SEPARABLE EQUATIONS

A first order differential equation isseparableif it can be written as

h.y/y0 _{Dg.x/; (2.2.1)}

where the left side is a product ofy0 _{and a function ofyand the right side is a function ofx. Rewriting}

a separable differential equation in this form is calledseparation of variables. In Section 2.1 we used separation of variables to solve homogeneous linear equations. In this section we’ll apply this method to nonlinear equations.

To see how to solve (2.2.1), let’s first assume thatyis a solution. LetG.x/andH.y/be antiderivatives ofg.x/andh.y/; that is,

H0.y/Dh.y/ and G0.x/Dg.x/: (2.2.2)

Then, from the chain rule, d dxH.y.x//DH 0_.y.x//y0_.x/Dh.y/y0_.x/: Therefore (2.2.1) is equivalent to d dxH.y.x// D d dxG.x/:

Integrating both sides of this equation and combining the constants of integration yields

H.y.x//_DG.x/_Cc: (2.2.3)

Although we derived this equation on the assumption thatyis a solution of (2.2.1), we can now view it differently: Any differentiable functionythat satisfies (2.2.3) for some constantcis a solution of (2.2.1). To see this, we differentiate both sides of (2.2.3), using the chain rule on the left, to obtain

H0_.y.x//y0_.x/DG0_.x/;

which is equivalent to

h.y.x//y0_.x/Dg.x/

because of (2.2.2).

In conclusion, to solve (2.2.1) it suffices to find functionsG _D G.x/andH _D H.y/that satisfy (2.2.2). Then any differentiable functiony _Dy.x/that satisfies (2.2.3) is a solution of (2.2.1).

Example 2.2.1 Solve the equation

y0Dx.1Cy2/:

Solution Separating variables yields

y0 1_Cy2 Dx: Integrating yields tan 1yD x 2 2 Cc Therefore yDtan _x2 2 Cc :

Example 2.2.2

(a) Solve the equation

y0D x

y: (2.2.4)

(b) Solve the initial value problem

y0_D x

y; y.1/D1: (2.2.5)

(c) Solve the initial value problem

y0_D x

y; y.1/D 2: (2.2.6)

SOLUTION(a) Separating variables in (2.2.4) yields yy0_{D x:} Integrating yields y2 2 D x2 2 Cc; or, equivalently, x 2 Cy2 D2c:

The last equation shows thatcmust be positive ify is to be a solution of (2.2.4) on an open interval. Therefore we let2c_Da2(witha > 0) and rewrite the last equation as

x2Cy2Da2: (2.2.7)

This equation has two differentiable solutions foryin terms ofx:

yD pa2 _x2_{; a < x < a; (2.2.8)}

and

y_D pa2 _x2_{; a < x < a: (2.2.9)}

The solution curves defined by (2.2.8) are semicircles above thex-axis and those defined by (2.2.9) are semicircles below thex-axis (Figure2.2.1).

SOLUTION(b) The solution of (2.2.5) is positive whenx_D1; hence, it is of the form (2.2.8). Substituting x _D1andy _D1into (2.2.7) to satisfy the initial condition yieldsa2_D 2; hence, the solution of (2.2.5) is

y _Dp2 x2_; p_{2 < x <}p_2:

SOLUTION(c) The solution of (2.2.6) is negative when x D 1 and is therefore of the form (2.2.9). Substitutingx D 1andy D 2into (2.2.7) to satisfy the initial condition yieldsa2 D 5. Hence, the solution of (2.2.6) is

Section 2.2Separable Equations 47 x y 1 2 −1 −2 1 2 −1 −2 (a) (b) Figure 2.2.1 (a)y _Dp2 x2_, p_{2 < x <}p_{2; (b)yD} p_{5 x}2_, p_{5 < x <}p₅

Implicit Solutions of Separable Equations

In Examples2.2.1and2.2.2we were able to solve the equationH.y/ _D G.x/_Cc to obtain explicit formulas for solutions of the given separable differential equations. As we’ll see in the next example, this isn’t always possible. In this situation we must broaden our definition of a solution of a separable equation. The next theorem provides the basis for this modification. We omit the proof, which requires a result from advanced calculus called as theimplicit function theorem.

Theorem 2.2.1 Supposeg_Dg.x/is continous on.a; b/andh_Dh.y/are continuous on.c; d /:LetG

be an antiderivative ofgon.a; b/and letHbe an antiderivative ofhon.c; d /:Letx0be an arbitrary point in.a; b/;lety0be a point in.c; d /such thath.y0/_¤0;and define

cDH.y0/ G.x0/: (2.2.10)

Then there’s a functiony_Dy.x/defined on some open interval.a1; b1/;whereaa1< x0< b1b;

such thaty.x0/_Dy0and

H.y/_DG.x/_Cc (2.2.11)

fora1< x < b1. Thereforeyis a solution of the initial value problem

h.y/y0_{Dg.x/; y.x0/Dx0: (2.2.12)}

It’s convenient to say that (2.2.11) withcarbitrary is animplicit solutionofh.y/y0 _{D g.x/. Curves}

defined by (2.2.11) are integral curves ofh.y/y0 _{Dg.x/. Ifcsatisfies (2.2.10), we’ll say that (2.2.11) is}

animplicit solution of the initial value problem(2.2.12). However, keep these points in mind:

The functionyin (2.2.11) (not (2.2.11) itself) is a solution ofh.y/y0 _Dg.x/.

Example 2.2.3

(a) Find implicit solutions of

y0_D 2xC1

5y4_C1: (2.2.13)

(b) Find an implicit solution of

y0_D 2xC1

5y4_C1; y.2/D1: (2.2.14)

SOLUTION(a) Separating variables yields

.5y4C1/y0 D2xC1: Integrating yields the implicit solution

y5_Cy_Dx2_Cx_Cc: (2.2.15)

of (2.2.13).

SOLUTION(b) Imposing the initial conditiony.2/D1in (2.2.15) yields1C1D4C2Cc, socD 4. Therefore

y5_Cy_Dx2_Cx 4

is an implicit solution of the initial value problem (2.2.14). Although more than one differentiable func- tiony _D y.x/ satisfies2.2.13) nearx _D 1, it can be shown that there’s only one such function that satisfies the initial conditiony.1/_D2.

Figure2.2.2shows a direction field and some integral curves for (2.2.13).

Constant Solutions of Separable Equations

An equation of the form

y0_Dg.x/p.y/

is separable, since it can be rewritten as 1 p.y/y

0_Dg.x/:

However, the division byp.y/is not legitimate ifp.y/_D0for some values ofy. The next two examples show how to deal with this problem.

Example 2.2.4 Find all solutions of

y0_D2xy2_{: (2.2.16)}

Solution Here we must divide byp.y/_Dy2to separate variables. This isn’t legitimate ifyis a solution of (2.2.16) that equals zero for some value ofx. One such solution can be found by inspection:y 0. Now supposeyis a solution of (2.2.16) that isn’t identically zero. Sinceyis continuous there must be an interval on whichyis never zero. Since division byy2is legitimate forxin this interval, we can separate variables in (2.2.16) to obtain

y0 y2 D2x:

Section 2.2Separable Equations 49 1 1.5 2 2.5 3 3.5 4 −1 −0.5 0 0.5 1 1.5 2 x y

Figure 2.2.2 A direction field and integral curves fory0 _D 2xC1 5y4_C1

Integrating this yields

1 y Dx 2 Cc; which is equivalent to yD 1 x2_Cc: (2.2.17)

We’ve now shown that ifyis a solution of (2.2.16) that is not identically zero, theny must be of the form (2.2.17). By substituting (2.2.17) into (2.2.16), you can verify that (2.2.17) is a solution of (2.2.16). Thus, solutions of (2.2.16) arey 0and the functions of the form (2.2.17). Note that the solutiony0 isn’t of the form (2.2.17) for any value ofc.

Figure2.2.3shows a direction field and some integral curves for (2.2.16) Example 2.2.5 Find all solutions of

y0 _D 1 2x.1 y

2_{/: (2.2.18)}

Solution Here we must divide byp.y/ _D 1 y2_{to separate variables. This isn’t legitimate ify is a} solution of (2.2.18) that equals_˙1for some value ofx. Two such solutions can be found by inspection: y1andy 1. Now supposeyis a solution of (2.2.18) such that1 y2_{isn’t identically zero. Since} 1 y2is continuous there must be an interval on which1 y2is never zero. Since division by1 y2is legitimate forxin this interval, we can separate variables in (2.2.18) to obtain

2y0

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x

Figure 2.2.3 A direction field and integral curves fory0_D2xy2

A partial fraction expansion on the left yields ₁ y 1 1 y_C1 y0_{D x;}

and integrating yields

ln ˇ ˇ ˇ ˇ y 1 y_C1 ˇ ˇ ˇ ˇD x2 2 CkI hence, ˇ ˇ ˇ ˇ y 1 y_C1 ˇ ˇ ˇ ˇD eke x2=2:

Sincey.x/¤ ˙1forxon the interval under discussion, the quantity.y 1/=.yC1/can’t change sign in this interval. Therefore we can rewrite the last equation as

y 1

y_C1 Dce x2₌₂

;

wherec_{D ˙}ek, depending upon the sign of.y 1/=.y_C1/on the interval. Solving foryyields y _D 1Cce

x2₌₂

1 ce x2₌₂: (2.2.19)

We’ve now shown that ifyis a solution of (2.2.18) that is not identically equal to_˙1, thenymust be as in (2.2.19). By substituting (2.2.19) into (2.2.18) you can verify that (2.2.19) is a solution of (2.2.18). Thus, the solutions of (2.2.18) arey 1,y 1and the functions of the form (2.2.19). Note that the

Section 2.2Separable Equations 51 constant solutiony 1can be obtained from this formula by takingcD0; however, the other constant solution,y 1, can’t be obtained in this way.

Figure2.2.4shows a direction field and some integrals for (2.2.18).

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −3 −2 −1 0 1 2 3 x y

Figure 2.2.4 A direction field and integral curves fory0_D x.1 y 2_/ 2

Differences Between Linear and Nonlinear Equations

Theorem2.1.2states that ifpandf are continuous on.a; b/then every solution of y0_{Cp.x/y Df .x/}

on.a; b/can be obtained by choosing a value for the constantcin the general solution, and ifx0is any point in.a; b/andy0is arbitrary, then the initial value problem

y0_{Cp.x/yDf .x/; y.x}

0/Dy0 has a solution on.a; b/.

The not true for nonlinear equations. First, we saw in Examples 2.2.4and 2.2.5 that a nonlinear equation may have solutions that can’t be obtained by choosing a specific value of a constant appearing in a one-parameter family of solutions. Second, it is in general impossible to determine the interval of validity of a solution to an initial value problem for a nonlinear equation by simply examining the equation, since the interval of validity may depend on the initial condition. For instance, in Example2.2.2

we saw that the solution of

dy dx D

x

y; y.x0/Dy0 is valid on. a; a/, whereaDqx₀2Cy₀2.

Example 2.2.6 Solve the initial value problem

y0D2xy2; y.0/Dy0 and determine the interval of validity of the solution.

Solution First supposey0¤0. From Example2.2.4, we know thatymust be of the form

yD 1

x2_Cc: (2.2.20)

Imposing the initial condition shows thatc _D 1=y0. Substituting this into (2.2.20) and rearranging terms yields the solution

y_D y0 1 y0x2

:

This is also the solution ify0 D 0. Ify0 < 0, the denominator isn’t zero for any value ofx, so the the solution is valid on. ₁;₁/. Ify0> 0, the solution is valid only on. 1=py0; 1=py0/.

2.2 Exercises

In Exercises1–6find all solutions.

1. y0_D 3x

2_C2xC1

y 2 2. .sinx/.siny/C.cosy/y 0 _D0

3. xy0_Cy2_{CyD0 4. y}0_{lnjyj Cx}2_yD0 5. .3y3_C3ycosy_C1/y0_C .2xC1/y

1_Cx2 D0 6. x2_yy0 _D.y2 _1/3=2

In Exercises7–10find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region. 7. C/G y0_Dx2_.1 Cy2/_{I f} 1x1; 1y1_g 8. C/G y0_.1Cx2_{/CxyD0I f 2x2; 1y 1g} 9. C/G y0_{D.x 1/.y 1/.y 2/I f 2x2; 3y 3g} 10. C/G .y 1/2y0 _{D2xC3I f 2x2; 2y5g}

In Exercises11and12solve the initial value problem.

11. y0_D x

2_C3xC2

y 2 ; y.1/D4 12. y0_Cx.y2_{Cy/D0; y.2/D1}

In Exercises13-16solve the initial value problem and graph the solution.

Section 2.2Separable Equations 53 14. C/G y0_C.yC1/.y 1/.y 2/ xC1 D0; y.1/D0 15. C/G y0_{C2x.yC1/D0; y.0/D2} 16. C/G y0_D2xy.1Cy2_{/; y.0/} D1

In Exercises17–23solve the initial value problem and find the interval of validity of the solution.

17. y0_.x2 C2/_C4x.y2_C2y_C1/_D0; y.1/_D 1 18. y0_{D 2x.y}2 _3y C2/; y.0/_D3 19. y0_D 2x 1_C2y; y.2/D0 20. y 0_{D2y y}2_{; y.0/D1} 21. x_Cyy0_{D0; y.3/D 4} 22. y0_Cx2_{.yC1/.y 2/}2_{D0; y.4/D2} 23. .x_C1/.x 2/y0_{CyD0; y.1/D 3} 24. Solvey0D .1Cy 2_/

.1_Cx2_/ explicitly. HINT:Use the identitytan.ACB/D

tanA_CtanB 1 tanAtanB. 25. Solvey0p1 x2_Cp

1 y2_{D0explicitly. HINT:Use the identitysin.A B/DsinAcosB} cosAsinB.

26. Solvey0 D cosx

siny; y. /D

2 explicitly. HINT:Use the identitycos.xC=2/D sinx and

the periodicity of the cosine.

27. Solve the initial value problem

y0_{Day by}2_{; y.0/}

Dy0: Discuss the behavior of the solution if(a)y00;(b)y0 < 0. 28. The populationP _DP .t /of a species satisfies the logistic equation

P0_{DaP .1 ˛P /}

andP .0/_DP0> 0. FindP fort > 0, and find limt!1P .t /.

29. An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if Sdenotes the total population of susceptible people andI _DI.t /denotes the number of infected people at timet, then

I0_{DrI.S I /;}

wherer is a positive constant. Assuming thatI.0/ _D I0, findI.t / fort > 0, and show that limt!1I.t /DS.

30. L The result of Exercise29is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise29, but there’s a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes

I0_{DrI.S I / qI .A/}

(a) ChooserandSpositive. By plotting direction fields and solutions of (A) on suitable rectan- gular grids

RD f0t T; 0I dg

in the.t; I /-plane, verify that ifIis any solution of (A) such thatI.0/ > 0, then limt!1I.t /D S q=rifq < rS and limt!1I.t /D0ifqrS.

(b) To verify the experimental results of(a), use separation of variables to solve (A) with initial conditionI.0/_D I0 > 0, and find limt!1I.t /. HINT:There are three cases to consider:

(i)q < rS;(ii)q > rS;(iii)q_DrS.

31. L Consider the differential equation

y0Day by2 q; .A/

wherea,bare positive constants, andqis an arbitrary constant. Supposeydenotes a solution of this equation that satisfies the initial conditiony.0/_Dy0.

(a) Chooseaandbpositive andq < a2=4b. By plotting direction fields and solutions of (A) on suitable rectangular grids

R_{D f}0t T; cy d_g .B/

in the .t; y/-plane, discover that there are numbersy1 andy2 withy1 < y2 such that if y0> y1then limt!1y.t /Dy2, and ify0< y1theny.t /D 1for some finite value oft. (What happens ify0Dy1?)

(b) Chooseaandbpositive andq _D a2=4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that there’s a numbery1 such that if y0 y1then limt!1y.t / Dy1, while ify0 < y1theny.t /D 1for some finite value oft.

(c) Choose positivea,bandq > a2_{=4b. By plotting direction fields and solutions of (A) on} suitable rectangular grids of the form (B), discover that no matter whaty0 is,y.t / D 1 for some finite value oft.

(d) Verify your results experiments analytically. Start by separating variables in (A) to obtain y0

ay by2 _q D1:

To decide what to do next you’ll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there!

Because of its role in the transition between these three cases, q0 D a2=4b is called a bifurcation valueofq. In general, ifqis a parameter in any differential equation,q0is said to be a bifurcation value ofqif the nature of the solutions of the equation withq < q0is qualitatively different from the nature of the solutions withq > q0.

32. L By plotting direction fields and solutions of

y0_{Dqy y}3_;

convince yourself thatq0 D 0is a bifurcation value ofqfor this equation. Explain what makes you draw this conclusion.

33. Suppose a disease spreads according to the model of Exercise29, but there’s a medication that cures the infected population at a constant rate ofqindividuals per unit time, whereq > 0. Then the equation for the number of infected individuals becomes

I0_{DrI.S I / q:}

Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 55 34. Assuming thatp60, state conditions under which the linear equation

y0Cp.x/y Df .x/

is separable. If the equation satisfies these conditions, solve it by separation of variables and by the method developed in Section 2.1.

Solve the equations in Exercises35–38using variation of parameters followed by separation of variables.

35. y0_{Cy D} 2xe x 1_Cyex 36. xy 0 _2yD x 6 y_Cx2 37. y0 _{y D} .xC1/e 4x .yCex_/2 38. y0 2yD xe2x 1 ye 2x

39. Use variation of parameters to show that the solutions of the following equations are of the form yDuy1, whereusatisfies a separable equationu0Dg.x/p.u/. Findy1andgfor each equation. (a)xy0_{CyDh.x/p.xy/ (b)xy}0 _yDh.x/py x (c)y0_{Cy Dh.x/p.e}x_{y/ (d)xy}0_CryDh.x/p.xr_y/ (e)y0_Cv0.x/ v.x/y Dh.x/p .v.x/y/

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