5.6.1: Reduction of Order (Exercises)

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Jan 7, 2020
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- 5.6: Reduction of Order
- 5.7: Variation of Parameters

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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. $(2 x + 1) y^{″} - 2 y^{'} - (2 x + 3) y = {(2 x + 1)}^{2}; y_{1} = e^{- x}$

2. $x^{2} y^{″} + x y^{'} - y = \frac{4}{x^{2}}; y_{1} = x$

3. $x^{2} y^{″} - x y^{'} + y = x; y_{1} = x$

4. $y^{″} - 3 y^{'} + 2 y = \frac{1}{1 + e^{- x}}; y_{1} = e^{2 x}$

5. $y^{″} - 2 y^{'} + y = 7 x^{3 / 2} e^{x}; y_{1} = e^{x}$

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

7. $y^{″} - 2 y^{'} + 2 y = e^{x} \sec x; y_{1} = e^{x} \cos x$

8. $y^{″} + 4 x y^{'} + (4 x^{2} + 2) y = 8 e^{- x (x + 2)}; y_{1} = e^{- x^{2}}$

9. $x^{2} y^{″} + x y^{'} - 4 y = - 6 x - 4; y_{1} = x^{2}$

10. $x^{2} y^{″} + 2 x (x - 1) y^{'} + (x^{2} - 2 x + 2) y = x^{3} e^{2 x}; y_{1} = x e^{- x}$

11. $x^{2} y^{″} - x (2 x - 1) y^{'} + (x^{2} - x - 1) y = x^{2} e^{x}; y_{1} = x e^{x}$

12. $(1 - 2 x) y^{″} + 2 y^{'} + (2 x - 3) y = (1 - 4 x + 4 x^{2}) e^{x}; y_{1} = e^{x}$

13. $x^{2} y^{″} - 3 x y^{'} + 4 y = 4 x^{4}; y_{1} = x^{2}$

14. $2 x y^{″} + (4 x + 1) y^{'} + (2 x + 1) y = 3 x^{1 / 2} e^{- x}; y_{1} = e^{- x}$

15. $x y^{″} - (2 x + 1) y^{'} + (x + 1) y = - e^{x}; y_{1} = e^{x}$

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

17. $x^{2} y^{″} - 5 x y^{'} + 8 y = 4 x^{2}; y_{1} = x^{2}$

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. $x y^{″} + (2 - 2 x) y^{'} + (x - 2) y = 0; y_{1} = e^{x}$

19. $x^{2} y^{″} - 4 x y^{'} + 6 y = 0; y_{1} = x^{2}$

20. $x^{2} {(\ln | x |)}^{2} y^{″} - (2 x \ln | x |) y^{'} + (2 + \ln | x |) y = 0; y_{1} = \ln | x |$

21. $4 x y^{″} + 2 y^{'} + y = 0; y_{1} = \sin \sqrt{x}$

22. $x y^{″} - (2 x + 2) y^{'} + (x + 2) y = 0; y_{1} = e^{x}$

23. $x^{2} y^{″} - (2 a - 1) x y^{'} + a^{2} y = 0; y_{1} = x^{a}$

24. $x^{2} y^{″} - 2 x y^{'} + (x^{2} + 2) y = 0; y_{1} = x \sin x$

25. $x y^{″} - (4 x + 1) y^{'} + (4 x + 2) y = 0; y_{1} = e^{2 x}$

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

27. $4 x^{2} y^{″} - 4 x y^{'} + (3 - 16 x^{2}) y = 0; y_{1} = x^{1 / 2} e^{2 x}$

28. $(2 x + 1) x y^{″} - 2 (2 x^{2} - 1) y^{'} - 4 (x + 1) y = 0; y_{1} = 1 / x$

29. $(x^{2} - 2 x) y^{″} + (2 - x^{2}) y^{'} + (2 x - 2) y = 0; y_{1} = e^{x}$

30. $x y^{″} - (4 x + 1) y^{'} + (4 x + 2) y = 0; y_{1} = e^{2 x}$

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. $x^{2} y^{″} - 3 x y^{'} + 4 y = 4 x^{4}, y (- 1) = 7, y^{'} (- 1) = - 8; y_{1} = x^{2}$

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

33. ${(x + 1)}^{2} y^{″} - 2 (x + 1) y^{'} - (x^{2} + 2 x - 1) y = {(x + 1)}^{3} e^{x}, y (0) = 1, y^{'} (0) = - 1; y_{1} = (x + 1) e^{x}$

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. $x^{2} y^{″} + 2 x y^{'} - 2 y = x^{2}, y (1) = \frac{5}{4}, y^{'} (1) = \frac{3}{2}; y_{1} = x$

35. $(x^{2} - 4) y^{″} + 4 x y^{'} + 2 y = x + 2, y (0) = - \frac{1}{3}, y^{'} (0) = - 1; y_{1} = \frac{1}{x - 2}$

Q5.6.5

36. Suppose $p_{1}$ and $p_{2}$ are continuous on $(a, b)$ . Let $y_{1}$ be a solution of

$\begin{matrix} (A) & y^{″} + p_{1} (x) y^{'} + p_{2} (x) y = 0 \end{matrix}$

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

$\begin{matrix} y_{2} (x) = y_{1} (x) \int_{x_{0}}^{x} \frac{1}{y_{1}^{2} (t)} \exp (- \int_{x_{0}}^{t} p_{1} (s) d s) d t \end{matrix}$

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

37. The nonlinear first order equation

$\begin{matrix} (A) & y^{'} + y^{2} + p (x) y + q (x) = 0 \end{matrix}$

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 $\begin{matrix} (B) & z^{″} + p (x) z^{'} + q (x) z = 0. \end{matrix}$
Show that the general solution of (A) is $\begin{matrix} (C) & y = \frac{c_{1} z_{1}^{'} + c_{2} z_{2}^{'}}{c_{1} z_{1} + c_{2} z_{2}}, \end{matrix}$ where ${z_{1}, z_{2}}$ is a fundamental set of solutions of (B) and $c_{1}$ and $c_{2}$ 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.

$y^{'} + y^{2} + k^{2} = 0$
$y^{'} + y^{2} - 3 y + 2 = 0$
$y^{'} + y^{2} + 5 y - 6 = 0$
$y^{'} + y^{2} + 8 y + 7 = 0$
$y^{'} + y^{2} + 14 y + 50 = 0$
$6 y^{'} + 6 y^{2} - y - 1 = 0$
$36 y^{'} + 36 y^{2} - 12 y + 1 = 0$

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.

$x^{2} (y^{'} + y^{2}) - x (x + 2) y + x + 2 = 0; y_{1} = 1 / x$
$y^{'} + y^{2} + 4 x y + 4 x^{2} + 2 = 0; y_{1} = - 2 x$
$(2 x + 1) (y^{'} + y^{2}) - 2 y - (2 x + 3) = 0; y_{1} = - 1$
$(3 x - 1) (y^{'} + y^{2}) - (3 x + 2) y - 6 x + 8 = 0; y_{1} = 2$
$x^{2} (y^{'} + y^{2}) + x y + x^{2} - \frac{1}{4} = 0; y_{1} = - \tan x - \frac{1}{2 x}$
$x^{2} (y^{'} + y^{2}) - 7 x y + 7 = 0; y_{1} = 1 / x$

40. The nonlinear first order equation

$\begin{matrix} (A) & y^{'} + r (x) y^{2} + p (x) y + q (x) = 0 \end{matrix}$

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 $y = z^{'} / r z$ , where $\begin{matrix} (B) & z^{″} + [p (x) - \frac{r^{'} (x)}{r (x)}] z^{'} + r (x) q (x) z = 0. \end{matrix}$
Show that the general solution of (A) is $\begin{matrix} y = \frac{c_{1} z_{1}^{'} + c_{2} z_{2}^{'}}{r (c_{1} z_{1} + c_{2} z_{2})}, \end{matrix}$ where ${z_{1}, z_{2}}$ is a fundamental set of solutions of (B) and $c_{1}$ and $c_{2}$ are arbitrary constants.

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