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=7cosx−sinx
2. y″+3y′+y=(2−6x)cosx−9sinx
3. y″+2y′+y=ex(6cosx+17sinx)
4. y″+3y′−2y=−e2x(5cos2x+9sin2x)
5. y″−y′+y=ex(2+x)sinx
6. y″+3y′−2y=e−2x[(4+20x)cos3x+(26−32x)sin3x]
7. y″+4y=−12cos2x−4sin2x
8. y″+y=(−4+8x)cosx+(8−4x)sinx
9. 4y″+y=−4cosx/2−8xsinx/2
10. y″+2y′+2y=e−x(8cosx−6sinx)
11. y″−2y′+5y=ex[(6+8x)cos2x+(6−8x)sin2x]
12. y″+2y′+y=8x2cosx−4xsinx
13. y″+3y′+2y=(12+20x+10x2)cosx+8xsinx
14. y″+3y′+2y=(1−x−4x2)cos2x−(1+7x+2x2)sin2x
15. y″−5y′+6y=−ex[(4+6x−x2)cosx−(2−4x+3x2)sinx]
16. y″−2y′+y=−ex[(3+4x−x2)cosx+(3−4x−x2)sinx]
17. y″−2y′+2y=ex[(2−2x−6x2)cosx+(2−10x+6x2)sinx]
Q5.5.2
In Exercises 5.5.18-5.5.21 find a particular solution and graph it.
18. y″+2y′+y=e−x[(5−2x)cosx−(3+3x)sinx]
19. y″+9y=−6cos3x−12sin3x
20. y″+3y′+2y=(1−x−4x2)cos2x−(1+7x+2x2)sin2x
21. y″+4y′+3y=e−x[(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. y″−7y′+6y=−ex(17cosx−7sinx),y(0)=4,y′(0)=2
23. y″−2y′+2y=−ex(6cosx+4sinx),y(0)=1,y′(0)=4
24. y″+6y′+10y=−40exsinx,y(0)=2,y′(0)=−3
25. y″−6y′+10y=−e3x(6cosx+4sinx),y(0)=2,y′(0)=7
26. y″−3y′+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. y″−2y′−3y=4e3x+ex(cosx−2sinx)
28. y″+y=4cosx−2sinx+xex+e−x
29. y″−3y′+2y=xex+2e2x+sinx
30. y″−2y′+2y=4xexcosx+xe−x+1+x2
31. y″−4y′+4y=e2x(1+x)+e2x(cosx−sinx)+3e3x+1+x
32. y″−4y′+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=−e−2x[(4−7x)cosx+(2−4x)sinx],y(0)=3,y′(0)=1
34. y″+4y′+4y=2cos2x+3sin2x+e−x,y(0)=−1,y′(0)=2
35. y″+4y=ex(11+15x)+8cos2x−12sin2x,y(0)=3,y′(0)=5
Q5.5.6
36.
- Verify that if yp=A(x)cosωx+B(x)sinωx where A and B are twice differentiable, then y′p=(A′+ωB)cosωx+(B′−ωA)sinωx andy″p=(A″+2ωB′−ω2A)cosωx+(B″−2ωA′−ω2B)sinωx.
- Use the results of (a) to verify that ay″p+by′p+cyp=[(c−aω2)A+bωB+2aωB′+bA′+aA″]cosωx+[−bωA+(c−aω2)B−2aωA′+bB′+aB″]sinωx.
- Use the results of (a) to verify that y″p+ω2yp=(A″+2ωB′)cosωx+(B″−2ωA′)sinωx.
- Prove Theorem 5.5.2.
37. Let a, b, c, and ω be constants, with a≠0 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.
- 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 (c−aω2)A+bωB+2aωB′+bA′+aA″=P.−bωA+(c−aω2)B−2aωA′+bB′+aB″=Q, where (Ak,Bk), (Ak−1,Bk−1), …,(A0,B0) can be computed successively by solving the systems −(c−aω2)Ak+bωBk=pk.−bωAk+(c−aω2)Bk=qk, and, if 1≤r≤k, −(c−aω2)Ak−r+bωBk−r=pk−r+⋯.−bωAk−r+(c−aω2)Bk−r=qk−r+⋯, where the terms indicated by “⋯” depend upon the previously computed coefficients with subscripts greater than k−r. 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.
- 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(B″−2ωA′)=Q, where the pairs (Ak,Bk), (Ak−1,Bk−1), …, (A0,B0) can be computed successively as follows: Ak=−qk2aω(k+1)Bk=−pk2aω(k+1), and, if k≥1, Ak−j=−12ω[qk−ja(k−j+1)−(k−j+2)Bk−j+1]Bk−j=−12ω[pk−ja(k−j+1)−(k−j+2)Ak−j+1] for 1≤j≤k. 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:
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[y″−2λy′+(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 ω≠0 and
P(x)=p0+p1x+⋯+pkxk,Q(x)=q0+q1x+⋯+qkxk.
- Show that y=eλxu, where u′+λu=P(x)cosωx+Q(x)sinωx.
- 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), (Ak−1,Bk−1), …,(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, k−1, …, 0.
- 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.
- ∫x2cosxdx
- ∫x2excosxdx
- ∫xe−xsin2xdx
- ∫x2e−xsinxdx
- ∫x3exsinxdx
- ∫ex[xcosx−(1+3x)sinx]dx
- ∫e−x[(1+x2)cosx+(12x)sinx]dx