m (m− 1)xm−2 . Substituting into the differential equation:

ax2m (m− 1)xm−2 + bxmxm−1 + cxm = 0

am (m− 1)xm + bmxm + cxm = 0

Since we have x > 0 we can divide by xm to get the characteristic equation

am (m− 1) + bm + c = 0

am2 + (b− a) m + c = 0

As with constant coefficient equations we have three cases.

Case 1: Two real solutions, m1 and m2. The general solution is y = c1xm1 + c2x

m2 .

47

Example: Find the general solution of 6x2y′′ + 17xy′ + 3y = 0 on (0,∞).

The characteristic equation is

6m2 + (17− 6)m + 3m = 0

6m2 + 11m + 3m = 0

(3m + 1) (2m + 3) = 0

Then y1 = x−13 and y2 = x−

32 and the general solution is y = c1x

− 13 + c2x

− 32

Case 2: One real solution, m. One solution is y1 = xm. Using the technique reduction of orderwhich is covered in lesson 15 you can show that a second linearly independent solution is given byy2 = xm lnx . The general solution is then

y = c1xm + c2x

m ln x.

Example: Find the general solution of 4x2y′′ + 8xy′ + y = 0 on (0,∞).

The characteristic equation is

4m (m− 1) + 8m + 1 = 0

4m2 + 4m + 1 = 0

(2m + 1)2 = 0

Then y1 = x−12 and y2 = x−

12 ln x . The general solution is y =

c1 + c2 lnx√x

.

48

Case 3: Two complex solutions m1 = α + iβ and m2 = α− iβ .

y = xα+iβ

y = xαxiβ

To deal with xiβ we use Euler’s identity:

xiβ = eiβ ln x = cos (β lnx) + i sin (β ln x)

From this we can see that the general solution is given by

y = xα (c1 cos (β lnx) + c2 sin (β lnx))

.

Example: Find the general solution for x2y′′ + 3xy′ + 10y = 0 on (0,∞).

The characteristic equation is m2 + 2m + 10 = 0. From the quadratic formula

m =−2±

√22 − 4 (10)2

= −1± 3i

So the general solution is then given by y =c1 cos (3 lnx) + c2 sin (3 lnx)

x


Read Section 7.4, Pages 324 (Start with EULER EQUATIONS) - 327Section 7.4 - { 1 - 10 }

49

Lesson 13

Nonhomogeneous Linear Equations

Section 5.3

Let’s consider the non-homogenous case where y′′ + p(x)y′ + q(x)y = f(x). A single solutionto this equation is called a particular solution. The equation with f(x) = 0 is called thecomplementary equation.

If p, q, f are continuous on (a, b) and x0 is a point in (a, b) then for any constants k0 and k1 theIVP

y′′ + p(x)y′ + q(x) = f(x), y(x0) = k0, y′(x1) = k1

has a unique solution on (a, b).

Furthermore, if yp is a particular solution to y′′ + p(x)y′ + q(x)y = f(x) on (a, b) and c1y1 + c2y2

is the general solution to the complementary equation y′′ + p(x)y′ + q(x)y = 0 on (a, b) then y is asolution to y′′+p(x)y′+q(x)y = f(x) on (a, b) if and only if y = yp +c1y1 +c2y2 for some constantsc1, c2.

Example: Let’s solve the following IVP: y′′ + 9y = 9, y(0) = 0, y′(0) = 4.

The characteristic polynomial of the complementary equation y′′+9y = 0 is r2 +9 which has rootsr = ±3i. So c1 cos (3x) + c2 sin (3x) is the general solution of y′′ + 9y = 0.

Obviously y1 = 1 is a particular solution of y′′ + 9y = 9. Hence y = 1 + c1 cos (3x) + c2 sin (3x) isthe general solution of y′′ + 9y = 9.

Imposing the initial condition y(0) = 0 implies 0 = 1+c1, so c1 = −1. Imposing the initial condition

y′(0) = 4 implies 4 = 3c2, so c2 =43.

So the solution to the IVP isy = 1− cos (3x) +

43

sin (3x).

WARNING: Do not use the initial conditions to find the constants in the general solution of thecomplementary equation and then add a particular solution. Doing so yields an incorrect result.

50

Superposition Principle: Suppose p, q, f1, f2 are continuous on (a, b), yp1 is a particular solutionto y′′+p(x)y′+q(x)y = f1(x) on (a, b), and yp2 is a particular solution to y′′+p(x)y′+q(x)y = f2(x)on (a, b). Then y = yp1 + yp2 is a solution to y′′ + p(x)y′ + q(x)y = f1(x) + f2(x) on (a, b).

Example: Let’s find the general solution to y′′ − 7y′ + 12y = 5e−2x + 12x2 − 2x + 5.

Clearly the general solution to the complementary homogeneous equation is y = c1e3x + c2e

4x.

We can try to find a pair of particular solutions to y′′ − 7y′ + 12y = 5e−2x and y′′ − 7y′ + 12y =12x2 − 2x + 5 respectively.

Let’s begin with the former. Let’s judiciously guess that there is a solution of the form y = Ae−2x

and substitute it into y′′ − 7y′ + 12y = 5e−2x:

4Ae−2x − 7(−2Ae−2x) + 12Ae−2x = 5e−2x

4A + 14A + 12A = 5

A =16

So yp1 =16e−2x is a particular solution.

Now we proceed with the latter. Let’s judiciously guess that there is a solution of the formy = ax2 + bx + c and substitute it into y′′ − 7y′ + 12y = 12x2 − 2x + 5:

2a− 7(2ax + b) + 12(ax2 + bx + c) = 12x2 − 2x + 5

2a− 7b + 12c = 5, 12b− 14a = −2, and 12a = 12

a = 1, b = 1, c = 0.

So yp2 = x2 + x is a particular solution.

By superposition, yp =16e−2x + x2 + x, is a particular solution of

y′′ − 7y′ + 12y = 5e−2x + 12x2 − 2x + 5.

So the general solution is y =16e−2x + x2 + x + c1e

3x + c2e4x.


Read Section 5.3, Pages 209 214Section 5.3 { 1, 4, 16, 17 }

51

Lesson 14

The Method of Undetermined Coefficients

Section 5.4, 5.5

Consider the nonhomogeneous differential equation y′′+p (t) y′+ q (t) y = g (t). To find the generalsolution:

1) First find the general solution to the homogeneous differential equation, called the complementaryequation

y′′ + p (t) y′ + q (t) y = 0

yh = c1y1 + c2y2

We call yh the homogeneous solution or the complementary solution.

2) Next find a particular solution yp to y′′ + p (t) y′ + q (t) y = g (t).

3) The general solution to y′′ + p (t) y′ + q (t) y = g (t) is y = yh + yp = c1y1 + c2y2.

To find yp we will look at two methods:

i) The Method of Undetermined Coefficientsii) Variation of Parameters

For this lesson we will cover the method of undetermined coefficients.

The method of undetermined coefficients is a guess and check method that will only use for constantcoefficient equations and is only feasible for certain forms of the forcing functions g (t).

The method of undetermined coefficients works if the forcing function is a linear combination offunctions of the forms:

1, tn, n = 1, 2, 3, · · · ,eat, tneat, n = 1, 2, 3, · · ·cos (βt) , sin (βt), eat cos (βt) , eat sin (βt)

tn cos (βt) , tn sin (βt) , n = 1, 2, 3, · · ·tneat cos (βt) , tneat sin (βt) , n = 1, 2, 3, · · ·

52

In a general sense (we will see an exception later) you guess at a form of the solution that is similarto the forcing function. Your guess will contain unknown coefficients that you will need to solvefor. If the term contains tn you must include all lower order terms of the polynomial.

Examples (without the exception):

g (t) = 2t3 − 7, you would take yp = At3 + Bt2 + Ct + D where A,B,C, and D are constants thatyou will need to evaluate.

g (t) = 4t2e−5t , you would take yp = At2e−5t + Bte−5t + Ce−5t

If the term includes either a cosine or a sine you must include both cosine and sine in the form ofyour solution.

Examples (without the exception):

g (t) = 2 cos (3t) , you would take yp = A cos (3t) + B sin (3t)

g(t) = 2te−t sin t , you would take yp = Ate−t cos t + Be−t cos t + Cte−t sin t + De−t sin t

The Exception: If part of the forcing function is a solution to the homogeneous equation you mustuse repeated root techniques and multiply by t (or t2 if the root is doubled)

Examples:

y′′ − y = et, yh = c1et + c2e

−t . g (t) = et is a homogeneous solution take yp = Atet

y′′ − y = t2et, yh = c1et + c2e

−t . g (t) = et is a homogeneous solution takeyp = t

(At2 + Bt + C

)et or the more useful form yp = At3et + Bt2et + Ctet

Important Note: You can tell if you have the correct form of the particular solution when you goto solve for the unknown coefficients. If you do not have enough terms you will not get a system ofequations that will allow you to uniquely solve for all the coefficients. If you have too many termsyou will not get any useful equations for solving for some of the terms.

53

Example: y′′ − y′ + y = 2 sin (3t)

The complementary solution is yh = e12t[c1 cos

(√3

2 t)

+ c2 sin(√

32 t

)].

Since sin (3t) is not a solution to the homogeneous equation we can take yp = A cos (3t)+B sin (3t).

To solve for A and B we substitute yp into the differential equation.

yp = A cos (3t) + B sin (3t)

yp′ = −3A sin (3t) + 3B cos (3t)

yp′′ = −9A cos (3t)− 9B sin (3t)

Theny′′ − y′ + y = 2 sin (3t)

(−9A cos (3t)− 9B sin (3t))− (−3A sin (3t) + 3B cos (3t)) + (A cos (3t) + B sin (3t)) = 2 sin (3t)

(−9A− 3B + A) cos (3t) + (−9B + 3A + B) sin (3t) = 2 sin (3t)

Equating coefficients gives

−8A− 3B = 0

3A− 8B = 2

A =673

, B = −1673

The particular solution is yp = 673 cos (3t)− 16

73 sin (3t) which gives the general solution:

yh = e12t[c1 cos

(√3

2 t)

+ c2 sin(√

32 t

)]+ 6

73 cos (3t)− 1673 sin (3t)

54

Example: y′′ − 6y′ + 9y = −12e3t

The complementary solution is yh = c1e3t + c2te

3t . Since 3 is a double root to the characteristicequation we must take yp = At2e3t.

yp = At2e3t

yp′ = 2Ate3t + 3At2e3t

yp′′ = 2Ae3t + 12Ate3t + 9At2e3t

Then

y′′ − 6y′ + 9y = −12e3t

(2Ae3t + 12Ate3t + 9At2e3t

)− 6(2Ate3t + 3At2e3t

)+ 9At2e3t = −12e3t

2Ae3t = −12e3t

A = −6

yp = −6t2e3t

y = c1e3t + c2te

3t − 6t2e3t

The textbook have numerous examples and additional examples will be given in lecture.

Suggested Homework:Read Sections 5.4 and 5.5, Pages 217 232Section 5.4 { 1, 2, 3, 4, 8, 15, 16 } ansSection 5.5 { 1, 2, 3, 4, 5, 7, 8, 32 }

55

Lesson 15

Reduction of Order

Section 5.6

The textbook does reduction of order for nonhomogeneous equations. We are only going to doreduction of order for homogeneous equations.

Somehow we have found a nonzero solution y1 to the linear homogeneous linear differential equationy′′ + p(t)y′ + q(t)y = 0. This means that y′′1 + py′1 + qy1 = 0.

To find a second linearly independent solution assume y2 = uy1 where u is an as yet unknownfunction of t. We will substitute this into the differential equation and see if we can find a non-constant solution for u. First we take derivatives:

y′2 = u′y1 + uy′1

y′′2 = u′′y1 + u′y′1 + u′y′1 + uy′′1= u′′y1 + 2u′y′1 + uy′′1

Then y′′ + py′ + qy = 0 becomes u′′y1 + 2u′y1′ + uy1

′′ + p [u′y1 + uy′1] + quy1 = 0.

This rearranges to u [y1′′ + py1

′ + qy1] + u′′y1 + 2u′y1′ + pu′y1 = 0.

Recalling that y1′′ + py1

′ + qy1 = 0 we see that u′′y1 + 2u′y1′ + pu′y1 = 0.

This is a second order linear differential equation for u, but we can reduce it to a first order differ-ential equation by the substitution w = u′ (thus the title Reduction of Order).

w′y1 + (2y′1 + py1)w = 0.

This is a first order differential equation that is both linear and separable (remember y1 is a knownfunction.)

56

Example: If the second order linear differential equation with constant coefficients has a charac-teristic equation with repeated roots, (r − a)2 = 0 then the differential equation is of the formy′′ − 2ay′ + a2y = 0. One solution is y1 = eat. To find a second linearly independent solution weassume y2 = ueat . We begin by finding the first and second derivative of y2.

y2′ = u′eat + aueat

y2′′ = u′′eat + 2au′eat + a2ueat

Then we substitute these into the differential equation:

u′′eat + 2au′eat + a2ueat − 2a(u′eat + aueat

)+ a2ueat = 0.

This simplifies to u′′ = 0 . The simplest non-constant solution is u = t . Then y2 = teat and thegeneral solution is y = Ateat + Beat.

Example: Given that y1 = ex is a solution to xy′′ − (2x + 1) y′ + (x + 1) y = 0 find the generalsolution.

First you should verify that y1 is a solution. We will leave that as an exercise for the reader. Alsoit is worth noting that for these problems there is little need to put the differential equation instandard form if you are going to use the process. However, if you are going to jump straight tothe derived equation w′y1 + (2y′1 + py1) w = 0 then the equation needs to be in standard form toproperly identify p.

Assume y2 = uex. We begin by taking derivatives.

y2′ = u′ex + uex

y2′′ = u′′ex + 2u′ex + uex

Then substitute these into the differential equation:

x (u′′ex + 2u′ex + uex)− (2x + 1) (u′ex + uex) + (x + 1)uex = 0

Which simplifies to xu′′ − u′ = 0. Make the substitution w = u′ to reduce this to a first orderdifferential equation xw′ − w = 0.

We separate,w′

w=

1x

, integrate ln |w| = ln |x| + Constant. We can take w = x. Then u′ = x and

u =12x2. Remembering that any constant times a solution to a homogeneous linear differential

equation is also a solution we can take u = x2 which gives y2 = x2ex.

The general solution is then y = Aex + Bx2ex.57


Read this sectin of the study guideSection 5.6 { 18, 19, 20, 21, 25, 32 }NOTE: Skip problems 1 - 17, 31, 33, 34, 35, 37 - 40

58

Lesson 16

Variation of Parameters

Section 5.7

In this lesson we cover a method called variation of parameters for finding a particular solutionof

y′′ + p(t)y′ + q(t)y = f(t).

The first step is to find the general solution yh = c1y1 + c2y2 to the complementary equation

y′′ + p(t)y′ + q(t)y = 0

which we will call the homogenous solution.

We assume a particular solution has the form yp = u1y1 + u2y2.

With the assumption that u′1y1 + u′2y2 = 0 one is able to derive that

u′1 =−fy2

y1y′2 − y′1y2

u′2 =fy1

y1y′2 − y′1y2

Note: These require that the differential equation is in the standard form y′′+p(t)y′+q(t)y = f(t).

By rewriting these equations in determinant form we accomplish two goals. The first is that it’seasier to remember; the second is that it is easier to see how to extend these formulas to higherdegree equations.

The first thing to notice is that in each denominator is the Wronskian W =∣∣∣∣

y1 y2

y′1 y′2

∣∣∣∣ of y1 andy2.

We define W1 =∣∣∣∣

0 y2

1 y′2

∣∣∣∣ and W2 =∣∣∣∣

y1 0y′1 1

∣∣∣∣.

Then the above equations for u′1 and u′2 become

u′1 =fW1

W

u′2 =fW2

W59

Example: Find the general solution to x2y′′ + 8xy′ + 10y = 4x−2.

The reader is left to check that the homogeneous solution is yh = c1x−2 + c2x

−5. We will usey1 = x−2 and y2 = x−5.

First we put the equation into standard form y′′+8x−1y′+10x−2y = 4x−4 to identify f(x) = 4x−4.

Then

u′1 =fW1

W=

4x−4

∣∣∣∣0 x−5

1 −5x−6

∣∣∣∣∣∣∣∣

x−2 x−5

−2x−3 −5x−6

∣∣∣∣=−4x−9

−3x−8=

43x−1

u′2 =fW2

W=

4x−4

∣∣∣∣x−2 0−2x−3 1

∣∣∣∣∣∣∣∣

x−2 x−5

−2x−3 −5x−6

∣∣∣∣=

4x−6

−3x−8= −4

3x2

So we can choose u1 =43

ln x and u2 = −49x3.

Then yp = u1y1 + u2y2 =43x−2 ln x − 4

9x−2 is a particular solution of x2y′′ + 8xy′ + 10y = 4x−2.

Since −49x−2 is a solution to the homogeneous equation we can simply take yp =

43x−2 ln x as our

particular solution.

So the general solution to x2y′′ + 8xy′ + 10y = 4x−2 is

y =c1

x2+

c2

x5+

4 lnx

3x2

60

To see how to generalize variation of parameters to higher orders let’s look at the equations forthird order differential equation y′′′ + p (t) y′′ + q (t) y′ + r (t) y = f (t). As before we first find thecomplementary solution, yh = c1y1 + c2y2 + c3y3 . The Wronskian is

W =

∣∣∣∣∣∣

y1 y2 y3

y1′ y2

′ y3′

y1′′ y2

′′ y3′′

∣∣∣∣∣∣

To find a particular solution we assume yp = u1y1 + u2y2 + u3y3 .

Thenu1′ =

fW1

W, u2

′ =fW2

W, u3

′ =fW3

W

Where

W1 =

∣∣∣∣∣∣

0 y2 y3

0 y2′ y3

′

1 y2′′ y3

′′

∣∣∣∣∣∣, W2 =

∣∣∣∣∣∣

y1 0 y3

y1′ 0 y3

′

y1′′ 1 y3

′′

∣∣∣∣∣∣, W3 =

∣∣∣∣∣∣

y1 y2 0y1′ y2

′ 0y1′′ y2

′′ 1

∣∣∣∣∣∣


Read Section 5.7, Pages 242 249Section 5.7 { 1, 3, 5, 6, 7, 10, 11, 12, 24, 28 }

61

Lesson 17

Spring Problems I

Section 6.1

Consider an object of mass m suspended from a spring (of negligible mass). We say that thespring-mass system is at equilibrium when the object is at rest and the sum of the forces acting onit is 0. The corresponding position is called the equilibrium position.

Let y be the displacement of the mass from the equilibrium where y > 0 is upward.

The following forces may or will act on the object:

• Gravity: Fg = −mg where g is the gravitational constant.

• Spring Force: Fs = k∆L where k is the spring constant (Hooke’s Law) and ∆L is the changein length (stretching is positive) from the spring equilibrium position with no mass attached.

• Damping Force: Fd = −cy′ where c is the damping constant. Simple harmonic motion occurswhen c = 0.

• An external force F , independent of the motion, not accounted for above.

By Newton’s second law we can write an o.d.e. to model the spring-mass system described above:

my′′ = F −mg − cy′ + k∆L.

Let’s relate ∆L and y. Imagine the spring without a mass attached. When a mass m is attached,the spring stretches by some amount ` and at equilibrium, the sum of forces on the spring is 0, soFg + Fs = 0. Hence −mg + k` = 0, or ` = mg/k.

If at this point, a displacement of y occurs then we would have ∆L = `− y = mg/k − y.

So my′′ = F −mg − cy′ + k∆L becomes my′′ = F − cy′ − ky or equivalently my′′ + cy′ + ky = F .

If F = 0 we say the motion is free. Otherwise, it is said to be forced.

62

Consider the free undamped system modeled by y′′ +k

my = 0. We know that a general solution is

y = c1 cos (ωt) + c2 sin (ωt), where ω =√

k/m.

Let A =√

c21 + c2

2. Then c1 = A cos (φ) and c2 = A sin (φ) for some angle φ in (−π, π].

Using the identity, cos (ωt− φ) = cos (ωt) cos (φ) + sin (ωt) sin (φ) we get

y = c1 cos (ωt) + c2 sin (ωt) = A cos (φ) cos (ωt) + A sin (φ) sin (ωt) = A cos (ωt− φ).

The A is the amplitude (motion goes from −A to A). T = 2π/ω is the period (time it takesto go through a full cycle, such as from A back to A). φ is the phase angle. ω is the naturalangular frequency, and f = ω/(2π) = 1/T is the frequency.

Example: Suppose an object is hung from a free undamped spring whose natural length is 98 cmand attains an equilibrium position of 1 m. Suppose the object is displaced downward 5 cm, butgiven an initial upward velocity of 11 cm/s. Find the displacement for time t > 0.

Since g/` = k/m, we have that ω =√

k/m =√

g/` =√

9.8/0.02 ≈ 22.

So y ≈ c1 cos (22t) + c2 sin (22t). Imposing y(0) = −0.05 yields c1 = −0.05. Imposing y′(0) = 0.11yields 22c2 = 0.11, so c2 = 0.005.

Hence y ≈ −0.05 cos (22t) + 0.005 sin (22t) ≈ 0.0503 cos (22t− 3.24).


Read Sections 6.1 and 6.2, Pages 253 271Section 6.1 { 2, 3, 6, 7 } andSection 6.2 { 1, 4, 5, 24, 25, 26 }

63

Lesson 18

Spring Problems II and the RLC Circuit


Let’s return to the general case of a spring with a damping constant c, a spring constant k, amass m hanging from it and using a non-zero forcing function F (t). Such a system satisfies thedifferential equation

my′′ + cy′ + ky = F (t),

where y is the displacement from equilibrium.

Example: Suppose a 1 kg mass hangs from an undamped (c = 0) of spring constant k = 900 N/m.Suppose we have a forcing function of F (t) = 644 cos 16t N. Find the IVP solution where y(0) = 0,y′(0) = 0.

First we solve y′′ + 900y = 644 cos (16t).

The complementary equation is y′′ + 900y = 0 and has the general solution

y = c1 cos (30t) + c2 sin (30t).

By using the method of undetermined coefficients with yP = A cos (16t) + B sin (16t) we get aparticular solution to y′′ + 900y = 644 cos (16t) work out as yP = cos (16t). The reader is left tocheck the details.

So the general solution of y′′ + 900y = 644 cos (16t) is y = cos (16t) + c1 cos (30t) + c2 sin (30t).Imposing the initial conditions we see that c1 = −1 and c2 = 0.

64

So the solution is y = cos (16t)− cos (30t), which looks like this:

1 2 3 4 5

-2

-1

1

2

In this example the periodic forcing function F (t) = 644 cos (16t) has an angular frequency of8π

which doesn’t match the natural frequency of the spring which is15π

. In such cases, the solutionto any IVP with this equation must be bounded as it’s a finite sum of sine and cosine functions.

When a periodic forcing function matches the natural frequency of the spring the spring system issaid to be in resonance.

Example: Suppose we want to solve the same mass-spring system IVP as described in the previ-ous example, except that the periodic forcing function is F (t) = 120 sin (30t) (and thus we haveresonance).

The difference comes when using the method of undetermined coefficients. We guess that a partic-ular solution looks like yP = t(A cos (30t) + B sin (30t)).

We can then work out that A = −2 and B = 0. The reader is left to check the details.

65

The general solution to y′′ + 900y = 120 sin (30t) is y = −2t cos (30t) + c1 cos (30t) + c2 sin (30t).

Imposing the initial conditions yields c1 = 0 and c2 = 2.

This gives a solution of y = −2t cos (30t) + 2 sin (30t), which looks like this:

1 2 3 4 5

-10

-5

5

10

As can be seen the amplitude goes towards ∞. Of course, in a physical mass-spring system theamplitude growth eventually causes the spring to fail.

For coverage of the effects of a non-zero damping constant on a spring, see the Trench textbook,which does a nice job describing the underdamped, critically damped and overdamped cases.

66

An RLC circuit is one that includes a voltage source of voltage V (t) (SI units of volts) as a functionof time, a resistor of resistance R > 0 (SI units ohms), a coiled wire inductor of inductance L > 0(SI units henries), and a plate capacitor of capacitance C > 0 (SI units farads).

Let Q(t) be the charge (SI units of coulombs) on the capacitor at time t. Then I(t) =dQ

dtis the

electrical current (SI units of amps) in the circuit. We have the following basic laws for the changein voltage across each circuit component:

• Resistor: V = IR

• Inductor: V = LdI

dt

• Capacitor: V =Q

C

Kirchhoffs law says that the voltage change across any two points has to be independent of thepath used to travel between the two points. Hence

LdI

dt+ IR +

Q(t)C

= V (t).

By differentiating both sides and using I =dQ

dtwe get

LI ′′ + RI ′ +1C

I = V ′(t).

Note: One could also get a second order linear constant coefficient equation in terms of Q(t) withoutdifferentiating.

67

Example: For an alternating current voltage source V (t) = E sin (ωt) the differential equationbecomes

LI ′′(t) + RI ′(t) +1C

I(t) = ωE cos (ωt).

Let’s find the general solution of this differential equation. The complementary equation has acharacteristic polynomial of Lx2 + Rx + 1/C = 0 which has roots at

x =−R±

√R2 − 4L/C

2L= r1, r2.

Then the general solution to the complementary equation is either I = c1er1t + c2e

r2t if

R > 2√

L/C (underdamped) or I = (c1 + c2t)e−R2L

t if R = 2√

L/C (critically damped)

or I = e−R2L

t

[c1 cos

(√4L/C −R2

2Lt

)+ c2 sin

(√4L/C −R2

2Lt

)]if R < 2

√L/C (overdamped).

In each case the complementary solution decays towards 0 as t →∞.

The method of undetermined coefficients, where we guess a particular solution of the form IP =A cos (ωt) + B sin (ωt), yields (the reader is left to check the details)

A = − ωE(Lω2 − 1/C)R2ω2 + (Lω2 − 1/C)2

and B =Rω2E

R2ω2 + (Lω2 − 1/C)2.

Putting the solutions together and imposing initial conditions allows us to solve for the currentgiven any RLC circuit with any voltage source and initial conditions.

For example, suppose we have an RLC circuit where the resistance is R = 600 ohms, the capacitanceis C = 0.000001 farads, the inductance is L = 0.25 henries, and the voltage source is V (t) =120 sin (120πt) volts.

With these values R < 2√

L/C, A ≈ 0.0445 and B ≈ 0.0104. So the general solution is

I(t) ≈ 0.0445 cos (120πt) + 0.0104 sin (120πt) + e−1200t [c1 cos (1600t) + c2 sin (1600t)] .

Then given initial conditions we can find c1 and c2 for a solution of the above form, called atransient solution, since the part of the solution involving c1 and c2 decays away quickly ast →∞. In the long run the current approaches the steady state solution ofI(t) ≈ 0.0445 cos (120πt) + 0.0104 sin (120πt).

68


Read Section 6.3, Pages 273 278Section 6.1 { 11, 13, 15 } andSection 6.2 { 13, 17, 18 } andSection 6.3 { 1, 2, 6, 7 }

69

Lesson 19

Introduction to the Laplace Transform

Section 8.1

Let f(t) be defined for t ≥ 0 and s represent a real number. Then the Laplace transform of f is

F (s) =∫ ∞

0e−stf(t) dt.

The functions f and F are called a transform pair.

Notation: L(f) = F or f ↔ F .

Note: Technically the improper integral must be turned in a limit, but we will be lazy and neverformally do this. We will allow “∞” to appear as though it were a real number (but with theunderstanding that we are taking a limit).

Let’s calculate some basic Laplace transforms:

(i) f(t) = 1 (a constant function). Then F (0) diverges (to ∞) and for s 6= 0,

F (s) =∫ ∞

0e−st dt = −1

se−st

∣∣∣∣∞

0

=

1s

if s > 0

∞ if s < 0

. So we write L(1) =1s, s > 0.

(ii) f(t) = t2. Then F (0) diverges (to ∞) and for s 6= 0,

F (s) =∫ ∞

0t2e−st dt

= − t2

se−st

∣∣∣∣∞

0

+2s

∫ ∞

0te−st dt

= − t2

se−st

∣∣∣∣∞

0

+2s

[− t

se−st

∣∣∣∣∞

0

+1s

∫ ∞

0e−st dt

]

= − t2

se−st

∣∣∣∣∞

0

+2s

[− t

se−st

∣∣∣∣∞

0

+1s

(−1

se−st

∣∣∣∣∞

0

)]

=

2s3

if s > 0

∞ if s < 0

So we write L(t2) =2s3

, s > 0.70

(iii) f(t) = eat. F (a) diverges to ∞.

F (s) =∫ ∞

0e−steat dt =

∫ ∞

0e(a−s)t dt =

1a− s

e(a−s)t

∣∣∣∣∞

0

=

1s− a

if s > a

∞ if s < a

.

So we write L(eat) =1

s− a, s > a.

(iv) f(t) =

2− t if 0 ≤ t < 1

t if t ≥ 1

F (s) =∫ ∞

0f(t)e−st dt =

∫ 1

0(2− t)e−st dt +

∫ ∞

1te−st dt

=

[−2− t

se−st

∣∣∣∣1

0

− 1s

∫ 1

0e−st dt

]+

[− t

se−st

∣∣∣∣∞

1

+1s

∫ ∞

1e−st dt

]

=[2s− 1

se−s +

1s2

(e−s − 1)]

+[1se−s +

1s2

e−s

]=

2s− 1

s2+

2s2

e−s.

A table of Laplace transforms is included at the beginning of this study guide.

Linearity of Laplace transforms: If L(f) and L(g) both exist for s > s0 then for any constantsc1, c2 we have

L(c1f(t) + c2g(t)) = c1L(f) + c2L(g) for s > s0.

The proof is very straightforward.

Let’s find the Laplace transform of sinh (at) =eat − e−at

2where a > 0.

We know that L(eat) =1

s− afor s > a and L(e−at) =

1s + a

for s > −a.

By linearity, L(sinh (t)) =12

(1

s− a− 1

s + a

)=

a

s2 − a2for s > a.

71

Suppose L(f) = F . Let’s determine L(eatf(t)) where a is a constant.

F (s) = L(f(t)) =∫ ∞

0e−stf(t) dt

and thus L(eatf(t)) =∫ ∞

0e−st(eatf(t)) dt =

∫ ∞

0e−(s−a)tf(t) dt = F (s− a).

This property is referred to as the first shifting property of the Laplace transform.

Given that L(cos (ωt)) =s

s2 + ω2(from the table) we use the first shifting property to find

L(eat cos (ωt)):

L(eat cos (ωt)) =s− a

(s− a)2 + ω2.

A function f is of real exponential order s0 if there are constants M and t0 such that

|f(t)| ≤ Mes0t for all t ≥ t0.

Often we say a function is of exponential order without a reference to a particular s0. The followingproposition is straightforward to prove.

Proposition: Suppose f is of exponential order s0. Then L(f) is defined for s > s0.


Read Section 8.1, Pages 394 - 402Section 8.1 - { 1, 2, 6, 8, 12 }

72

Lesson 20

The Inverse Laplace Transform

Section 8.2

Suppose F = L(f). Then we call f the inverse Laplace transform of F . Notation: f = L−1(F ).

While there is a formula for computing L−1(F ) we will not use it as it requires the theory offunctions of a complex variable (complex analysis). We can however, make use the table of Laplacetransforms to find inverse Laplace transforms.

Suppose we need L−1

(s

s2 + 4

). By looking through the table of Laplace transforms we find

L(cos (kt)) =s

s2 + k2.

So L−1

(s

s2 + 4

)= cos (2t).

Linearity also holds for the inverse transform. Let’s find L−1

(2s + 3

s2 + 4s + 13

):

L−1

(2s + 3

s2 + 4s + 13

)= L−1

(2(s + 2)− 1(s + 2)2 + 9

)= 2L−1

(s + 2

(s + 2)2 + 9

)− 1

3L−1

(3

(s + 2)2 + 9

)

= 2e−2t cos (3t)− 13e−2t sin (3t) = e−2t

(2 cos (3t)− 1

3sin (3t)

).

In the last example, the denominator is irreducible over the reals. Our method changes when thedenominator factors nicely:

Let’s find L−1

(2s− 1

s2 − 5s + 6

).

Using a P.F.D. we get2s− 1

s2 − 5s + 6=

2s− 1(s− 2)(s− 3)

=5

s− 3− 3

s− 2.

Then L−1

(2s− 1

s2 − 5s + 6

)= 5L−1

(1

s− 3

)− 3L−1

(1

s− 2

)= 5e3t − 3e2t.

73

Let’s find L−1

(s

s2 − 8s + 16

).

We can setup a P.F.D. as follows:

s

s2 − 8s + 16=

s

(s− 4)2=

A

s− 4+

4(s− 4)2

.

Clearing fractions and equating common-degree coefficients, we get A = 1.

So L−1

(s

s2 − 8s + 16

)= L−1

(1

s− 4

)+ 4L−1

(1

(s− 4)2

)= e4t + 4te4t = e4t(1 + 4t).

Proposition: Suppose f is continuous on [0,∞) and of exponential order s0, and f ′ is piecewisecontinuous on [0,∞). Then L(f) and L(f ′) are defined for s > s0 and

L(f ′) = sL(f)− f(0).

Proof : We will only prove it in the case where f ′ is continuous on [0,∞). You can see the proof inthe other case in the textbook.

Assume f ′ is continuous on [0,∞). By integration by parts,

L(f ′) =∫ T

0e−stf ′(t) dt = e−stf(t)

∣∣T0

+ s

∫ T

0e−stf(t) dt.

Since f is of exponential order s0, for s > s0 we have limT→∞

e−sT f(T ) = 0 and L(f) =∫ ∞

0e−stf(t) dt

converges. Hence L(f ′) converges for s > s0 and

L(f ′) = −f(0) + sL(f) ¤


Read Section 8.2, Pages 405 - 412Section 8.2 - { 2, 4, 6, 9 }

74

Lesson 21

Solution of Initial Value Problems

Section 8.3

Consider the basic first order IVP y′ = ay, y(0) = y0. We already know the solution is y(t) = y0eat.

Let’s see if we can find the same result using our new tool! Let Y (s) = L(y) be the Laplacetransform of the solution y(t).

Let’s transform both sides of the o.d.e. y′ = ay.

L(y′) = L(ay) → sL(y)− y(0) = aL(y) → (s− a)L(y) = y0 → Y (s) = L(y) =y0

s− a.

Then y(t) = L−1

(y0

s− a

)= y0e

at. Neat!

What about a second order IVP? Well, we can use that

L(y′′) = sL(y′)− y′(0) = s(sL(y)− y(0))− y′(0) = s2L(y)− y′(0)− sy(0).

Let’s derive a simplified form of the Laplace transform of the solution of the IVP

ay′′ + by′ + cy = f(t), y(0) = k0, y′(0) = k1.

We begin with aL(y′′) + bL(y′) + cL(y) = L(f(t)).

Then by substitution, this becomes a(s2L(y)− k1 − sk0) + b(sL(y)− k0) + cL(y) = L(f(t)).

Letting Y (s) = L(y) we get (as2 + bs + c)Y (s) = L(f(t)) + ak1 + ak0s + bk0 and hence

Y (s) =L(f(t)) + a(k1 + k0s) + bk0

as2 + bs + c.

One can use the above as a formula to solve such an IVP, or alternatively, just transform bothsides...

75

Let’s solve the IVP y′′ + 4y′ − 5y = 6et, y(0) = 1, y′(0) = 0 using Laplace transforms.

We would have that Y (s) =6

s−1 + (0 + 1s) + 4s2 + 4s− 5

is the Laplace transform of the solution.

Simplifying this yields Y (s) =6 + (s + 4)(s− 1)(s− 1)2(s + 5)

=s2 + 3s + 2

(s− 1)2(s + 5).

We can begin our P.F.D. by having Y (s) =s2 + 3s + 2

(s− 1)2(s + 5)=

A

s− 1+

1(s− 1)2

+1/3

s + 5.

By clearing fractions,

s2+3s+2 = A(s−1)(s+5)+(s+5)+(1/3)(s−1)2 = As2+4As−5A+s+5+(1/3)s2−(2/3)s+(1/3).

So A = 2/3 and Y (s) =2/3

s− 1+

1(s− 1)2

+1/3

s + 5.

Then y(t) = (2/3)L−1

(1

s− 1

)+ L−1

(1

(s− 1)2

)+ (1/3)L−1

(1

s + 5

)=

23et + tet +

13e−5t.



76

Lesson 22

The Unit Step Function

Section 8.4

The unit step function u(t) =

0 if t < 0

1 if t ≥ 0is a basic piecewise continuous function.

Consider a piecewise continuous function f(t) =

f0(t) if 0 ≤ t < t1

f1(t) if t ≥ t1

.

We can rewrite f(t) in a convenient form using the unit step function appropriately:

f(t) = f0(t) + u(t− t1)[f1(t)− f0(t)].

The proof of the following fact is in the textbook:

L(u(t− r)g(t)) = e−srL(g(t + r)).

Let’s use this fact to find the Laplace transform of f(t) =

2− t if 0 ≤ t < 1

t if t ≥ 1.

First we rewrite it as f(t) = 2− t + u(t− 1)[2t− 2] = 2− t + 2u(t− 1)[t− 1].

Then L(f(t)) =2s− 1

s2+ 2L(u(t− 1)[t− 1]) =

2s− 1

s2+ 2e−sL(t) =

2s− 1

s2+

2s2

e−s.

Wasn’t that nicer than how we did it before (example (iv) on page 2)?

77

Consider a piecewise continuous function f(t) =

f0(t) if 0 ≤ t < t1

f1(t) if t1 ≤ t < t2

f2(t) if t ≥ t2

.

We can rewrite f(t) in a convenient form using the unit step function appropriately:

f(t) = f0(t) + u(t− t1)[f1(t)− f0(t)] + u(t− t2)[f2(t)− f1(t)].

Then we find the Laplace transform of f(t), using the above and L(u(t− r)g(t)) = e−srL(g(t+ r)).

Of course this generalizes to any number of “pieces,” but we will focus on functions that are either2 or 3 “pieces.”

Since L(u(t− r)g(t)) = e−srL(g(t + r)) we get

L(u(t− r)g(t− r)) = e−srL(g(t)),

which is called the second shifting property.

Use linearity and the second shifting property to find L−1

(1 + e−s

s2

).

First we use linearity and then apply the second shifting property:

L−1

(1 + e−s

s2

)= L−1

(1s2

)+ L−1

(e−s 1

s2

)= t + L−1 (e−sL(t)) = t + u(t− 1)[t− 1].

Of course we can rewrite this in the more familiar form L−1

(1 + e−s

s2

)=

t if 0 ≤ t < 1

2t− 1 if t ≥ 1.



78

Lesson 23

Constant Coefficient Equations with Piecewise Continuous Forcing Functions

Section 8.5

Let’s solve the IVP y′′ + y =

t if 0 ≤ t < 2π

−2t if t ≥ 2π, y(0) = 1, y′(0) = 2.

Let’s first find L(f) where f(t) =

t if 0 ≤ t < 2π

−2t if t ≥ 2π.

We can rewrite this function as f(t) = t− u(t− 2π)[3t].

The L(f) = L(t)− 3L (u(t− 2π)[t]) =1s2− 3e−2πsL(t + 2π) =

1s2− 3e−2πs

(1s2

+2π

s

).

Recall: Y (s) =L(f(t)) + a(k1 + k0s) + bk0

as2 + bs + c.

So

Y (s) =

1s2− 3e−2πs

(1s2

+2π

s

)+ (2 + s)

s2 + 1=

1 + (s + 2)(s2)s2(s2 + 1)

− 3e−2πs

(1 + 2πs

s2(s2 + 1)

).

So Y (s) = Y1(s)− 3Y2(s) where Y1(s) =s3 + 2s2 + 1s2(s2 + 1)

and Y2(s) = e−2πs

(1 + 2πs

s2(s2 + 1)

).

Let’s find L−1(Y1(s)) = L−1

(s3 + 2s2 + 1s2(s2 + 1)

).

We can begin a P.F.D. of Y1(s) =s3 + 2s2 + 1s2(s2 + 1)

=A

s+

1s2

+Bs + C

s2 + 1.

Clearing fractions yields s3+2s2+1 = As(s2+1)+s2+1+(Bs+C)s2 = As3+As+s2+1+Bs3+Cs2.

Thus A = 0 and B = C = 1. So Y1(s) =1s2

+s

s2 + 1+

1s2 + 1

. Hence L−1(Y1(s)) = t+cos (t)+sin (t).

79

Now let’s find L−1(Y2(s)) = L−1

(e−2πs

(1 + 2πs

s2(s2 + 1)

)). We actually begin with

L−1

(1 + 2πs

s2(s2 + 1)

).

We can begin with the P.F.D.1 + 2πs

s2(s2 + 1)=

A

s+

1s2

+Bs + C

s2 + 1.

Clearing fractions yields 1+2πs = As(s2 +1)+s2 +1+(Bs+C)s2 = As3 +As+s2 +1+Bs3 +Cs2.

Thus A = 2π, B = −2π, and C = −1. So1 + 2πs

s2(s2 + 1)=

2π

s+

1s2− 2πs + 1

s2 + 1.

Then L−1

(1 + 2πs

s2(s2 + 1)

)= 2π + t− 2π cos (t)− sin (t).

So by the second shifting property,

L−1

(e−2πs 1 + 2πs

s2(s2 + 1)

)= L−1

(e−2πsL (2π + t− 2π cos (t)− sin (t))

)

= u(t− 2π)[t− 2π cos (t− 2π)− sin (t− 2π)].

Therefore the solution is y(t) = L−1(Y (s)) = t+cos (t)+ sin (t)− 3u(t− 2π)[t− 2π cos (t)− sin (t)].

In a more familiar form,

y(t) =

t + cos (t) + sin (t) if 0 ≤ t < 2π

−2t + (1 + 6π) cos (t) + 4 sin (t) if t ≥ 2π.


Read Section 8.5, Pages 431 - 437Section 8.5 - { 4, 10, 18 }

80

Lesson 24

Convolution

Section 8.6

Let f(t), g(t) be functions. The convolution of f and g is the function (f∗g)(t) =∫ t

0f(r)g(t−r) dr.

Let’s compute the convolution of f(t) = et and g(t) = e−t:

(f ∗ g)(t) =∫ t

0ere−t+r dr =

∫ t

0e−te2r dr = e−t 1

2e2r

∣∣∣∣t

0

=e−t

2(e2t − 1

)=

et − e−t

2= sinh (t).

The following theorem is called convolution theorem (the proof may be covered in class):

Proposition: If f(t), g(t) are functions such that L(f) = F and L(g) = G then L(f ∗ g) = FG.

Let’s find a formula for the solution of the IVP y′′ + 4y′ − 5y = f(t), y(0) = k0, y′(0) = k1.

By taking the Laplace transform of both sides we get (s2 + 4s− 5)Y (s) = F (s) + (k1 + k0s) + 4k0,where Y (s) = L(y(t)) and F (s) = L(f(t)).

Then Y (s) = F (s)G(s) +k1 + 4k0 + k0s

(s + 5)(s− 1)= F (s)G(s) +

k1 + 5k0

6(s− 1)− k1 − k0

6(s + 5), where

G(s) = (s2 + 4s− 5)−1.

Then by taking the inverse Laplace transform we get

y(t) =k1 + 5k0

6et +

k0 − k1

6e−5t +

∫ t

0f(r)g(t− r) dr, where L(g) = G.

In particular, g(t) = L−1

(1

s2 + 4s− 5

)= L−1

(1/6

s− 1− 1/6

s + 5

)=

16

(et − e−5t

).

So y(t) =k1 + 5k0

6et +

k0 − k1

6e−5t +

∫ t

0

f(r)6

(et−r − e−5t+5r

)dr

Let’s apply this formula to f(t) = 6et, k0 = 1, k1 = 0:

y(t) =56et +

16e−5t +

∫ t

0

6er

6(et−r − e−5t+5r

)dr =

56et +

16e−5t +

∫ t

0et − e−5t+6r dr =

=56et +

16e−5t +

(ret − 1

6e−5t+6r

∣∣∣∣r=t

r=0

)=

56et +

16e−5t + tet − 1

6et +

16e−5t =

23et + tet +

13e−5t.

Compare to the solution we got earlier (page 7).81

Let’s find a formula for the solution of the IVP y′′ + 2y′ + 2y = f(t), y(0) = k0, y′(0) = k1.

Taking the Laplace transform of both sides yields ((s + 1)2 + 1)Y (s) = F (s) + (k1 + sk0) + 2k0,where Y (s) = L(y(t)) and F (s) = L(f(t)).

Thus Y (s) = F (s)G(s) +k0(s + 1)

(s + 1)2 + 1+

k1 + k0

(s + 1)2 + 1, where G(s) = [(s + 1)2 + 1]−1.

Then by taking the inverse Laplace transform (using the table of Laplace transforms) we get

y(t) = k0e−t cos (t) + (k0 + k1)e−t sin (t) +

∫ t

0f(r)g(t− r) dr, where L(g) = G.

In particular, g(t) = L−1

(1

(s + 1)2 + 1

)= e−t sin (t).

So y(t) = k0e−t cos (t) + (k0 + k1)e−t sin (t) +

∫ t

0f(r)e−t+r sin (t− r) dr.

One can also use this theory to evaluate convolution integrals. For example, let’s evaluate∫ t

0e−r sin (t− r) dr.

This is the convolution of f(t) = e−t and g(t) = sin (t). Then F (s) =1

s + 1and G(s) =

1s2 + 1

are the Laplace transforms respectively. All we need to do is find the inverse Laplace transform of

F (s)G(s) =1

(s + 1)(s2 + 1).

We begin with the P.F.D.1

(s + 1)(s2 + 1)=

1/2s + 1

+As + B

s2 + 1. From this we get A = −1/2 and

B = 1/2.

So1

(s + 1)(s2 + 1)=

1/2s + 1

− (1/2)s− 1s2 + 1

=12

[1

s + 1+

1s2 + 1

− s

s2 + 1

]and the inverse Laplace

transform, and therefore the integral, is

12

[e−t + sin (t)− cos (t)

].


Read Section 8.6, Pages 441 - 449Section 8.6 - { 2, 3, 5 }

82

Lesson 25

Constant Coefficient Equations with Impulses

Section 8.7

The rigorous definition of the dirac delta function may be covered in class, but here we will justuse the heuristic definition:

δ(t) =

∞ if t = 0

0 if t 6= 0, such that

∫

Iδ(t) dt = 1 on any interval I containing 0.

For our purposes, it suffices to use the following result:

If t0 > 0 then the solution to the IVP ay′′ + by′ + cy = δ(t− t0), y(0) = 0, y′(0) = 0 is

y = u(t− t0)w(t− t0),

where w = L−1

(1

as2 + bs + c

).

The solution the IVP ay′′ + by′ + cy = f(t) + αδ(t− t0), y(0) = k0, y′(0) = k1 is

y = y + αu(t− t0)w(t− t0),

where w = L−1

(1

as2 + bs + c

)and y is the solution to ay′′+by′+cy = f(t), y(0) = k0, y′(0) = k1.

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Example: Let’s solve y′′ − 7y′ + 6y = 2e−7t + 2δ(t− 1), y(0) = 1, y′(0) = −1.

(ELFY) Use Laplace transforms to show that

y = −2465

e6t +2720

et +152

e−7t,

is a solution to y′′ − 7y′ + 6y = 2e−7t, y(0) = 1, y′(0) = −1.

Next we find that w = L−1

(1

(s− 6)(s− 1)

)=

15

(e6t − et

).

Thus the solution is y = y+2u(t−1) [w(t− 1)] = −2465

e6t+2720

et+152

e−7t+2u(t−1)[15

(e6(t−1) − et−1

)].

In a more familiar form,

y(t) =

−2465

e6t +2720

et +152

e−7t if 0 ≤ t < 1

(−24

65+

25e6

)e6t +

(2720− 2

5e

)et +

152

e−7t if t ≥ 1

.

Example: Let’s solve y′′ + y = 1 + 2δ(t− π) + 3δ(t− 2π), y(0) = 1, y′(0) = −2.

(ELFY) Use Laplace transforms to show that

y = 1− 2 sin (t),

is a solution to y′′ + y = 1, y(0) = 1, y′(0) = −2.

Next we find that w = L−1

(1

s2 + 1

)= sin (t).

Thus the solution is

y = y+2u(t−π) [w(t− π)]+3u(t−2π) [w(t− 2π)] = 1−2 sin (t)+2u(t−π) [sin (t− π)]+3u(t−2π) [sin (t− 2π)] .

Using the basic identities sin (t− π) = − sin (t) and sin (t− 2π) = sin (t), this becomes

y = 1− 2 sin (t)− 2u(t− π) [sin (t)] + 3u(t− 2π) [sin (t)] .

Or in a more familiar form,

y(t) =

1− 2 sin (t) if 0 ≤ t < π

1− 4 sin (t) if π ≤ t < 2π

1− sin (t) if t ≥ 2π

.

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Here is the graph of the solution:

2 4 6 8

-1

1

2

3

4

5



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Lesson 26

Euler’s Method and Runge-Kutta Method


Consider a first order differential equation dydx = f(x, y) . We have learned many methods for find-

ing analytic solutions. However there is no one method that will always work and there are manypractical differential equation for which there is no known analytic solution. When thus occurswe can resort to a numerical approximation. There are numerous numerical methods that can beapplied to differential equations. Indeed numerical differential equations is a topic of study itself.We will present just a bare bones introduction to this topic. One aspect of this field that we will notspend enough time on is the idea that an approximation is only as good as the maximum possibleerror of the method. There are two main sources of errors, the method itself and roundoff errorsfrom the computing machine being used.

The study of roundoff errors is way beyond the scope of this course. We can tell you that roundofferrors are cumulative. One way to reduce roundoff errors is to reduce the number of calculationsbeing done.

The first method we will look at in this course is Euler’s method. Euler’s method is too crudeto be of much practical use. However it is easy to understand. It is a simple tangent line approxi-mation. Euler’s method is equivalent to the left hand rule for numerical integration. For straightlines Euler’s method will give an exact answer. This tells that the error from the method is withiny′′ and the step size. The formulas and examples for Euler’s method are well presented in thetextbook in section 3.1 and will not be repeated here.

A second method that will be discussed in this course is called the Runge Kutta Method. Itis a weighted average of values of f(x, y) at different values of x and y. The Runge Kutta Methodis accurate enough that it is often used in practice. Although compared to most modern methodsit is still relatively crude. The Runge Kutta Mehtod is equivilant to Simpon’s Rule for numericalintegration. Runge Kutta will give an exact answer if the result is a cubic. This tells us that theerror from the method is within y(4) and the step size. Again the formulas and examples for theRunge Kutta Method are given in section 3.3.

Section 3.2, which we will skip, presents other forms of Euler’s Method and a method called theImproved Euler. The Improved Euler is equivalent to the Trapezoid rule for numerical integration.As such it gives little advantage over Euler’s Method for it’s extra complexity.

But to reiterate, a numerical method is only as good as the maximum known error. The methodspresented here have well known and proven error analysis methods. Way too often these errors areignored by beginners, often to great cost and danger.


Read Section 3.1, Pages 88 100 and Section 3.3, Pages 113 118Section 3.1 { 1, 2, 6, 7 } andSection 3.3 { 1, 2, 6, 7, compare your answers with those found using Euler’s Method }

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