5.7E: Variation of Parameters (Exercises)
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- Jul 20, 2020
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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. y″−3y′+2y=41+e−x
4. y″−2y′+2y=3exsecx
5. y″−2y′+y=14x3/2ex
6. y″−y=4e−x1−e−2x
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″+xy′−y=2x2+2;y1=x,y2=1x
8. xy″+(2−2x)y′+(x−2)y=e2x;y1=ex,y2=exx
9. 4x2y″+(4x−8x2)y′+(4x2−4x−1)y=4x1/2ex,x>0; y1=x1/2ex,y2=x−1/2ex
10. y″+4xy′+(4x2+2)y=4e−x(x+2);y1=e−x2,y2=xe−x2
11. x2y″−4xy′+6y=x5/2,x>0;y1=x2,y2=x3
12. x2y″−3xy′+3y=2x4sinx;y1=x,y2=x3
13. (2x+1)y″−2y′−(2x+3)y=(2x+1)2e−x;y1=e−x,y2=xex
14. 4xy″+2y′+y=sin√x;y1=cos√x,y2=sin√x
15. xy″−(2x+2)y′+(x+2)y=6x3ex;y1=ex,y2=x3ex
16. x2y″−(2a−1)xy′+a2y=xa+1;y1=xa,y2=xalnx
17. x2y″−2xy′+(x2+2)y=x3cosx;y1=xcosx,y2=xsinx
18. xy″−y′−4x3y=8x5;y1=ex2,y2=e−x2
19. (sinx)y″+(2sinx−cosx)y′+(sinx−cosx)y=e−x;y1=e−x,y2=e−xcosx
20. 4x2y″−4xy′+(3−16x2)y=8x5/2;y1=√xe2x,y2=√xe−2x
21. 4x2y″−4xy′+(4x2+3)y=x7/2;y1=√xsinx,y2=√xcosx
22. x2y″−2xy′−(x2−2)y=3x4;y1=xex,y2=xe−x
23. x2y″−2x(x+1)y′+(x2+2x+2)y=x3ex;y1=xex,y2=x2ex
24. x2y″−xy′−3y=x3/2;y1=1/x,y2=x3
25. x2y″−x(x+4)y′+2(x+3)y=x4ex;y1=x2,y2=x2ex
26. x2y″−2x(x+2)y′+(x2+4x+6)y=2xex;y1=x2ex,y2=x3ex
27. x2y″−4xy′+(x2+6)y=x4;y1=x2cosx,y2=x2sinx
28. (x−1)y″−xy′+y=2(x−1)2ex;y1=x,y2=ex
29. 4x2y″−4x(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. (3x−1)y″−(3x+2)y′−(6x−8)y=(3x−1)2e2x,y(0)=1,y′(0)=2; y1=e2x,y2=xe−x
31. (x−1)2y″−2(x−1)y′+2y=(x−1)2,y(0)=3,y′(0)=−6;
y1=x−1, y2=x2−1
32. (x−1)2y″−(x2−1)y′+(x+1)y=(x−1)3ex,y(0)=4,y′(0)=−6;
y1=(x−1)ex,y2=x−1
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. (x2−1)y″+4xy′+2y=2x,y(0)=0,y′(0)=−2;y1=1x−1,y2=1x+1
34. x2y″+2xy′−2y=−2x2,y(1)=1,y′(1)=−1;y1=x,y2=1x2
35. (x+1)(2x+3)y″+2(x+2)y′−2y=(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,y′1(x0)=0,y2(x0)=0,y′2(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 u′1 and u′2 from x0 to x.
Show also that
y′(x)=k0y′1(x)+k1y′2(x)+∫xx0(y1(t)y′2(x)−y′1(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
y″−y=f(x),y(0)=k0,y′(0)=k1.
39. Suppose f is continuous on (a,∞), where a<0, so x0=0 is in (a,∞).
- 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)=cosAcosB−sinAsinB For the rest of this exercise assume that the improper integral ∫∞0f(t)dt is absolutely convergent.
- Show that if y is a solution of y″+y=f(x) on (a,∞), then limx→∞(y(x)−A0cosx−A1sinx)=0 and limx→∞(y′(x)+A0sinx−A1cosx)=0, where A0=k0−∫∞0f(t)sintdtandA1=k1+∫∞0f(t)costdt. HINT: Recall from calculus that if ∫∞0f(t)dt converges absolutely, then limx→∞∫∞x|f(t)|dt=0.
- 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).