17Differential EquationsMany physical phenomena can be modeled using the language of calculus. For example,observa tional evidence suggests that the temperature of a cup of tea (or some other liqui d)in a room of constant temperature will cool over time at a rate proportional to the differencebetween the room temperature and the temperature o f the tea.In symbols, if t is the time, M is the roo m temperature, and f(t) is the temperatureof the tea at time t then f′(t) = k(M − f(t)) where k > 0 is a constant which will dependon the kind of tea (or more generally the kind o f liquid) but not on the room temperatureor the temperature of the tea. This is Newton’s law of cooling and the equation thatwe just wr o te down is an example of a differential equati on. Ideally we would like tosolve this equation, namely, find the function f(t) that describes the temperature overtime, though t hi s often turns out to be impossible, in which case various a pproximationtechniques must be used. The use and solution of different ial equations is an impo r tantfield of mathematics; here we see how to solve some simple but useful types of di fferentialequation.Informally, a differential equation is an equation in which one or more of the derivativesof some function appear. Typical ly, a scientific theory will produce a differential equation(or a system of differential equations) that describes or g overns some physical process, butthe theory will not produce the desired function or functions directl y.Recall from section 6.2 that when the variabl e is time the derivative of a function y(t)is sometimes written as ˙y instead of y′; this is quite common in the study of differentialequations.455456 Chapter 17 Differential Equations17.1 First Order Differential EquationsWe start by considering equations in which only the first derivative of the function appears.DEFINITION 17.1.1 A first order differential equation is an equation of the formF (t, y, ˙y) = 0. A solution of a first order differential equation is a function f(t) that makesF (t, f(t), f′(t)) = 0 for every value of t .Here, F is a function of three variables w hi ch we label t, y, and ˙y. It is understoodthat ˙y will explicitly appear in the equation although t and y need not. The term “firstorder” means that the first derivative of y appears, but no higher order derivatives do.EXAMPLE 17.1.2 The equation from Newton’s law of cooling, ˙y = k(M −y) is a firstorder differential equatio n; F (t, y, ˙y) = k(M −y) − ˙y.EXAMPLE 17.1.3 ˙y = t2+1 is a first order differential equation; F (t, y, ˙y) = ˙y −t2−1.All solutions to this equat ion are of the form t3/3 + t + C.DEFINITION 17.1.4 A first order initial value problem is a sy st em of equationsof the form F (t, y, ˙y) = 0, y(t0) = y0. Here t0is a fixed time and y0is a number. Asolution of an initial value problem is a solution f (t) of the differential equation that alsosatisfies the initial condition f(t0) = y0.EXAMPLE 17.1.5 The initial value problem ˙y = t2+ 1, y(1) = 4 has solution f (t) =t3/3 + t + 8/3.The general first order equation is rather too general, that is, we can’t describe methodsthat will work on them all, o r even a large portion of t hem. We can make progress withspecific kinds of first order differential equations. For example, much can be said aboutequations of the form ˙y = φ( t, y) where φ is a function of the two va riables t and y. Underreasonable conditions on φ, such an equation has a solution and the corresponding initialvalue problem has a unique solution. However, in general, these equations can be verydifficult or impossible to solve explicitly.EXAMPLE 17.1.6 Consider this specific example of an initial value problem for New-ton’s law of cooling: ˙y = 2(25 − y), y(0) = 40. We first note that if y(t0) = 2 5, the righthand side of the di fferential equation is zero, and so the constant function y(t) = 25 is asolution to the differential equation. It is not a solution to the initial value problem, sincey( 0) 6= 40. (The physical interpretation of this constant solutio n is that if a liquid is atthe same temperature as its surroundings, then the liquid wi ll stay at that temperature.)17.1 First Order Differential Equations 457So long as y is not 25, we can rewrite the differential equation asdydt125 − y= 2125 − ydy = 2 dt,soZ125 − ydy =Z2 dt,that is, the two anti-derivatives must be the same except for a constant difference. We cancalculate these anti-derivatives and rearrange the results:Z125 − ydy =Z2 dt(−1) ln |25 − y| = 2t + C0ln |25 − y| = −2t − C0= −2t + C|25 −y| = e−2t+C= e−2teCy − 25 = ±eCe−2ty = 25 ± eCe−2t= 25 + Ae−2t.Here A = ±eC= ±e−C0is some non-zero constant. Since we want y(0) = 40, wesubstitute and solve for A:40 = 25 + Ae015 = A,and so y = 25+15e−2tis a solution to t he initia l value probl em. Note that y is never 25, sothis makes sense for all values of t. However, if we allow A = 0 we get the solution y = 25to the different ial equation, which would be the soluti on to the initial value problem if wewere to require y(0) = 25. Thus, y = 25 + Ae−2tdescrib es all solutions to the differentialequation ˙y = 2(25 − y), and all solutions to the associated initial value problems.Why could we solve this problem? Our solution depended on rewriting the equationso that a ll instances of y were on one side of the equation and all instances of t were on theother; of course, in this case the only t was originally hidden, since we didn’t write dy/dtin the original equation. This is not required, however.458 Chapter 17 Differential EquationsEXAMPLE 17.1.7 Solve the differentia l equatio n ˙y = 2 t(25 − y). This is almostidentical to the previous exampl e. As before, y(t) = 25 is a solution. If y 6= 25,Z125 − ydy =Z2t dt(−1) ln |25 − y| = t2+ C0ln |25 − y| = −t2− C0= −t2+ C|25 −y| = e−t2+C= e−t2eCy − 25 = ±eCe−t2y = 25 ± eCe−t2= 25 + Ae−t2.As before, all solutions a r e represented by y = 2 5 + Ae−t2, allowing A t o be zero.DEFINITION 17.1.8 A first order differential equation is separable if it can be writtenin the form ˙y = f (t)g(y).As in the examples, we can attempt to solve a separable equation by converting to theformZ1g(y)dy =Zf(t) dt.This technique is called separation of variabl es. The simplest (in principle) sort ofseparable equation is one in which g(y) = 1, in which case we attempt to solveZ1 dy =Zf(t) dt.We can do this if