5.6E: Reduction of Order (Exercises)
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Q5.6.1
In Exercises 5.6.1-5.6.17 find the general solution, given that y1 satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation.
1. (2x+1)y″−2y′−(2x+3)y=(2x+1)2;y1=e−x
2. x2y″+xy′−y=4x2;y1=x
3. x2y″−xy′+y=x;y1=x
4. y″−3y′+2y=11+e−x;y1=e2x
5. y″−2y′+y=7x3/2ex;y1=ex
6. 4x2y″+(4x−8x2)y′+(4x2−4x−1)y=4x1/2ex(1+4x);y1=x1/2ex
7. y″−2y′+2y=exsecx;y1=excosx
8. y″+4xy′+(4x2+2)y=8e−x(x+2);y1=e−x2
9. x2y″+xy′−4y=−6x−4;y1=x2
10. x2y″+2x(x−1)y′+(x2−2x+2)y=x3e2x;y1=xe−x
11. x2y″−x(2x−1)y′+(x2−x−1)y=x2ex;y1=xex
12. (1−2x)y″+2y′+(2x−3)y=(1−4x+4x2)ex;y1=ex
13. x2y″−3xy′+4y=4x4;y1=x2
14. 2xy″+(4x+1)y′+(2x+1)y=3x1/2e−x;y1=e−x
15. xy″−(2x+1)y′+(x+1)y=−ex;y1=ex
16. 4x2y″−4x(x+1)y′+(2x+3)y=4x5/2e2x;y1=x1/2
17. x2y″−5xy′+8y=4x2;y1=x2
Q5.6.2
In Exercises 5.6.18-5.6.30 find a fundamental set of solutions, given that y1 is a solution.
18. xy″+(2−2x)y′+(x−2)y=0;y1=ex
19. x2y″−4xy′+6y=0;y1=x2
20. x2(ln|x|)2y″−(2xln|x|)y′+(2+ln|x|)y=0;y1=ln|x|
21. 4xy″+2y′+y=0;y1=sin√x
22. xy″−(2x+2)y′+(x+2)y=0;y1=ex
23. x2y″−(2a−1)xy′+a2y=0;y1=xa
24. x2y″−2xy′+(x2+2)y=0;y1=xsinx
25. xy″−(4x+1)y′+(4x+2)y=0;y1=e2x
26. 4x2(sinx)y″−4x(xcosx+sinx)y′+(2xcosx+3sinx)y=0;y1=x1/2
27. 4x2y″−4xy′+(3−16x2)y=0;y1=x1/2e2x
28. (2x+1)xy″−2(2x2−1)y′−4(x+1)y=0;y1=1/x
29. (x2−2x)y″+(2−x2)y′+(2x−2)y=0;y1=ex
30. xy″−(4x+1)y′+(4x+2)y=0;y1=e2x
Q5.6.3
In Exercises 5.6.31-5.6.33 solve the initial value problem, given that y1 satisfies the complementary equation.
31. x2y″−3xy′+4y=4x4,y(−1)=7,y′(−1)=−8;y1=x2
32. (3x−1)y″−(3x+2)y′−(6x−8)y=0,y(0)=2,y′(0)=3;y1=e2x
33. (x+1)2y″−2(x+1)y′−(x2+2x−1)y=(x+1)3ex,y(0)=1,y′(0)= −1;y1=(x+1)ex
Q5.6.4
In Exercises 5.6.34 and 5.6.35 solve the initial value problem and graph the solution, given that y1 satisfies the complementary equation.
34. x2y″+2xy′−2y=x2,y(1)=54,y′(1)=32;y1=x
35. (x2−4)y″+4xy′+2y=x+2,y(0)=−13,y′(0)=−1;y1=1x−2
Q5.6.5
36. Suppose p1 and p2 are continuous on (a,b). Let y1 be a solution of
y″+p1(x)y′+p2(x)y=0
that has no zeros on (a,b), and let x0 be in (a,b). Use reduction of order to show that y1 and
y2(x)=y1(x)∫xx01y21(t)exp(−∫tx0p1(s)ds)dt
form a fundamental set of solutions of (A) on (a,b).
37. The nonlinear first order equation
y′+y2+p(x)y+q(x)=0
is a Riccati equation. (See Exercise 2.4.55.) Assume that p and q are continuous.
- Show that y is a solution of (A) if and only if y=z′/z, where z″+p(x)z′+q(x)z=0.
- Show that the general solution of (A) is y=c1z′1+c2z′2c1z1+c2z2, where {z1,z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.
- Does the formula (C) imply that the first order equation (A) has a two–parameter family of solutions? Explain your answer.
38. Use a method suggested by Exercise 5.6.37 to find all solutions. of the equation.
- y′+y2+k2=0
- y′+y2−3y+2=0
- y′+y2+5y−6=0
- y′+y2+8y+7=0
- y′+y2+14y+50=0
- 6y′+6y2−y−1=0
- 36y′+36y2−12y+1=0
39. Use a method suggested by Exercise 5.6.37 and reduction of order to find all solutions of the equation, given that y1 is a solution.
- x2(y′+y2)−x(x+2)y+x+2=0;y1=1/x
- y′+y2+4xy+4x2+2=0;y1=−2x
- (2x+1)(y′+y2)−2y−(2x+3)=0;y1=−1
- (3x−1)(y′+y2)−(3x+2)y−6x+8=0;y1=2
- x2(y′+y2)+xy+x2−14=0;y1=−tanx−12x
- x2(y′+y2)−7xy+7=0;y1=1/x
40. The nonlinear first order equation
y′+r(x)y2+p(x)y+q(x)=0
is the generalized Riccati equation. (See Exercise 2.4.55.) Assume that p and q are continuous and r is differentiable.
- Show that y is a solution of (A) if and only if y=z′/rz, where z″+[p(x)−r′(x)r(x)]z′+r(x)q(x)z=0.
- Show that the general solution of (A) is y=c1z′1+c2z′2r(c1z1+c2z2), where {z1,z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.