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5.7E: Variation of Parameters (Exercises)

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Q5.7.1

In Exercises 5.7.1–5.7.6 use variation of parameters to find a particular solution.

1. y+9y=tan3x

2. y+4y=sin2xsec22x

3. y3y+2y=41+ex

4. y2y+2y=3exsecx

5. y2y+y=14x3/2ex

6. yy=4ex1e2x

Q5.7.2

In Exercises 5.7.7-5.7.29 use variation of parameters to find a particular solution, given the solutions y1,y2 of the complementary equation.

7. x2y+xyy=2x2+2;y1=x,y2=1x

8. xy+(22x)y+(x2)y=e2x;y1=ex,y2=exx

9. 4x2y+(4x8x2)y+(4x24x1)y=4x1/2ex,x>0; y1=x1/2ex,y2=x1/2ex

10. y+4xy+(4x2+2)y=4ex(x+2);y1=ex2,y2=xex2

11. x2y4xy+6y=x5/2,x>0;y1=x2,y2=x3

12. x2y3xy+3y=2x4sinx;y1=x,y2=x3

13. (2x+1)y2y(2x+3)y=(2x+1)2ex;y1=ex,y2=xex

14. 4xy+2y+y=sinx;y1=cosx,y2=sinx

15. xy(2x+2)y+(x+2)y=6x3ex;y1=ex,y2=x3ex

16. x2y(2a1)xy+a2y=xa+1;y1=xa,y2=xalnx

17. x2y2xy+(x2+2)y=x3cosx;y1=xcosx,y2=xsinx

18. xyy4x3y=8x5;y1=ex2,y2=ex2

19. (sinx)y+(2sinxcosx)y+(sinxcosx)y=ex;y1=ex,y2=excosx

20. 4x2y4xy+(316x2)y=8x5/2;y1=xe2x,y2=xe2x

21. 4x2y4xy+(4x2+3)y=x7/2;y1=xsinx,y2=xcosx

22. x2y2xy(x22)y=3x4;y1=xex,y2=xex

23. x2y2x(x+1)y+(x2+2x+2)y=x3ex;y1=xex,y2=x2ex

24. x2yxy3y=x3/2;y1=1/x,y2=x3

25. x2yx(x+4)y+2(x+3)y=x4ex;y1=x2,y2=x2ex

26. x2y2x(x+2)y+(x2+4x+6)y=2xex;y1=x2ex,y2=x3ex

27. x2y4xy+(x2+6)y=x4;y1=x2cosx,y2=x2sinx

28. (x1)yxy+y=2(x1)2ex;y1=x,y2=ex

29. 4x2y4x(x+1)y+(2x+3)y=x5/2ex;y1=x,y2=xex

Q5.7.3

In Exercises 5.7.30–5.7.32 use variation of parameters to solve the initial value problem, given y1, y2 are solutions of the complementary equation.

30. (3x1)y(3x+2)y(6x8)y=(3x1)2e2x,y(0)=1,y(0)=2; y1=e2x,y2=xex

31. (x1)2y2(x1)y+2y=(x1)2,y(0)=3,y(0)=6;

y1=x1, y2=x21

32. (x1)2y(x21)y+(x+1)y=(x1)3ex,y(0)=4,y(0)=6;

y1=(x1)ex,y2=x1

Q5.7.4

In Exercises 5.7.33-5.7.35 use variation of parameters to solve the initial value problem and graph the solution, given that y1, y2 are solutions of the complementary equation.

33. (x21)y+4xy+2y=2x,y(0)=0,y(0)=2;y1=1x1,y2=1x+1

34. x2y+2xy2y=2x2,y(1)=1,y(1)=1;y1=x,y2=1x2

35. (x+1)(2x+3)y+2(x+2)y2y=(2x+3)2,y(0)=0,y(0)=0; y1=x+2,y2=1x+1

Q5.7.5

36. Suppose

yp=¯y+a1y1+a2y2

is a particular solution of

P0(x)y+P1(x)y+P2(x)y=F(x),

where y1 and y2 are solutions of the complementary equation

P0(x)y+P1(x)y+P2(x)y=0.

Show that ¯y is also a solution of (A).

37. Suppose p, q, and f are continuous on (a,b) and let x0 be in (a,b). Let y1 and y2 be the solutions of

y+p(x)y+q(x)y=0

such that

y1(x0)=1,y1(x0)=0,y2(x0)=0,y2(x0)=1.

Use variation of parameters to show that the solution of the initial value problem

y+p(x)y+q(x)y=f(x),y(x0)=k0,y(x0)=k1,

is

y(x)=k0y1(x)+k1y2(x)+xx0(y1(t)y2(x)y1(x)y2(t))f(t)exp(tx0p(s)ds)dt.

HINT: Use Abel's formula for the Wronskian of {y1,y2}, and integrate u1 and u2 from x0 to x.

Show also that

y(x)=k0y1(x)+k1y2(x)+xx0(y1(t)y2(x)y1(x)y2(t))f(t)exp(tx0p(s)ds)dt.

38. Suppose f is continuous on an open interval that contains x0=0. Use variation of parameters to find a formula for the solution of the initial value problem

yy=f(x),y(0)=k0,y(0)=k1.

39. Suppose f is continuous on (a,), where a<0, so x0=0 is in (a,).

  1. Use variation of parameters to find a formula for the solution of the initial value problem y+y=f(x),y(0)=k0,y(0)=k1. HINT: You will need the addition formulas for the sine and cosine. sin(A+B)=sinAcosB+cosAsinBcos(A+B)=cosAcosBsinAsinB For the rest of this exercise assume that the improper integral 0f(t)dt is absolutely convergent.
  2. Show that if y is a solution of y+y=f(x) on (a,), then limx(y(x)A0cosxA1sinx)=0 and limx(y(x)+A0sinxA1cosx)=0, where A0=k00f(t)sintdtandA1=k1+0f(t)costdt. HINT: Recall from calculus that if 0f(t)dt converges absolutely, then limxx|f(t)|dt=0.
  3. Show that if A0 and A1 are arbitrary constants, then there’s a unique solution of y+y=f(x) on (a,) that satisfies (B) and (C).

This page titled 5.7E: Variation of Parameters (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.

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