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5.5E: The Method of Undetermined Coefficients II (Exercises)

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Q5.5.1

In Exercises 5.5.1-5.5.17 find a particular solution.

1. y+3y+2y=7cosxsinx

2. y+3y+y=(26x)cosx9sinx

3. y+2y+y=ex(6cosx+17sinx)

4. y+3y2y=e2x(5cos2x+9sin2x)

5. yy+y=ex(2+x)sinx

6. y+3y2y=e2x[(4+20x)cos3x+(2632x)sin3x]

7. y+4y=12cos2x4sin2x

8. y+y=(4+8x)cosx+(84x)sinx

9. 4y+y=4cosx/28xsinx/2

10. y+2y+2y=ex(8cosx6sinx)

11. y2y+5y=ex[(6+8x)cos2x+(68x)sin2x]

12. y+2y+y=8x2cosx4xsinx

13. y+3y+2y=(12+20x+10x2)cosx+8xsinx

14. y+3y+2y=(1x4x2)cos2x(1+7x+2x2)sin2x

15. y5y+6y=ex[(4+6xx2)cosx(24x+3x2)sinx]

16. y2y+y=ex[(3+4xx2)cosx+(34xx2)sinx]

17. y2y+2y=ex[(22x6x2)cosx+(210x+6x2)sinx]

Q5.5.2

In Exercises 5.5.18-5.5.21 find a particular solution and graph it.

18. y+2y+y=ex[(52x)cosx(3+3x)sinx]

19. y+9y=6cos3x12sin3x

20. y+3y+2y=(1x4x2)cos2x(1+7x+2x2)sin2x

21. y+4y+3y=ex[(2+x+x2)cosx+(5+4x+2x2)sinx]

Q5.5.3

In Exercises 5.5.22-5.5.26 solve the initial value problem.

22. y7y+6y=ex(17cosx7sinx),y(0)=4,y(0)=2

23. y2y+2y=ex(6cosx+4sinx),y(0)=1,y(0)=4

24. y+6y+10y=40exsinx,y(0)=2,y(0)=3

25. y6y+10y=e3x(6cosx+4sinx),y(0)=2,y(0)=7

26. y3y+2y=e3x[21cosx(11+10x)sinx],y(0)=0,y(0)=6

Q5.5.4

In Exercises 5.5.27-5.5.32 use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem.

27. y2y3y=4e3x+ex(cosx2sinx)

28. y+y=4cosx2sinx+xex+ex

29. y3y+2y=xex+2e2x+sinx

30. y2y+2y=4xexcosx+xex+1+x2

31. y4y+4y=e2x(1+x)+e2x(cosxsinx)+3e3x+1+x

32. y4y+4y=6e2x+25sinx,y(0)=5,y(0)=3

Q5.5.5

In Exercises 5.5.33-5.5.35 solve the initial value problem and graph the solution.

33. y+4y=e2x[(47x)cosx+(24x)sinx],y(0)=3,y(0)=1

34. y+4y+4y=2cos2x+3sin2x+ex,y(0)=1,y(0)=2

35. y+4y=ex(11+15x)+8cos2x12sin2x,y(0)=3,y(0)=5

Q5.5.6

36.

  1. Verify that if yp=A(x)cosωx+B(x)sinωx where A and B are twice differentiable, then yp=(A+ωB)cosωx+(BωA)sinωx andyp=(A+2ωBω2A)cosωx+(B2ωAω2B)sinωx.
  2. Use the results of (a) to verify that ayp+byp+cyp=[(caω2)A+bωB+2aωB+bA+aA]cosωx+[bωA+(caω2)B2aωA+bB+aB]sinωx.
  3. Use the results of (a) to verify that yp+ω2yp=(A+2ωB)cosωx+(B2ωA)sinωx.
  4. Prove Theorem 5.5.2.

37. Let a, b, c, and ω be constants, with a0 and ω>0, and let

P(x)=p0+p1x++pkxkandQ(x)=q0+q1x++qkxk,

where at least one of the coefficients pk, qk is nonzero, so k is the larger of the degrees of P and Q.

  1. Show that if cosωx and sinωx are not solutions of the complementary equation ay+by+cy=0, then there are polynomials A(x)=A0+A1x++AkxkandB(x)=B0+B1x++Bkxk such that (caω2)A+bωB+2aωB+bA+aA=P.bωA+(caω2)B2aωA+bB+aB=Q, where (Ak,Bk), (Ak1,Bk1), …,(A0,B0) can be computed successively by solving the systems (caω2)Ak+bωBk=pk.bωAk+(caω2)Bk=qk, and, if 1rk, (caω2)Akr+bωBkr=pkr+.bωAkr+(caω2)Bkr=qkr+, where the terms indicated by “” depend upon the previously computed coefficients with subscripts greater than kr. Conclude from this and Exercise 5.5.36b that yp=A(x)cosωx+B(x)sinωx is a particular solution of ay+by+cy=P(x)cosωx+Q(x)sinωx.
  2. Conclude from Exercise 5.5.36c that the equation a(y+ω2y)=P(x)cosωx+Q(x)sinωx does not have a solution of the form (B) with A and B as in (A). Then show that there are polynomials A(x)=A0x+A1x2++Akxk+1andB(x)=B0x+B1x2++Bkxk+1 such that a(A+2ωB)=Pa(B2ωA)=Q, where the pairs (Ak,Bk), (Ak1,Bk1), …, (A0,B0) can be computed successively as follows: Ak=qk2aω(k+1)Bk=pk2aω(k+1), and, if k1, Akj=12ω[qkja(kj+1)(kj+2)Bkj+1]Bkj=12ω[pkja(kj+1)(kj+2)Akj+1] for 1jk. Conclude that (B) with this choice of the polynomials A and B is a particular solution of (C).

38. Show that Theorem 5.5.1 implies the next theorem:

Theorem 5.5E.1

Suppose ω is a positive number and P and Q are polynomials. Let k be the larger of the degrees of P and Q. Then the equation

ay+by+cy=eλx(P(x)cosωx+Q(x)sinωx)

has a particular solution

yp=eλx(A(x)cosωx+B(x)sinωx),

where

A(x)=A0+A1x++AkxkandB(x)=B0+B1x++Bkxk,

provided that eλxcosωx and eλxsinωx are not solutions of the complementary equation. The equation

a[y2λy+(λ2+ω2)y]=eλx(P(x)cosωx+Q(x)sinωx)

(for which eλxcosωx and eλxsinωx are solutions of the complementary equation) has a particular solution of the form (A), where

A(x)=A0x+A1x2++Akxk+1andB(x)=B0x+B1x2++Bkxk+1.

39. This exercise presents a method for evaluating the integral

y=eλx(P(x)cosωx+Q(x)sinωx)dx

where ω0 and

P(x)=p0+p1x++pkxk,Q(x)=q0+q1x++qkxk.

  1. Show that y=eλxu, where u+λu=P(x)cosωx+Q(x)sinωx.
  2. Show that (A) has a particular solution of the form up=A(x)cosωx+B(x)sinωx, where A(x)=A0+A1x++Akxk,B(x)=B0+B1x++Bkxk, and the pairs of coefficients (Ak,Bk), (Ak1,Bk1), …,(A0,B0) can be computed successively as the solutions of pairs of equations obtained by equating the coefficients of xrcosωx and xrsinωx for r=k, k1, …, 0.
  3. Conclude that eλx(P(x)cosωx+Q(x)sinωx)dx=eλx(A(x)cosωx+B(x)sinωx)+c, where c is a constant of integration.

40. Use the method of Exercise 5.5.39 to evaluate the integral.

  1. x2cosxdx
  2. x2excosxdx
  3. xexsin2xdx
  4. x2exsinxdx
  5. x3exsinxdx
  6. ex[xcosx(1+3x)sinx]dx
  7. ex[(1+x2)cosx+(12x)sinx]dx

This page titled 5.5E: The Method of Undetermined Coefficients II (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

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