ordinary di erential equations lecture notes for math 133a

Ordinary Di↵erential Equations

Lecture notes for Math 133A

Slobodan N. Simic

c� Slobodan N. Simic 2016

Contents

Chapter 1. First order di↵erential equations 11.1. What is a di↵erential equation? 11.2. Basic examples and modeling 21.3. The notion of a solution 111.4. Separable equations 131.5. Linear equations 181.6. Existence and uniqueness of solutions 251.7. The phase line and classification of equilibria 321.8. Linearization 351.9. Numerical methods 38

Chapter 2. First order systems 452.1. Examples of systems of di↵erential equations 452.2. Vector fields and solutions curves 492.3. The phase plane, equilibria and periodic solutions 552.4. Existence and uniqueness of solutions 58

Chapter 3. Linear systems 613.1. Definition and examples 613.2. Linear algebra preliminaries 633.3. Properties of linear systems 733.4. Phase planes for planar linear systems 773.5. The trace-determinant plane 1033.6. Second order linear equations 1063.7. Forced harmonic oscillators 112

Chapter 4. Nonlinear systems 1214.1. Equilibria and periodic solutions 1214.2. Linearization 1284.3. Hamiltonian systems 1314.4. Gradient systems 131

Chapter 5. Laplace transform 1335.1. Why another method? 133

vii

viii CONTENTS

5.2. Definition and basic properties 1345.3. Discontinuous forcing 1405.4. Impulse forcing 1485.5. Convolution and the Laplace transform 150

Preface

This is a set of lecture notes for Math 133A: Ordinary Differential Equations taughtby the author at San Jose State University in the Fall 2014 and 2015. The only prerequisite forthe course is multivariable calculus. The notes focus on qualitative analysis of di↵erential equationsin dimensions one and two. Since the class is mainly intended for SJSU engineering majors, theLaplace transform is also covered in some detail.

Sections 4.3 and 4.4 (on Hamiltonian and gradient systems) remain to be added.Corrections, comments, and suggestions would be greatly appreciated and should be emailed to

the author at [email protected] authors thanks the Textbook Alternatives Project at SJSU for their support.

Slobodan N. SimicDepartment of Mathematics and Statistics

San Jose State University

ix

CHAPTER 1

First order di↵erential equations

Newton’s fundamental discovery, the one which he considerednecessary to keep secret and published only in the form of ananagram, consists of the following: Data aequatione quotcunque

fluentes quantitae involvente fluxiones invenire et vice versa. Incontemporary mathematical language this means: “It is useful tosolve di↵erential equations”.

Vladimir Arnold

1.1. What is a di↵erential equation?

Di↵erential equations are arguably the most important types of equations in mathematics,the natural sciences and engineering. They are one of the basic tools of mathematics. Modelingnatural phenomena leads straight to di↵erential equations. The first person to systematically studydi↵erential equations was Isaac Newton. He considered his discovery that di↵erential equations areuseful for modeling physical phenomena one of his greatest, so he kept it secret and published itonly in the form of an anagram (see the quote above).

But what is a di↵erential equation? Here’s an example:

(1.1)dy

dt= y.

Here’s another:

(1.2)@2u

@x2+

@2u

@y2= 0.

And here’s an equation that’s not a di↵erential equation:

(1.3) x2 + 3x + 2 = 0.

In each example there’s an unknown we are supposed to solve for, but unlike the last equation (whichis just a simple quadratic equation), where the unknown is a number, in the first two equations theunknown is a function. In (1.1) the unknown is the function y which depends on a single real variablet; in (1.2), the unknown is a function u which depends on two real variables, x and y. The firsttwo equations involve derivatives of the unknown function, which is why we call them di↵erential

1

2 1. FIRST ORDER DIFFERENTIAL EQUATIONS

equations. Equation (1.1) contains an “ordinary” derivativedy

dtof the unknown function y, so we

call it an ordinary di↵erential equation. This equation arises in modeling exponential growth (e.g.,population dynamics). Equation (1.2) contains partial derivatives of the unknown function u, so wecall it a partial di↵erential equation. This equation (called Laplace’s equation) arises in many areas,such as astronomy, electromagnetism, and fluid dynamics.

In these lecture notes we will only be concerned with ordinary di↵erential equations.An ordinary di↵erential equation is an equation in which the unknown is a function of one inde-

pendent variable and the equation involves derivatives of that function. The order of an equation isthe order of the highest derivative appearing in it.

We will always denote the independent variable by t and think of it as time. Usually the unknownfunction of t will be denoted by the letter y. Thus a first-order ordinary di↵erential equation isalways of the form

F (t, y, y0) = 0,

where F is some function of three variables and y0 =dy

dt. It is often possible to solve for y0 and

rewrite the equation in the more useful form

(1.4) y0 = f(t, y).

For instance, in (1.1), f(t, y) = y. If the right-hand side in (1.4) does not depend on t, we call theequation autonomous. Otherwise, it is called non-autonomous. Thus (1.1) is autonomous, whereasy0 = t + y is not.

Similarly, most second-order ordinary di↵erential equations can be written in the form

y00 = g(t, y, y0),

where g is some function of three variables and y00 =d2y

dt2.

In Chapters 2, 3 and 4, we will be studying systems of di↵erential equations, where there is notjust one, but several unknown functions, all depending on the same independent variable t. Wepostpone the discussion of systems to Chapter 2.

Since the unknown in a di↵erential equation is a function, solving a di↵erential equation means(at least in principle) finding infinitely many unknown numbers. This is in general much harderthen solving for just one or two numbers (like in numerical equations such as (1.3)). As a rule ofthumb, di↵erential equations are harder to solve than numerical equations, and among di↵erentialequations, partial di↵erential equations are harder to solve than ordinary di↵erential equations.In fact, it turns out that most di↵erential equations cannot be solved explicitly (i.e., there’s noequivalent of the quadratic formula). But more about this later.

Notation. If y is a di↵erentiable function of an independent variable t, we will denote itsderivative either by y0 (when it’s clear with respect to what variable the derivative is taken) or bydy

dt. Sometimes (especially if t denotes time) we will use Newton’s notation y. So remember:

y0 = y =dy

dt.

1.2. Basic examples and modeling

In this section we will introduce three di↵erential equations which arise in the modeling ofpopulation growth and harmonic oscillators. Before we do that, let us first discuss what we meanby modeling.

1.2. BASIC EXAMPLES AND MODELING 3

Essentially, all of science is a search for models for how the Universe works. But to quote GeorgeE. P. Box, a British-American statistician:

Essentially, all models are wrong, but some are useful.

That’s of course because the Universe (or almost any subset of it) is too complicated to bemodeled by our puny mathematical tools, since a model is just a mathematical representation ofthe real world (whatever that is). We can only hope to partially represent some aspects of the realphenomenon we wish to understand. So to obtain a reasonable model we need to make simplifyingassumptions. That allows us to identify a finite (and ideally small) number of relevant variablesand parameters, and then write down a set (also ideally small) of equations comprising our model.Once we have it, we can temporarily (but only temporarily) forget the “real thing” and analyze themodel using any tools and methods that mathematics can o↵er. And that means calculus, geometry,topology, etc. After we have understood our mathematical model, we can make predictions aboutits future behavior and compare those predictions with real data (which we can presumably obtainby observation and measurement). This can give us some idea about the validity and limitations ofour model, which we can use to improve it, say by including more relevant variables or parameters,choosing a better set of equations, etc.

We usually start by building a model that is as simple as possible yet reflecting the basic featuresof the real phenomenon we are trying to represent. If the model is not good enough, we make itmore complex. The goal is to have a good balance between simplicity (which often translates intotractability) and faithfulness in representing the real world.

Let’s see how this works in the following examples.

1.2.1. Unlimited population growth. Assume we have the population of some species (sayfruit flies), which has the following remarkable properties:

• Its members never die.• The population has access to an unlimited supply of food.• The species spends all of its time reproducing itself.

How do we go about modeling the change of this population with time? First, let us denotethe number of individual members of the population by P and time by t. So P is a function oft. Of course, in reality P is a positive integer, but since we are hoping to study its rate of changewith respect to t, we choose to treat it as a continuous variable, i.e., a real number. In fact, we will

assume that P is a di↵erentiable function of t; in other words, we assume that P 0 =dP

dtexists for

all values of t.What do our standing assumptions tell us about the rate of change P 0 of P with respect to t? It

is clear that the larger the size P of the population, the faster it reproduces, i.e., the larger its rateof change P 0. Thus it makes sense to assume that P 0 is proportional to P , which means that thereexists a positive parameter k1 such that P 0 = kP . Thus the simplest di↵erential equation modelingthe growth of the population satisfying the above properties is

(1.5)dP

dt= kP.

This is the simplest non-trivial example of a di↵erential equation (in which the right-hand sideis not just a function of the independent variable). We will study it in two ways: qualitatively andanalytically (i.e., by solving it).

1This parameter is related to the food supply, but we will not elaborate on that connection here.

Qualitative approach: First of all, what is qualitative analysis of a di↵erential equation?The short answer is: it is a type of analysis in which we do not explicitly solve the equation(because we usually can’t) but instead try to understand what happens to solutions ast ! +1 and t ! �1.

Let P (t) be a solution to (1.5) and set P0 = P (0). This is the initial value of thepopulation when t = 0. Let’s see what we can say about P (t) for large t for various valuesof P0. Assume first that P0 = 0. If there is no population to speak of at the beginning,there will obviously be no population at any time in the future or past, so P (t) = 0, forall values of t. The solution is constant; we will call such solutions equilibrium solutions orequilibria.2 Finding and understanding such solutions is an important part of qualitativeanalysis.

If P0 > 0, then P 0(0) > 0, so by calculus P must be an increasing function near t = 0.Therefore P (t) is positive at least for small values of t > 0. But for any such value of t,P (t) > 0 implies that P 0(t) is also positive, since P 0(t) = kP (t) and k > 0. Therefore, P (t)keeps increasing as t gets bigger and it is not hard to see that it never stops increasing.That is, if P0 > 0, then P (t) ! 1, as t ! 1, which is not surprising.

Forgetting that P denotes population for a moment, let’s see what happens if P0 < 0.We leave it as an exercise to the reader to show, in exactly the same way as in the previousparagraph, that P (t) ! �1, as t ! 1.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2.4

-1.6

-0.8

0.8

1.6

2.4

Figure 1.1. Three types of solutions of equation (1.5).

Therefore, qualitatively speaking, there are only three types of solutions: the equilib-rium solution (corresponding to the situation when P0 = 0), those solutions that tend to+1 and those that tend to �1 (both as t ! 1).

Analytic approach: Now let us solve (1.5). It is easy to see that one solution is P (t) ⌘ 03:just plug it into both sides of the equation and check that it “works”. Assume P 6= 0.Divide both sides of the equation by P and, pretending that dP and dt have a meaning

2Equilibria is the plural of equilibrium.3The symbol ⌘ means that the left-hand side equals the right-hand side for all values of the independent variable.


independent of each other, “multiply” both sides by dt. We obtain

dP

P= k dt.

Now integrate both sides:Z

dP

P=

Z

k dt.

Observe that P is now a “dummy variable” in the indefinite integral on the left and t playsthe same role in the integral on the right. Solve these integrals. (If you don’t rememberhow to do that, you need to review basic integration techniques.) This gives us

ln |P | = kt + C,

where C is a constant and ln denotes the natural logarithm. Exponentiating both sides weget

|P | = ekt+C = ekteC = Bekt,

where B = eC , just another constant (for now). Since |P | = ±P , taking A = ±B (justanother arbitrary constant), we obtain

P = Aekt.

So every non-zero solution appears to be a constant multiple of the exponential functionekt! How does the constant A relate to the initial value P0 of P? Let’s see: set t = 0 inthe previous equation and get

P0 = Aek·0 = A.

So A equals the initial value P0 of P (t)! Therefore, given P0 = P (0), our solution must be

(1.6) P (t) = P0ekt,

for all values of t. Note that we obtained this formula assuming that P0 6= 0, but (1.6)does work even if P0 = 0.

The method we used for finding the solution is called separation of variables. We’ll learnmore about it in Section 1.4.

Observe that the explicit solution given by (1.6) is consistent with the conclusions ofthe qualitative analysis: even though there are infinitely many solutions (one for eachinitial value of P (t)), they fall into one of the three categories discussed above. ⇤

1.2.2. Logistic population model. It’s pretty clear that the assumptions we made in theprevious model are extremely unrealistic. Indeed, if we compared the predictions of this model toreal data (say, the US census) we would see that they agree only for small values of t. When tbecomes large, the predictions have nothing to do with reality whatsoever.

So how can we improve our model and keep it not too complicated? Recall that the right-handside of equation (1.5) is a very simple function of P : it is linear in P , which is why (1.5) is anexample of a linear di↵erential equation (with constant coe�cients); more on those in Section 1.5.So one way to improve our model would be to use a slightly more sophisticated, non-linear functionon the right-hand side of the equation. Of course, we don’t want to make it too complicated either,so the most natural choice would be a quadratic function of P , i.e., something that looks likethis: f(P ) = AP 2 + BP + C, for some suitably chosen parameters A, B and C. But how do wechoose these parameters? One way is to make better assumptions. For starters, no more unlimited

quantities of food. How do we incorporate this assumption in the new model? Here’s one way todo that.

(1) For small P , we still want P 0 to be nearly proportional to P , i.e., when the population issmall we would like P 0 to be approximately equal to kP , where k is a parameter as in theprevious model.

(2) However, if P gets too large, we will assume that P 0 becomes negative, that is, the popula-tion decreases. How large is too large? We assume that there is a sort of “ideal population”N – also called the carrying capacity – such that if P < N , then P 0 > 0 and if P > N ,then P 0 < 0.

What is the simplest quadratic function f(P ) which incorporates these assumptions? Assump-tion (1) suggests that we should try f(P ) = kP · (something), where “something” is close to 1 whenP is small. But we need to make sure that assumption (2) is also satisfied. After thinking aboutthis for a while, we are inescapably lead to the following solution:

something = 1 � P

N.

Thus our new model for population growth becomes

(1.7)dP

dt= f(P ), where f(P ) = kP

✓

1 � P

N

◆

.

The graph of the function f (with k = 3 and N = 4) is given in Figure 1.4.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 1.2. The graph of f with k = 3 and N = 4.

It is not hard to see that assumptions (1) and (2) are satisfied. Equation (1.7) is called alogistic equation. It is an example of a non-linear di↵erential equation. Even though we could againseparate the variables and find an explicit formula for all solutions, we will not do that here (seeSection 1.4 for that approach). Instead, we employ only the qualitative approach to show its powerand simplicity.

As in the previous example, let us first find the equilibrium solutions. Recall that a solutionP (t) is an equilibrium solution if it is constant, i.e., if P 0(t) = 0 for all t. But if P 0 = 0, then

f(P ) = 0, so we can find equilibrium solutions by solving the algebraic equation f(P ) = 0, whichin our case is

kP

✓

1 � P

N

◆

= 0.

And that’s a pretty easy task: the only solutions are P = 0 and P = N . Therefore, there are onlytwo equilibrium solutions: the trivial one (when the population is always zero) and the “ideal one”when the population equals the carrying capacity: P⇤(t) = N , for all t.

What if the initial population P0 is di↵erent from 0 and N? Here’s how to approach thisquestion. Suppose that at some time t the population is P . If 0 0, soP 0 = f(P ) > 0, which means that P is increasing. If P > N , then P 0 = f(P ) < 0, which meansthat P is decreasing, exactly as we wanted. Thus solutions whose value is in the interval (0, N) areincreasing and those in the interval (N, 1) are decreasing. Here’s a crucial question: if 0 N , can it decrease below N? Let’s see.

Assume we have a solution P (t) such that 0 N . Solutions are continuous (in fact, di↵erentiable!) functions, so there must be someintermediate time t0 such that P (t0) = N . (What theorem of calculus guarantees this?) Therefore,when t = t0, our solution and the equilibrium solution P⇤(t) ⌘ N cross! This would mean that aftertime t = t0 our population could “choose” to follow the solution P or it could “decide” to stay at theequilibrium N . Population growth would not be a deterministic process! This is clearly nonsense,so solutions cannot cross. (This is related to uniqueness of solutions, which will be discussed inSection 1.6.) Thus if a solution P satisfies 0 N for some t, the same holds for all t.

x ’ = 1 y ’ = 3 y (1 − y/4)

−0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

0

1

2

3

4

5

6

x

y

Figure 1.3. Graphs of several solutions of the logistic equation (1.7) with k = 3and N = 4.

We draw the following conclusions (see Figure 1.3):

• There are two equilibrium solutions, P = 0 and P = N .• If 0 N , the solution decreases and converges to N , as t ! 1. (Where does it converge

as t ! �1?)• Finally, if P < 0 (ignoring the physical impossibility of this), then we similarly obtain that

the solution must decrease to �1, as t ! 1. ⇤

1.1. Example. Let us try to use the ideas utilized in studying the logistic population model tounderstand the following di↵erential equation:

(1.8) y0 = y2 � 10y + 21.

As before, y0 stands for dy/dt, the derivative of the unknown function y with respect to time t.Observe that this is not a logistic equation, but that the right-hand side f(y) = y2 � 10y + 21 isagain a quadratic function of y.

First of all, what are the equilibrium solutions? To answer this question, we factor the right-handside of the equation:

f(y) = (y � 3)(y � 7),

then set f(y) = 0. The solutions to this algebraic equation are y1 = 3 and y2 = 7. These solutionsgive us the equilibrium solutions of the original di↵erential equation, namely, y1(t) = 3 and y2(t) = 7(for all t). Let us verify that y1(t) and y2(t) are indeed solutions to the equation (1.8). Substitutingy1(t) = 3 into both sides of (1.8), we obtain

d

dt3 = 32 � 10 · 3 + 21,

which is clearly satisfied. Therefore, y1(t) is indeed a solution and since it is constant, it is anequilibrium solution. We can check that y2(t) is also an equilibrium solution in a completelyanalogous way. It is clear that there are no other equilibrium solutions.

y

x´ = 1

y´ = y²-10*y+21

-1

0

1

2

3

4

5

6

7

8

9

10

x

-1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4

Figure 1.4. Graphs of several solutions of the equation (1.8).

Now let us find all the values of y for which y(t) is increasing. By this we really mean to findall values y0 such that if y(t) is the solution satisfying y(0) = y0, then y(t) is increasing. Since y(t)is increasing if y0(t) > 0 and y0(t) = f(y(t)), all we need to do is find the values of y for whichf(y) > 0. Since f(y) = (y � 3)(y � 7), we immediately see that f(y) > 0 only if y > 7 or y < 3. Soall solutions y(t) in the interval (7, 1) are increasing as are all solutions in (�1, 3).4

Similarly, all solutions in the interval (3, 7) are decreasing. Graphs of several solutions aredepicted in Fig. 1.4. ⇤

4It can be shown that if y(t) starts in one of these intervals, it will remain in the same interval for all time.

1.2.3. The harmonic oscillator. There are many harmonic oscillators in nature. We willdescribe the so called mass-spring oscillator consisting of a mass m attached by a spring to a wall.If the mass is displaced from its natural equilibrium position, it will oscillate. Our job is to find amathematical model for this physical system and explore its properties. We will do the former inthis section and the latter in Chapter 3.

Denote by y(t) the displacement of the mass m from its equilibrium position at time t. We canchoose the coordinates on the real line so that the equilibrium is at the origin and assume that thedistance increases towards the right, assuming the wall is on the left as in Figure 1.5.

Figure 1.5. A mass-spring oscillator.

The motion of the oscillator is governed by Newton’s Second Law:

The net force on an object equals its mass times its acceleration.5

In mathematical language,

(1.9) F = ma.

We know that a is the second derivative of the position:

a =d2y

dt2= y00

Let’s figure out what the net force F is. To make the problem tractable, we need to make somesimplifying assumptions such as these: the only forces that matter are friction (or damping) F

f

,the restoring force of the spring, F

s

, and an externally applied force Fe

. Thus we assume

F = Ff

+ Fs

+ Fe

.

By Hooke’s Law, the force needed to extend a spring by a certain distance is proportional to thatdistance. Therefore,

Fs

= �ky,

where k > 0 is a constant called the sti↵ness of the spring. Note the negative sign; it’s there becausethe spring is trying to return the mass to its equilibrium position (which is at y = 0), so when y > 0,Fs

< 0 and when y < 0, then Fs

> 0.To characterize the frictional force F

f

, we will use its simplest approximation and assume thatit is proportional to the velocity of the object. In other words,

Ff

= �by0,

where b > 0 is the damping constant (or the coe�cient of friction).The external force F

e

cannot be specified in advance so we are only going to assume that itdepends on time, i.e., F

e

= f(t), where f is some function. (f(t) is often periodic though it canbe any kind of function, including a discontinuous one. We will learn how to deal with both kindslater.)

5The correct statement is: the net force F equals the rate of change of the momentum, p, but for small speeds,mass can be assumed to be constant, so p

0 = (mv)0 = mv

0 = ma, where v is the velocity and a is the acceleration.


Substituting back in (1.9), we obtain

�ky � by0 + f(t) = my00,

which is equivalent to

my00 + by0 + ky = f(t).

This is the mass-spring oscillator equation. Observe that it’s a second order equation; we willstudy second order linear equations in Chapter 3. ⇤

Exercises

1. Consider the population model

dP

dt= 0.2P

✓

1 � P

135

◆

,

where P (t) is the population at time t.(a) For what values of P is the population in equilibrium?(b) For what values of P is the population increasing?(c) What is the carrying capacity?(d) For which initial values of P does the population converge to the carrying capacity as

t ! 1?

2. Consider the di↵erential equation

y0 = y3 + 3y2 � 10y.

(a) What are the equilibrium solutions?(b) For which values of y0 is the solution y(t) starting at y0 increasing? (y(t) starts at y0

if y(0) = y0.)(c) For which values of y0 is the solution y(t) starting at y0 decreasing?

3. In this exercise your task is to model radioactive decay. Use the following notation:t = time (independent variable);I(t) = amount of the radioactive isotope at time t (dependent variable);�� = decay rate, where � > 0 (parameter).

Write the simplest equation modeling the decay of a radioactive isotope with decayrate ��. State the corresponding initial-value problem if the initial amount of the isotopeis I0.

4. The half-life of a radioactive isotope is the amount of time it takes for a quantity of thatisotope to decay to one half of its original value.(a) Express the half-life of a radioactive isotope in terms of its decay rate.(b) The half-life of radiocarbon or Carbon 14 (C-14) is 5230 years. Determine its decay

rate parameter �.(c) Carbon dating is a method of determining the age of an object using the properties

of radiocarbon. It was pioneered by Willard Libby and collaborators in 1949 to datearchaeological, geological, and other samples. Its main idea is that by measuring theamount of radiocarbon still found in the organic mater and comparing it to the amountnormally found in living matter, we can approximate the amount of time since death

1.3. THE NOTION OF A SOLUTION 11

occurred.6 Using the decay-rate parameter found in part (b), find the time since deathif 35% of radiocarbon is still in the sample.

1.3. The notion of a solution

So far we have been a little imprecise and have freely talked about solutions to various di↵erentialequations without ever defining what we meant by it. So suppose we have a first-order di↵erentialequation

(1.10) y0 = f(t, y).

What do we mean when we say that y(t) is a solution? The answer quite natural: y(t) is a solutionto (1.10) if when plugged into (1.10), y(t) satisfies it, i.e., the left-hand side becomes identicallyand trivially equal to the right-hand side:

y0(t) = f(t, y(t)),

for all values of t for which both sides make sense. Here are some examples.

1.2. Example. Consider y0 = 2y. Then y(t) = e2t is a solution as one can easily check. So arethe functions �e2t and 2014e2t, but not t2, e�t or sin t. (Note that this is just an equation modelingunlimited population growth as in Section 1.2.1.) ⇤

1.3. Example. The function y(t) = tan t is a solution to the equation y0 = 1+y2. This equationhas no equilibrium solutions. (Check this!) ⇤

1.4. Example. Define a function z(t) by

z(t) =

(

0 if t 0

t3 it t > 0.

Verify that z satisfies the di↵erential equation z0 = 3z2/3. ⇤

We have already encountered the notion of an equilibrium solution but because of its importancewe repeat it definition:

A solution is called an equilibrium solution if it is constant. A point y0 is called anequilibrium point of a di↵erential equation if y(t) ⌘ y0 is a solution.

How do we look for equilibrium solutions? Suppose that y(t) = y0 is an equilibrium solution ofthe di↵erential equation y0 = f(t, y). Plugging y(t) into the equation we obtain

f(t, y0) = f(t, y(t)) = y0(t) =d

dty0 = 0,

for all t. Conversely, suppose we found y0 such that f(t, y0) = 0, for all values of t. Then the abovecalculation also shows that y(t) = y0 is an equilibrium solution of the equation y0 = f(t, y). Weconclude:

y0 is an equilibrium point of the equation y0 = f(t, y) if and only if f(t, y0) = 0, forall values of t.

6This is based on the assumption that radiocarbon makes a constant proportion of the carbon ingested by livingmatter and that once the matter dies no new carbon is added to it.


In particular, if the equation is autonomous, y0 = f(y), then finding equilibrium points amountsto solving the equation f(y) = y.

Observe that checking whether a given function is a solution is easy, but finding solutions isusually not. For instance, does the equation

y0 = 1 + y4 sin2 y + 17earctan y

3

have a solution satisfying the condition y(0) = 0? The answer is yes, but there’s no way we can tellusing the tools we have available at this point.

This brings us to another notion of a solution. Often it is important to find a solution to adi↵erential equation which satisfies an initial condition of the type y(0) = y0 or more generallyy(t0) = y0. We call the problem of finding a solution to the pair of equations

y0 = f(t, y), y(t0) = y0

an initial-value problem, which we will sometimes abbreviate by IVP. A solution to an initial-value problem is sometimes called a particular solution. We will also use the term “general solution”:an expression with parameters which describes all possible solutions to a di↵erential equation iscalled the general solution.

1.5. Example. The general solution to the equationdP

dt= kP is P (t) = Aekt, where A is a

constant. (Recall that in fact A is just P (0), the initial value of the solution.) Let us prove this.Assume that Q(t) is any solution to the above equation. Define R(t) = Q(t)e�kt. Then by the

product rule,

R0(t) = Q0(t)e�kt � kQ(t)e�kt = {Q0(t) � kQ(t)}e�kt.

But Q(t) is a solution to P 0 = kP , so Q0(t) � kQ(t) = 0. Therefore, R0(t) = 0, so R(t) must beconstant, say R(t) ⌘ A. Multiplying both sides by ekt, we obtain Q(t) = Aekt, as claimed. Thusevery solution is of the form Aekt. ⇤

Now that we have defined the notion of a solution, it’s time to ask:

Does every di↵erential equation have a solution? Does every initial-value problemhave a solution? If so, is it unique?

We will discuss these questions in Section 1.6. But first, let’s learn how to solve some simple,though common types of di↵erential equations.

Exercises

1. Verify the claim in Example 1.4.

2. Find a function g(t) such that y(t) = tet2

is a solution to the di↵erential equation

y0 = g(t)y.

3. Find a function h(y) such that the function y(t) = e2t is a solution to the equation

y0 = 2y � t + h(y).

ordinary di erential equations lecture notes for math 133a

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