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# 5.6E: Reduction of Order (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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## Q5.6.1

In Exercises 5.6.1-5.6.17 find the general solution, given that $$y_1$$ satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation.

1. $$(2x+1)y''-2y'-(2x+3)y=(2x+1)^2; \quad y_1=e^{-x}$$

2. $$x^2y''+xy'-y={4\over x^2}; \quad y_1=x$$

3. $$x^2y''-xy'+y=x; \quad y_1=x$$

4. $$y''-3y'+2y={1\over1+e^{-x}}; \quad y_1=e^{2x}$$

5. $$y''-2y'+y=7x^{3/2}e^x; \quad y_1=e^x$$

6. $$4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x(1+4x); \quad y_1=x^{1/2}e^x$$

7. $$y''-2y'+2y=e^x\sec x; \quad y_1=e^x\cos x$$

8. $$y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}; \quad y_1=e^{-x^2}$$

9. $$x^2y''+xy'-4y=-6x-4; \quad y_1=x^2$$

10. $$x^2y''+2x(x-1)y'+(x^2-2x+2)y=x^3e^{2x}; \quad y_1=xe^{-x}$$

11. $$x^2y''-x(2x-1)y'+(x^2-x-1)y=x^2e^x; \quad y_1=xe^x$$

12. $$(1-2x)y''+2y'+(2x-3)y=(1-4x+4x^2)e^x; \quad y_1=e^x$$

13. $$x^2y''-3xy'+4y=4x^4; \quad y_1=x^2$$

14. $$2xy''+(4x+1)y'+(2x+1)y=3x^{1/2}e^{-x}; \quad y_1=e^{-x}$$

15. $$xy''-(2x+1)y'+(x+1)y=-e^x; \quad y_1=e^x$$

16. $$4x^2y''-4x(x+1)y'+(2x+3)y=4x^{5/2}e^{2x}; \quad y_1=x^{1/2}$$

17. $$x^2y''-5xy'+8y=4x^2; \quad y_1=x^2$$

## Q5.6.2

In Exercises 5.6.18-5.6.30 find a fundamental set of solutions, given that $$y_{1}$$ is a solution.

18. $$xy''+(2-2x)y'+(x-2)y=0; \quad y_1=e^x$$

19. $$x^2y''-4xy'+6y=0; \quad y_1=x^2$$

20. $$x^2(\ln |x|)^2y''-(2x \ln |x|)y'+(2+\ln |x|)y=0; \quad y_1=\ln |x|$$

21. $$4xy''+2y'+y=0; \quad y_1=\sin \sqrt{x}$$

22. $$xy''-(2x+2)y'+(x+2)y=0; \quad y_1=e^x$$

23. $$x^2y''-(2a-1)xy'+a^2y=0; \quad y_1=x^a$$

24. $$x^2y''-2xy'+(x^2+2)y=0; \quad y_1=x \sin x$$

25. $$xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}$$

26. $$4x^2(\sin x)y''-4x(x\cos x+\sin x)y'+(2x\cos x+3\sin x)y=0; \quad y_1=x^{1/2}$$

27. $$4x^2y''-4xy'+(3-16x^2)y=0; \quad y_1=x^{1/2}e^{2x}$$

28. $$(2x+1)xy''-2(2x^2-1)y'-4(x+1)y=0; \quad y_1=1/x$$

29. $$(x^2-2x)y''+(2-x^2)y'+(2x-2)y=0; \quad y_1=e^x$$

30. $$xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}$$

## Q5.6.3

In Exercises 5.6.31-5.6.33 solve the initial value problem, given that $$y_{1}$$ satisfies the complementary equation.

31. $$x^2y''-3xy'+4y=4x^4,\quad y(-1)=7,\quad y'(-1)=-8; \quad y_1=x^2$$

32. $$(3x-1)y''-(3x+2)y'-(6x-8)y=0, \quad y(0)=2,\; y'(0)=3; \quad y_1=e^{2x}$$

33. $$(x+1)^2y''-2(x+1)y'-(x^2+2x-1)y=(x+1)^3e^x, \quad y(0)=1,\quad y'(0)=~-1; \quad y_1=(x+1)e^x$$

## Q5.6.4

In Exercises 5.6.34 and 5.6.35 solve the initial value problem and graph the solution, given that $$y_{1}$$ satisfies the complementary equation.

34. $$x^2y''+2xy'-2y=x^2, \quad y(1)={5\over4},\; y'(1)={3\over2}; \quad y_1=x$$

35. $$(x^2-4)y''+4xy'+2y=x+2, \quad y(0)=-{1\over3},\quad y'(0)=-1; \quad y_1={1\over x-2}$$

## Q5.6.5

36. Suppose $$p_1$$ and $$p_2$$ are continuous on $$(a,b)$$. Let $$y_1$$ be a solution of

$y''+p_1(x)y'+p_2(x)y=0 \tag{A}$

that has no zeros on $$(a,b)$$, and let $$x_0$$ be in $$(a,b)$$. Use reduction of order to show that $$y_1$$ and

$y_2(x)=y_1(x)\int^x_{x_0}{1\over y^2_1(t)} \exp \left(-\int^t_{x_0}p_1(s)\, ds\right)\,dt$

form a fundamental set of solutions of (A) on $$(a,b)$$.

37. The nonlinear first order equation

$y'+y^2+p(x)y+q(x)=0 \tag{A}$

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 $z''+p(x)z'+q(x)z=0. \tag{B}$
2. Show that the general solution of (A) is $y={c_1z'_1+c_2z'_2\over c_1z_1+c_2z_2}, \tag{C}$ where $$\{z_1,z_2\}$$ is a fundamental set of solutions of (B) and $$c_1$$ and $$c_2$$ 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. $$y'+y^2+k^2=0$$
2. $$y'+y^2-3y+2=0$$
3. $$y'+y^2+5y-6=0$$
4. $$y'+y^2+8y+7=0$$
5. $$y'+y^2+14y+50=0$$
6. $$6y'+6y^2-y-1=0$$
7. $$36y'+36y^2-12y+1=0$$

39. Use a method suggested by Exercise 5.6.37 and reduction of order to find all solutions of the equation, given that $$y_1$$ is a solution.

1. $$x^2(y'+y^2)-x(x+2)y+x+2=0; \quad y_1=1/x$$
2. $$y'+y^2+4xy+4x^2+2=0; \quad y_1=-2x$$
3. $$(2x+1)(y'+y^2)-2y-(2x+3)=0; \quad y_1=-1$$
4. $$(3x-1)(y'+y^2)-(3x+2)y-6x+8=0; \quad y_1=2$$
5. $${x^2(y'+y^2)+xy+x^2- {1\over 4}=0; \quad y_1=-\tan x -{1\over 2x}}$$
6. $${x^2(y'+y^2)-7xy+7=0; \quad y_1=1/x}$$

40. The nonlinear first order equation

$y'+r(x)y^2+p(x)y+q(x)=0 \tag{A}$

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 $z''+\left[p(x)-{r'(x)\over r(x)}\right] z'+r(x)q(x)z=0. \tag{B}$
2. Show that the general solution of (A) is $y={c_1z'_1+c_2z'_2\over r(c_1z_1+c_2z_2)},$ where $$\{z_1,z_2\}$$ is a fundamental set of solutions of (B) and $$c_1$$ and $$c_2$$ are arbitrary constants.