LINEAR FIRST ORDER EQUATIONS

A first order differential equation is said to belinearif it can be written as

y0_{Cp.x/y Df .x/: (2.1.1)}

A first order differential equation that can’t be written like this is nonlinear. We say that (2.1.1) is homogeneousiff 0; otherwise it’snonhomogeneous. Sincey 0 is obviously a solution of the homgeneous equation

y0_Cp.x/yD0;

we call it thetrivial solution. Any other solution isnontrivial. Example 2.1.1 The first order equations

x2y0_{C3y D x}2 ; xy0 _8x2_y D sinx; xy0_{C.lnx/y D 0;} y0 _{D x}2 y 2; are not in the form (2.1.1), but they are linear, since they can be rewritten as

y0_C 3 x2y D 1; y0 8xy D sinx x ; y0_C lnx x y D 0; y0 x2y D 2: Example 2.1.2 Here are some nonlinear first order equations:

xy0_C3y2

D 2x (becauseyis squared); yy0 _{D 3 (because of the productyy}0_); y0_Cxey

D 12 (because ofey):

General Solution of a Linear First Order Equation

To motivate a definition that we’ll need, consider the simple linear first order equation y0_D 1

x2: (2.1.2)

From calculus we know thatysatisfies this equation if and only if y_D 1

x Cc; (2.1.3)

wherecis an arbitrary constant. We call caparameter and say that (2.1.3) defines aone–parameter familyof functions. For each real numberc, the function defined by (2.1.3) is a solution of (2.1.2) on

Section 2.1Linear First Order Equations 31 . 1; 0/and.0;1/; moreover, every solution of (2.1.2) on either of these intervals is of the form (2.1.3) for some choice ofc. We say that (2.1.3) isthe general solutionof (2.1.2).

We’ll see that a similar situation occurs in connection with any first order linear equation

y0Cp.x/y Df .x/I (2.1.4)

that is, ifpandf are continuous on some open interval.a; b/then there’s a unique formulayDy.x; c/ analogous to (2.1.3) that involvesxand a parametercand has the these properties:

For each fixed value ofc, the resulting function ofxis a solution of (2.1.4) on.a; b/.

If y is a solution of (2.1.4) on.a; b/, theny can be obtained from the formula by choosing c appropriately.

We’ll cally_Dy.x; c/thegeneral solutionof (2.1.4).

When this has been established, it will follow that an equation of the form

P0.x/y0CP1.x/y DF .x/ (2.1.5)

has a general solution on any open interval.a; b/on whichP0,P1, andF are all continuous andP0has no zeros, since in this case we can rewrite (2.1.5) in the form (2.1.4) withp _D P1=P0andf DF =P0, which are both continuous on.a; b/.

To avoid awkward wording in examples and exercises, we won’t specify the interval.a; b/when we ask for the general solution of a specific linear first order equation. Let’s agree that this always means that we want the general solution on every open interval on whichpandf are continuous if the equation is of the form (2.1.4), or on whichP0,P1, andF are continuous andP0has no zeros, if the equation is of the form (2.1.5). We leave it to you to identify these intervals in specific examples and exercises.

For completeness, we point out that ifP0,P1, andF are all continuous on an open interval.a; b/, but P0doeshave a zero in.a; b/, then (2.1.5) may fail to have a general solution on.a; b/in the sense just defined. Since this isn’t a major point that needs to be developed in depth, we won’t discuss it further; however, see Exercise44for an example.

Homogeneous Linear First Order Equations

We begin with the problem of finding the general solution of a homogeneous linear first order equation. The next example recalls a familiar result from calculus.

Example 2.1.3 Letabe a constant. (a) Find the general solution of

y0 ayD0: (2.1.6)

(b) Solve the initial value problem

y0 _{ayD0; y.x}

0/Dy0:

SOLUTION(a) You already know from calculus that ifcis any constant, theny_Dceaxsatisfies (2.1.6). However, let’s pretend you’ve forgotten this, and use this problem to illustrate a general method for solving a homogeneous linear first order equation.

We know that (2.1.6) has the trivial solutiony0. Now supposeyis a nontrivial solution of (2.1.6). Then, since a differentiable function must be continuous, there must be some open intervalI on whichy has no zeros. We rewrite (2.1.6) as

y0 y Da

x 0.2 0.4 0.6 0.8 1.0 y 0.5 1.0 1.5 2.0 2.5 3.0 a = 2 a = 1.5 a = 1 a = −1 a = −2.5 a = −4

Figure 2.1.1 Solutions ofy0 _ayD0,y.0/D1

forxinI. Integrating this shows that

ln_jy_{j D}ax_Ck; so _jy_{j D}ekeax;

wherekis an arbitrary constant. Sinceeaxcan never equal zero,yhas no zeros, soy is either always positive or always negative. Therefore we can rewriteyas

y_Dceax (2.1.7) where c_D ek ify > 0; ek ify < 0:

This shows that every nontrivial solution of (2.1.6) is of the formy_Dceax_{for some nonzero constantc.} Since settingc_D0yields the trivial solution,allsolutions of (2.1.6) are of the form (2.1.7). Conversely, (2.1.7) is a solution of (2.1.6) for every choice ofc, since differentiating (2.1.7) yieldsy0_Daceax_Day. SOLUTION(b) Imposing the initial conditiony.x0/Dy0yieldsy0Dceax0, socDy0e ax0and

y_Dy0e ax0eax Dy0ea.x x0/:

Figure2.1.1show the graphs of this function withx0D0,y0 D1, and various values ofa. Example 2.1.4 (a) Find the general solution of

xy0_Cy_D0: (2.1.8)

(b) Solve the initial value problem

Section 2.1Linear First Order Equations 33

SOLUTION(a) We rewrite (2.1.8) as

y0_C 1

xy D0; (2.1.10)

wherexis restricted to either. 1; 0/or.0;1/. Ifyis a nontrivial solution of (2.1.10), there must be some open interval I on whichyhas no zeros. We can rewrite (2.1.10) as

y0 y D

1 x forxinI. Integrating shows that

ln_jyj D ln_jxj Ck; so _jyj D e

k

jx_j:

Since a function that satisfies the last equation can’t change sign on either. ₁; 0/ or.0;₁/, we can rewrite this result more simply as

y _D c x (2.1.11) where c_D ek _{ify > 0;} ek ify < 0:

We’ve now shown that every solution of (2.1.10) is given by (2.1.11) for some choice ofc. (Even though we assumed thaty was nontrivial to derive (2.1.11), we can get the trivial solution by settingc _D 0in (2.1.11).) Conversely, any function of the form (2.1.11) is a solution of (2.1.10), since differentiating (2.1.11) yields

y0D c

x2; and substituting this and (2.1.11) into (2.1.10) yields

y0C 1 xy D c x2 C 1 x c x D c x2 C c x2 D0:

Figure2.1.2shows the graphs of some solutions corresponding to various values ofc

SOLUTION(b) Imposing the initial conditiony.1/_D3in (2.1.11) yieldsc _D3. Therefore the solution of (2.1.9) is

y _D 3 x: The interval of validity of this solution is.0;1/.

The results in Examples2.1.3(a)and2.1.4(b)are special cases of the next theorem.

Theorem 2.1.1 Ifpis continuous on.a; b/;then the general solution of the homogeneous equation

y0_{Cp.x/yD0 (2.1.12)} on.a; b/is yDce P.x/; where P .x/D Z p.x/ dx (2.1.13)

x y

c > 0 c < 0

c > 0 c < 0

Figure 2.1.2 Solutions ofxy0_{CyD0on.0;1/and. 1; 0/}

is any antiderivative ofpon.a; b/_Ithat is;

P0_{.x/Dp.x/; a < x < b: (2.1.14)}

Proof IfyDce P.x/, differentiatingyand using (2.1.14) shows that y0_{D P}0_.x/ce P.x/

D p.x/ce P.x/ _D p.x/y; soy0_{Cp.x/y D0; that is,yis a solution of (2.1.12), for any choice ofc.}

Now we’ll show that any solution of (2.1.12) can be written asy_Dce P.x/ _{for some constantc. The} trivial solution can be written this way, withc_D0. Now supposeyis a nontrivial solution. Then there’s an open subintervalIof.a; b/on whichyhas no zeros. We can rewrite (2.1.12) as

y0

y D p.x/ (2.1.15)

forxinI. Integrating (2.1.15) and recalling (2.1.13) yields lnjyj D P .x/Ck; wherekis a constant. This implies that

jy_{j D}eke P.x/:

SinceP is defined for allxin.a; b/and an exponential can never equal zero, we can takeI _D.a; b/, so yhas zeros on.a; b/ .a; b/, so we can rewrite the last equation asy_Dce P.x/_{, where}

c_D

ek ify > 0on.a; b/; ek ify < 0on.a; b/:

Section 2.1Linear First Order Equations 35

REMARK: Rewriting a first order differential equation so that one side depends only onyandy0_{and the} other depends only onxis calledseparation of variables. We did this in Examples 2.1.3and2.1.4, and in rewriting (2.1.12) as (2.1.15).We’llapply this method to nonlinear equations in Section 2.2.

Linear Nonhomogeneous First Order Equations

We’ll now solve the nonhomogeneous equation

y0_{Cp.x/y Df .x/: (2.1.16)}

When considering this equation we call

y0_Cp.x/yD0

thecomplementary equation.

We’ll find solutions of (2.1.16) in the formy _D uy1, wherey1 is a nontrivial solution of the com- plementary equation anduis to be determined. This method of using a solution of the complementary equation to obtain solutions of a nonhomogeneous equation is a special case of a method calledvariation of parameters, which you’ll encounter several times in this book. (Obviously,ucan’t be constant, since if it were, the left side of (2.1.16) would be zero. Recognizing this, the early users of this method viewed uas a “parameter” that varies; hence, the name “variation of parameters.”)

If

y _Duy1; then y0 Du0y1Cuy10: Substituting these expressions foryandy0_{into (2.1.16) yields}

u0y1Cu.y10 Cp.x/y1/Df .x/; which reduces to

u0_y

1Df .x/; (2.1.17)

sincey1is a solution of the complementary equation; that is, y0

1Cp.x/y1 D0:

In the proof of Theorem2.2.1we saw thaty1has no zeros on an interval wherepis continuous. Therefore we can divide (2.1.17) through byy1to obtain

u0_{Df .x/=y}

1.x/:

We can integrate this (introducing a constant of integration), and multiply the result byy1to get the gen- eral solution of (2.1.16). Before turning to the formal proof of this claim, let’s consider some examples. Example 2.1.5 Find the general solution of

y0_C2yDx3_e 2x_{: (2.1.18)}

By applying(a)of Example 2.1.3witha _D 2, we see thaty1 D e 2x is a solution of the com- plementary equationy0_{C2y D 0. Therefore we seek solutions of (2.1.18) in the formy D ue} 2x_{, so} that

y0 Du0e 2x 2ue 2x and y0C2yDu0e 2x 2ue 2xC2ue 2xDu0e 2x: (2.1.19) Thereforeyis a solution of (2.1.18) if and only if

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 x y

Figure 2.1.3 A direction field and integral curves fory0_C2yDx2_e 2x

Therefore u_D x 4 4 Cc; and y_Due 2x_De 2x _x4 4 Cc

is the general solution of (2.1.18).

Figure2.1.3shows a direction field and some integral curves for (2.1.18). Example 2.1.6

(a) Find the general solution

y0_{C.cotx/yDxcscx: (2.1.20)}

(b) Solve the initial value problem

y0C.cotx/yDxcscx; y.=2/D1: (2.1.21)

SOLUTION(a) Herep.x/_Dcotxandf .x/_Dxcscxare both continuous except at the pointsx_Dr , whereris an integer. Therefore we seek solutions of (2.1.20) on the intervals.r ; .r_C1/ /. We need a nontrival solutiony1of the complementary equation; thus,y1must satisfyy₁0 C.cotx/y1 D0, which we rewrite as y0 1 y1 D cotx_D cosx sinx: (2.1.22)

Integrating this yields

Section 2.1Linear First Order Equations 37 where we take the constant of integration to be zero since we need onlyonefunction that satisfies (2.1.22). Clearlyy1D1=sinxis a suitable choice. Therefore we seek solutions of (2.1.20) in the form

y _D u sinx; so that y0_D u0 sinx ucosx sin2x (2.1.23) and y0_{C.cotx/y D} u0 sinx ucosx sin2x C ucotx sinx D u 0 sinx ucosx sin2x C ucosx sin2x D u 0 sinx: (2.1.24)

Thereforeyis a solution of (2.1.20) if and only if

u0_{=sinxDxcscxDx=sinx or, equivalently, u}0_Dx:

Integrating this yields

u_D x 2 2 Cc; and yD u sinx D x2 2sinx C c sinx: (2.1.25)

is the general solution of (2.1.20) on every interval.r ; .r_C1/ /(r_Dinteger).

SOLUTION(b) Imposing the initial conditiony.=2/_D1in (2.1.25) yields 1_D 2 8 Cc or cD1 2 8 : Thus, yD x 2 2sinx C .1 2=8/ sinx

is a solution of (2.1.21). The interval of validity of this solution is.0; /; Figure2.1.4shows its graph.

1 2 3 − 15 − 10 − 5 5 10 15 x y

REMARK: It wasn’t necessary to do the computations (2.1.23) and (2.1.24) in Example2.1.6, since we showed in the discussion preceding Example 2.1.5 that if y D uy1 wherey0

1C p.x/y1 D 0, then

y0_Cp.x/yDu0_{y1. We did these computations so you would see this happen in this specific example. We} recommend that you include these “unnecesary” computations in doing exercises, until you’re confident that you really understand the method. After that, omit them.

We summarize the method of variation of parameters for solving

y0_{Cp.x/y Df .x/ (2.1.26)}

as follows:

(a) Find a functiony1such that

y0 1

y1 D p.x/:

For convenience, take the constant of integration to be zero.

(b) Write

y _Duy1 (2.1.27)

to remind yourself of what you’re doing.

(c) Writeu0_{y1Df and solve foru}0_{; thus,u}0_{Df =y1.}

(d) Integrateu0_{to obtainu, with an arbitrary constant of integration.}

(e) Substituteuinto (2.1.27) to obtainy.

To solve an equation written as

P0.x/y0CP1.x/yDF .x/;

we recommend that you divide through by P0.x/ to obtain an equation of the form (2.1.26) and then follow this procedure.

Solutions in Integral Form

Sometimes the integrals that arise in solving a linear first order equation can’t be evaluated in terms of elementary functions. In this case the solution must be left in terms of an integral.

Example 2.1.7

(a) Find the general solution of

y0 _2xyD1:

(b) Solve the initial value problem

y0 _{2xyD1; y.0/Dy}

0: (2.1.28)

SOLUTION(a) To apply variation of parameters, we need a nontrivial solutiony1of the complementary equation; thus,y0

1 2xy1D0, which we rewrite as y0

1 y1 D

Section 2.1Linear First Order Equations 39 Integrating this and taking the constant of integration to be zero yields

lnjy1j Dx2; so jy1j Dex

2 : We choosey1Dex

2

and seek solutions of (2.1.28) in the formy Duex2, where u0ex2 D1; so u0De x2:

Therefore

uDcC

Z

e x2dx;

but we can’t simplify the integral on the right because there’s no elementary function with derivative equal toe x2

. Therefore the best available form for the general solution of (2.1.28) is y Duex2 Dex2 cC Z e x2dx : (2.1.29)

SOLUTION(b) Since the initial condition in (2.1.28) is imposed atx0 D 0, it is convenient to rewrite (2.1.29) as y_Dex2 c_C Z x 0 e t2dt ; since Z 0 0 e t2dt_D0:

Settingx_D0andy _Dy0here shows thatcDy0. Therefore the solution of the initial value problem is y Dex2 y0C Z x 0 e t2dt : (2.1.30)

For a given value ofy0and each fixedx, the integral on the right can be evaluated by numerical methods. An alternate procedure is to apply the numerical integration procedures discussed in Chapter 3 directly to the initial value problem (2.1.28). Figure2.1.5shows graphs of of (2.1.30) for several values ofy0.

x y

An Existence and Uniqueness Theorem

The method of variation of parameters leads to this theorem.

Theorem 2.1.2 Supposepandf are continuous on an open interval.a; b/;and lety1be any nontrivial solution of the complementary equation

y0_Cp.x/y_D0

on.a; b/. Then_W

(a) The general solution of the nonhomogeneous equation

y0Cp.x/yDf .x/ (2.1.31) on.a; b/is yDy1.x/ cC Z f .x/=y1.x/ dx : (2.1.32)

(b) Ifx0is an arbitrary point in.a; b/andy0is an arbitrary real number;then the initial value problem

y0_{Cp.x/y Df .x/; y.x0/Dy0}

has the unique solution

y_Dy1.x/ _y0 y1.x0/C Z x x0 f .t / y1.t /dt on.a; b/:

Proof (a)To show that (2.1.32) is the general solution of (2.1.31) on.a; b/, we must prove that: (i) Ifcis any constant, the functionyin (2.1.32) is a solution of (2.1.31) on.a; b/.

(ii) Ifyis a solution of (2.1.31) on.a; b/thenyis of the form (2.1.32) for some constantc. To prove(i), we first observe that any function of the form (2.1.32) is defined on.a; b/, sincepandf are continuous on.a; b/. Differentiating (2.1.32) yields

y0Dy10.x/ cC Z f .x/=y1.x/ dx Cf .x/: Sincey0

1 D p.x/y1, this and (2.1.32) imply that y0 _{D p.x/y} 1.x/ c_C Z f .x/=y1.x/ dx Cf .x/

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