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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)y2y(2x+3)y=(2x+1)2;y1=ex

2. x2y+xyy=4x2;y1=x

3. x2yxy+y=x;y1=x

4. y3y+2y=11+ex;y1=e2x

5. y2y+y=7x3/2ex;y1=ex

6. 4x2y+(4x8x2)y+(4x24x1)y=4x1/2ex(1+4x);y1=x1/2ex

7. y2y+2y=exsecx;y1=excosx

8. y+4xy+(4x2+2)y=8ex(x+2);y1=ex2

9. x2y+xy4y=6x4;y1=x2

10. x2y+2x(x1)y+(x22x+2)y=x3e2x;y1=xex

11. x2yx(2x1)y+(x2x1)y=x2ex;y1=xex

12. (12x)y+2y+(2x3)y=(14x+4x2)ex;y1=ex

13. x2y3xy+4y=4x4;y1=x2

14. 2xy+(4x+1)y+(2x+1)y=3x1/2ex;y1=ex

15. xy(2x+1)y+(x+1)y=ex;y1=ex

16. 4x2y4x(x+1)y+(2x+3)y=4x5/2e2x;y1=x1/2

17. x2y5xy+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+(22x)y+(x2)y=0;y1=ex

19. x2y4xy+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=sinx

22. xy(2x+2)y+(x+2)y=0;y1=ex

23. x2y(2a1)xy+a2y=0;y1=xa

24. x2y2xy+(x2+2)y=0;y1=xsinx

25. xy(4x+1)y+(4x+2)y=0;y1=e2x

26. 4x2(sinx)y4x(xcosx+sinx)y+(2xcosx+3sinx)y=0;y1=x1/2

27. 4x2y4xy+(316x2)y=0;y1=x1/2e2x

28. (2x+1)xy2(2x21)y4(x+1)y=0;y1=1/x

29. (x22x)y+(2x2)y+(2x2)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. x2y3xy+4y=4x4,y(1)=7,y(1)=8;y1=x2

32. (3x1)y(3x+2)y(6x8)y=0,y(0)=2,y(0)=3;y1=e2x

33. (x+1)2y2(x+1)y(x2+2x1)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+2xy2y=x2,y(1)=54,y(1)=32;y1=x

35. (x24)y+4xy+2y=x+2,y(0)=13,y(0)=1;y1=1x2

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.

  1. Show that y is a solution of (A) if and only if y=z/z, where z+p(x)z+q(x)z=0.
  2. Show that the general solution of (A) is y=c1z1+c2z2c1z1+c2z2, where {z1,z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.
  3. 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.

  1. y+y2+k2=0
  2. y+y23y+2=0
  3. y+y2+5y6=0
  4. y+y2+8y+7=0
  5. y+y2+14y+50=0
  6. 6y+6y2y1=0
  7. 36y+36y212y+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.

  1. x2(y+y2)x(x+2)y+x+2=0;y1=1/x
  2. y+y2+4xy+4x2+2=0;y1=2x
  3. (2x+1)(y+y2)2y(2x+3)=0;y1=1
  4. (3x1)(y+y2)(3x+2)y6x+8=0;y1=2
  5. x2(y+y2)+xy+x214=0;y1=tanx12x
  6. 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.

  1. 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.
  2. Show that the general solution of (A) is y=c1z1+c2z2r(c1z1+c2z2), where {z1,z2} is a fundamental set of solutions of (B) and c1 and c2 are arbitrary constants.

This page titled 5.6E: Reduction of Order (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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