Ordinary Di erential Equations Lecture notes for Math 133A. Slobodan N. Simić

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1 Ordinary Di erential Equations Lecture notes for Math 133A Slobodan N. Simić

2 c Slobodan N. Simić 2016

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4 Contents Chapter 1. First order di erential equations What is a di erential equation? Basic examples and modeling The notion of a solution Separable equations Linear equations Existence and uniqueness of solutions The phase line and classification of equilibria Linearization Numerical methods 38 Chapter 2. First order systems Examples of systems of di erential equations Vector fields and solutions curves The phase plane, equilibria and periodic solutions Existence and uniqueness of solutions 58 Chapter 3. Linear systems Definition and examples Linear algebra preliminaries Properties of linear systems Phase planes for planar linear systems The trace-determinant plane Second order linear equations Forced harmonic oscillators 112 Chapter 4. Nonlinear systems Equilibria and periodic solutions Linearization Hamiltonian systems Gradient systems 131 Chapter 5. Laplace transform Why another method? 133 vii

5 viii CONTENTS 5.2. Definition and basic properties Discontinuous forcing Impulse forcing Convolution and the Laplace transform 150

6 Preface This is a set of lecture notes for Math 133A: Ordinary Differential Equations taught by the author at San José State University in the Fall 2014 and The only prerequisite for the course is multivariable calculus. The notes focus on qualitative analysis of di erential equations in dimensions one and two. Since the class is mainly intended for SJSU engineering majors, the Laplace 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 ed to the author at slobodan.simic@sjsu.edu. The authors thanks the Textbook Alternatives Project at SJSU for their support. Slobodan N. Simić Department of Mathematics and Statistics San José State University ix

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8 CHAPTER 1 First order di erential equations Newton s fundamental discovery, the one which he considered necessary to keep secret and published only in the form of an anagram, consists of the following: Data aequatione quotcunque fluentes quantitae involvente fluxiones invenire et vice versa. In contemporary mathematical language this means: It is useful to solve 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. Modeling natural phenomena leads straight to di erential equations. The first person to systematically study di erential equations was Isaac Newton. He considered his discovery that di erential equations are useful for modeling physical phenomena one of his greatest, so he kept it secret and published it only in the form of an anagram (see the quote above). But what is a di erential equation? Here s an example: (1.1) Here s another: (1.2) dy dt = =0. And here s an equation that s not a di erential equation: (1.3) x 2 +3x +2=0. In each example there s an unknown we are supposed to solve for, but unlike the last equation (which is just a simple quadratic equation), where the unknown is a number, in the first two equations the unknown is a function. In(1.1) the unknown is the function y which depends on a single real variable t; in(1.2), the unknown is a function u which depends on two real variables, x and y. The first two equations involve derivatives of the unknown function, which is why we call them di erential 1

9 2 1. FIRST ORDER DIFFERENTIAL EQUATIONS equations. Equation (1.1) contains an ordinary derivative dy of the unknown function y, sowe dt 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, sowe call 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 independent variable and the equation involves derivatives of that function. The order of an equation is the 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 unknown function of t will be denoted by the letter y. Thus a first-order ordinary di erential equation is always of the form F (t, y, y 0 )=0, where F is some function of three variables and y 0 = dy dt. It is often possible to solve for y0 and rewrite the equation in the more useful form (1.4) y 0 = 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 the equation autonomous. Otherwise, it is called non-autonomous. Thus (1.1) is autonomous, whereas y 0 = t + y is not. Similarly, most second-order ordinary di erential equations can be written in the form y 00 = g(t, y, y 0 ), where g is some function of three variables and y 00 = d2 y dt 2. In Chapters 2, 3 and 4, we will be studying systems of di erential equations, where there is not just one, but several unknown functions, all depending on the same independent variable t. We postpone 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 harder then solving for just one or two numbers (like in numerical equations such as (1.3)). As a rule of thumb, di erential equations are harder to solve than numerical equations, and among di erential equations, 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 no equivalent 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 its derivative either by y 0 (when it s clear with respect to what variable the derivative is taken) or by dy. Sometimes (especially if t denotes time) we will use Newton s notation ẏ. Soremember: dt y 0 =ẏ = dy dt Basic examples and modeling In this section we will introduce three di erential equations which arise in the modeling of population growth and harmonic oscillators. Before we do that, let us first discuss what we mean by modeling.

10 1.2. BASIC EXAMPLES AND MODELING 3 Essentially, all of science is a search for models for how the Universe works. But to quote George E. 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 be modeled by our puny mathematical tools, since a model is just a mathematical representation of the real world (whatever that is). We can only hope to partially represent some aspects of the real phenomenon we wish to understand. So to obtain a reasonable model we need to make simplifying assumptions. That allows us to identify a finite (and ideally small) number of relevant variables and 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 the model 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 about its future behavior and compare those predictions with real data (which we can presumably obtain by observation and measurement). This can give us some idea about the validity and limitations of our 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 features of the real phenomenon we are trying to represent. If the model is not good enough, we make it more complex. The goal is to have a good balance between simplicity (which often translates into tractability) and faithfulness in representing the real world. Let s see how this works in the following examples Unlimited population growth. Assume we have the population of some species (say fruit 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 denote the number of individual members of the population by P and time by t. So P is a function of t. Of course, in reality P is a positive integer, but since we are hoping to study its rate of change with 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 exists for dt 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 rate of change P 0. Thus it makes sense to assume that P 0 is proportional to P, which means that there exists a positive parameter k 1 such that P 0 = kp. Thus the simplest di erential equation modeling the 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 side is not just a function of the independent variable). We will study it in two ways: qualitatively and analytically (i.e., by solving it). 1 This parameter is related to the food supply, but we will not elaborate on that connection here.

11 4 1. FIRST ORDER DIFFERENTIAL EQUATIONS 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 as t! +1 and t! 1. Let P (t) be a solution to (1.5) and set P 0 = P (0). This is the initial value of the population when t = 0. Let s see what we can say about P (t) for large t for various values of P 0. Assume first that P 0 = 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, for all values of t. The solution is constant; we will call such solutions equilibrium solutions or equilibria. 2 Finding and understanding such solutions is an important part of qualitative analysis. If P 0 > 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 P 0 > 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 P 0 < 0. We leave it as an exercise to the reader to show, in exactly the same way as in the previous paragraph, that P (t)! 1, as t! Figure 1.1. Three types of solutions of equation (1.5). Therefore, qualitatively speaking, there are only three types of solutions: the equilibrium solution (corresponding to the situation when P 0 = 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) 0 3 : 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 2 Equilibria is the plural of equilibrium. 3 The symbol means that the left-hand side equals the right-hand side for all values of the independent variable.

12 1.2. BASIC EXAMPLES AND MODELING 5 independent of each other, multiply both sides by dt. We obtain dp P = kdt. Now integrate both sides: Z Z dp P = kdt. Observe that P is now a dummy variable in the indefinite integral on the left and t plays the same role in the integral on the right. Solve these integrals. (If you don t remember how 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 we get P = e kt+c = e kt e C = Be kt, where B = e C, just another constant (for now). Since P = ±P, taking A = ±B (just another arbitrary constant), we obtain P = Ae kt. So every non-zero solution appears to be a constant multiple of the exponential function e kt! How does the constant A relate to the initial value P 0 of P? Let s see: set t =0in the previous equation and get P 0 = Ae k 0 = A. So A equals the initial value P 0 of P (t)! Therefore, given P 0 = P (0), our solution must be (1.6) P (t) =P 0 e kt, for all values of t. Note that we obtained this formula assuming that P 0 6= 0, but (1.6) does work even if P 0 = 0. The method we used for finding the solution is called separation of variables. We ll learn more about it in Section 1.4. Observe that the explicit solution given by (1.6) is consistent with the conclusions of the qualitative analysis: even though there are infinitely many solutions (one for each initial value of P (t)), they fall into one of the three categories discussed above Logistic population model. It s pretty clear that the assumptions we made in the previous model are extremely unrealistic. Indeed, if we compared the predictions of this model to real data (say, the US census) we would see that they agree only for small values of t. When t becomes 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-hand side of equation (1.5) is a very simple function of P : it is linear in P,whichiswhy(1.5) is an example 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 function on 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 like this: f(p )=AP 2 + BP + C, for some suitably chosen parameters A, B and C. But how do we choose these parameters? One way is to make better assumptions. For starters, no more unlimited

13 6 1. FIRST ORDER DIFFERENTIAL EQUATIONS quantities of food. How do we incorporate this assumption in the new model? Here s one way to do that. (1) For small P, we still want P 0 to be nearly proportional to P, i.e., when the population is small we would like P 0 to be approximately equal to kp, wherek is a parameter as in the previous model. (2) However, if P gets too large, we will assume that P 0 becomes negative, that is, the population 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,thenP 0 > 0 and if P>N, then P 0 < 0. What is the simplest quadratic function f(p ) which incorporates these assumptions? Assumption (1) suggests that we should try f(p )=kp (something), where something is close to 1 when P is small. But we need to make sure that assumption (2) is also satisfied. After thinking about this for a while, we are inescapably lead to the following solution: something = 1 Thus our new model for population growth becomes P N. (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 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 a logistic equation. It is an example of a non-linear di erential equation. Even though we could again separate the variables and find an explicit formula for all solutions, we will not do that here (see Section 1.4 for that approach). Instead, we employ only the qualitative approach to show its power and simplicity. As in the previous example, let us first find the equilibrium solutions. Recall that a solution P (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

14 1.2. BASIC EXAMPLES AND MODELING 7 f(p ) = 0, so we can find equilibrium solutions by solving the algebraic equation f(p ) = 0, which in our case is P kp 1 =0. N And that s a pretty easy task: the only solutions are P = 0 and P = N. Therefore, there are only two 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 P 0 is di erent from 0 and N? Here s how to approach this question. Suppose that at some time t the population is P. If 0 <P <N,thenf(P ) > 0, so P 0 = f(p ) > 0, which means that P is increasing. If P>N,thenP 0 = f(p ) < 0, which means that P is decreasing, exactly as we wanted. Thus solutions whose value is in the interval (0,N) are increasing and those in the interval (N,1) are decreasing. Here s a crucial question: if 0 <P <N, can the solution increase beyond N? Or:ifP>N, can it decrease below N? Let s see. Assume we have a solution P (t) such that 0 <P(0) <Nbutat some point in the future P (t + ) >N. Solutions are continuous (in fact, di erentiable!) functions, so there must be some intermediate time t 0 such that P (t 0 )=N. (What theorem of calculus guarantees this?) Therefore, when t = t 0, our solution and the equilibrium solution P (t) N cross! This would mean that after time t = t 0 our population could choose to follow the solution P or it could decide to stay at the equilibrium 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 in Section 1.6.) Thus if a solution P satisfies 0 <P(t) <N for some t, satisfies the same inequality for all t. Similarly, if P (t) >N for some t, the same holds for all t Figure 1.3. Graphs of several solutions of the logistic equation (1.7) with k = 3 and N = 4. We draw the following conclusions (see Figure 1.3): There are two equilibrium solutions, P = 0 and P = N. If 0 <P <N, the solution increases and converges to N, as t!1. (Where does it converge as t! 1?) If P>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.

15 8 1. FIRST ORDER DIFFERENTIAL EQUATIONS 1.1. Example. Let us try to use the ideas utilized in studying the logistic population model to understand the following di erential equation: (1.8) y 0 = y 2 10y As before, y 0 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) =y 2 10y + 21 is again a quadratic function of y. First of all, what are the equilibrium solutions? To answer this question, we factor the right-hand side of the equation: f(y) =(y 3)(y 7), then set f(y) = 0. The solutions to this algebraic equation are y 1 = 3 and y 2 = 7. These solutions give us the equilibrium solutions of the original di erential equation, namely, y 1 (t) = 3 and y 2 (t) =7 (for all t). Let us verify that y 1 (t) and y 2 (t) are indeed solutions to the equation (1.8). Substituting y 1 (t) = 3 into both sides of (1.8), we obtain d dt 3= , which is clearly satisfied. Therefore, y 1 (t) is indeed a solution and since it is constant, it is an equilibrium solution. We can check that y 2 (t) is also an equilibrium solution in a completely analogous way. It is clear that there are no other equilibrium solutions 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 find all values y 0 such that if y(t) is the solution satisfying y(0) = y 0,theny(t) is increasing. Since y(t) is increasing if y 0 (t) > 0 and y 0 (t) =f(y(t)), all we need to do is find the values of y for which f(y) > 0. Since f(y) =(y 3)(y 7), we immediately see that f(y) > 0 only if y>7 or y<3. So all 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 are depicted in Fig It can be shown that if y(t) starts in one of these intervals, it will remain in the same interval for all time.

16 1.2. BASIC EXAMPLES AND MODELING The harmonic oscillator. There are many harmonic oscillators in nature. We will describe 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 a mathematical model for this physical system and explore its properties. We will do the former in this 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 can choose the coordinates on the real line so that the equilibrium is at the origin and assume that the distance 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 = d2 y dt 2 = y00 Let s figure out what the net force F is. To make the problem tractable, we need to make some simplifying 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 F e. Thus we assume F = F f + F s + F e. By Hooke s Law, the force needed to extend a spring by a certain distance is proportional to that distance. Therefore, F s = ky, where k>0 is a constant called the sti ness of the spring. Note the negative sign; it s there because the spring is trying to return the mass to its equilibrium position (which is at y = 0), so when y>0, F s < 0 and when y<0, then F s > 0. To characterize the frictional force F f, we will use its simplest approximation and assume that it is proportional to the velocity of the object. In other words, F f = by 0, 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 it depends on time, i.e., F e = f(t), where f is some function. (f(t) is often periodic though it can be any kind of function, including a discontinuous one. We will learn how to deal with both kinds later.) 5 The 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.

17 10 1. FIRST ORDER DIFFERENTIAL EQUATIONS Substituting back in (1.9), we obtain ky by 0 + f(t) =my 00, which is equivalent to my 00 + by 0 + ky = f(t). This is the mass-spring oscillator equation. Observe that it s a second order equation; we will study second order linear equations in Chapter Consider the population model dp dt =0.2P Exercises 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 y 0 = y 3 +3y 2 10y. (a) What are the equilibrium solutions? (b) For which values of y 0 is the solution y(t) starting at y 0 increasing? (y(t) starts at y 0 if y(0) = y 0.) (c) For which values of y 0 is the solution y(t) starting at y 0 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 decay rate. State the corresponding initial-value problem if the initial amount of the isotope is I The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope 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 date archaeological, geological, and other samples. Its main idea is that by measuring the amount of radiocarbon still found in the organic mater and comparing it to the amount normally found in living matter, we can approximate the amount of time since death

18 1.3. THE NOTION OF A SOLUTION 11 occurred. 6 Using the decay-rate parameter found in part (b), find the time since death if 35% of radiocarbon is still in the sample The notion of a solution So far we have been a little imprecise and have freely talked about solutions to various di erential equations without ever defining what we meant by it. So suppose we have a first-order di erential equation (1.10) y 0 = f(t, y). What do we mean when we say that y(t) is a solution? The answer quite natural: y(t) is a solution to (1.10) if when plugged into (1.10), y(t) satisfies it, i.e., the left-hand side becomes identically and trivially equal to the right-hand side: y 0 (t) =f(t, y(t)), for all values of t for which both sides make sense. Here are some examples Example. Consider y 0 =2y. Theny(t) =e 2t is a solution as one can easily check. So are the functions e 2t and 2014e 2t, but not t 2, e t or sin t. (Note that this is just an equation modeling unlimited population growth as in Section ) 1.3. Example. The function y(t) = tan t is a solution to the equation y 0 =1+y 2. This equation has no equilibrium solutions. (Check this!) 1.4. Example. Define a function z(t) by z(t) = ( 0 if t apple 0 t 3 it t>0. Verify that z satisfies the di erential equation z 0 =3z 2/3. We have already encountered the notion of an equilibrium solution but because of its importance we repeat it definition: A solution is called an equilibrium solution if it is constant. A point y 0 is called an equilibrium point of a di erential equation if y(t) y 0 is a solution. How do we look for equilibrium solutions? Suppose that y(t) =y 0 is an equilibrium solution of the di erential equation y 0 = f(t, y). Plugging y(t) into the equation we obtain f(t, y 0 )=f(t, y(t)) = y 0 (t) = d dt y 0 =0, for all t. Conversely, suppose we found y 0 such that f(t, y 0 ) = 0, for all values of t. Then the above calculation also shows that y(t) =y 0 is an equilibrium solution of the equation y 0 = f(t, y). We conclude: y 0 is an equilibrium point of the equation y 0 = f(t, y) if and only if f(t, y 0 )=0,for all values of t. 6 This is based on the assumption that radiocarbon makes a constant proportion of the carbon ingested by living matter and that once the matter dies no new carbon is added to it.

19 12 1. FIRST ORDER DIFFERENTIAL EQUATIONS In particular, if the equation is autonomous, y 0 = f(y), then finding equilibrium points amounts to solving the equation f(y) =y. Observe that checking whether a given function is a solution is easy, but finding solutions is usually not. For instance, does the equation y 0 =1+y 4 sin 2 arctan y3 y + 17e have a solution satisfying the condition y(0) = 0? The answer is yes, but there s no way we can tell using 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 a di erential equation which satisfies an initial condition of the type y(0) = y 0 or more generally y(t 0 )=y 0. We call the problem of finding a solution to the pair of equations y 0 = f(t, y), y(t 0 )=y 0 an initial-value problem, which we will sometimes abbreviate by IVP. A solution to an initialvalue 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 is called the general solution Example. The general solution to the equation dp dt = kp is P (t) =Aekt,whereA 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.thenbythe product rule, R 0 (t) =Q 0 (t)e kt kq(t)e kt = {Q 0 (t) kq(t)}e kt. But Q(t) is a solution to P 0 = kp, soq 0 (t) kq(t) = 0. Therefore, R 0 (t) = 0, so R(t) mustbe constant, say R(t) A. Multiplying both sides by e kt, we obtain Q(t) =Ae kt, as claimed. Thus every solution is of the form Ae kt. 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 problem have 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 Find a function g(t) such that y(t) =te t2 is a solution to the di erential equation y 0 = g(t)y. 3. Find a function h(y) such that the function y(t) =e 2t is a solution to the equation y 0 =2y t + h(y).