5.7E: Variation of Parameters (Exercises)

Q5.7.1

In Exercises 5.7.1–5.7.6 use variation of parameters to find a particular solution.

1. $y''+9y=\tan 3x$

2. $y''+4y=\sin 2x\sec^2 2x$

3. $y''-3y'+2y={4\over 1+e^{-x}}$

4. $y''-2y'+2y=3e^x \sec x$

5. $y''-2y'+y=14x^{3/2}e^x$

6. $y''-y={4e^{-x}\over 1-e^{-2x}}$

Q5.7.2

In Exercises 5.7.7-5.7.29 use variation of parameters to find a particular solution, given the solutions $y_{1}, y_{2}$ of the complementary equation.

7. $x^2y''+xy'- y=2x^2+2; \quad y_1=x, \quad y_2={1\over x}$

8. ${xy''+(2-2x)y'+(x-2)y=e^{2x}; \quad y_1=e^x, \quad y_2={e^x\over x}}$

9. $4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x, \quad x > 0$; $y_1=x^{1/2} e^x,\; y_2=x^{-1/2}e^x$

10. $y''+4xy'+(4x^2+2)y=4e^{-x(x+2)};\quad y_1=e^{-x^2}, \quad y_2=xe^{-x^2}$

11. $x^2y''-4xy'+6y=x^{5/2},\, x > 0;\quad y_1=x^2,\; y_2=x^3$

12. $x^2y''-3xy'+3y=2x^4\sin x; \quad y_1=x,\; y_2=x^3$

13. $(2x+1)y''-2y'-(2x+3)y=(2x+1)^2e^{-x}; \quad y_1=e^{-x}, \quad y_2=xe^x$

14. $4xy''+2y'+y=\sin\sqrt x; \quad y_1=\cos\sqrt x, \quad y_2=\sin\sqrt x$

15. $xy''-(2x+2)y'+(x+2)y=6x^3e^x;\quad y_1=e^x,\quad y_2=x^3e^x$

16. $x^2y''-(2a-1)xy'+a^2y=x^{a+1}; \quad y_1=x^a, \quad y_2=x^a \ln x$

17. $x^2y''-2xy'+(x^2+2)y=x^3\cos x; \quad y_1=x\cos x, \quad y_2=x\sin x$

18. $xy''-y'-4x^3y=8x^5;\quad y_1=e^{x^2},\; y_2=e^{-x^2}$

19. $(\sin x)y''+(2\sin x-\cos x)y'+(\sin x-\cos x)y=e^{-x}; \quad y_1=e^{-x},\quad y_2=e^{-x}\cos x$

20. $4x^2y''-4xy'+(3-16x^2)y=8x^{5/2}; \quad y_1=\sqrt xe^{2x},\; y_2=\sqrt xe^{-2x}$

21. $4x^2y''-4xy'+(4x^2+3)y=x^{7/2}; \quad y_1=\sqrt x\sin x,\; y_2=\sqrt x\cos x$

22. $x^2y''-2xy'-(x^2-2)y=3x^4;\quad y_1=xe^x,\; y_2=xe^{-x}$

23. $x^2y''-2x(x+1)y' +(x^2+2x+2)y=x^3e^x; \quad y_1=xe^x, \quad y_2=x^2e^x$

24. $x^2y''-xy'-3y=x^{3/2}; \quad y_1=1/x, \quad y_2=x^3$

25. $x^2y''-x(x+4)y'+2(x+3)y=x^4e^x; \quad y_1=x^2, \quad y_2=x^2e^x$

26. $x^2y''-2x(x+2)y'+(x^2+4x+6)y=2xe^x; \quad y_1=x^2e^x, \quad y_2=x^3e^x$

27. $x^2y''-4xy'+(x^2+6)y=x^4; \quad y_1=x^2\cos x, \quad y_2=x^2\sin x$

28. $(x-1)y''-xy'+y=2(x-1)^2e^x; \quad y_1=x, \quad y_2=e^x$

29. $4x^2y''-4x(x+1)y'+(2x+3)y=x^{5/2}e^x; \quad y_1=\sqrt x, \quad y_2=\sqrt xe^x$

Q5.7.3

In Exercises 5.7.30–5.7.32 use variation of parameters to solve the initial value problem, given $y_{1}$, $y_{2}$ are solutions of the complementary equation.

30. $(3x-1)y''-(3x+2)y'-(6x-8)y=(3x-1)^2e^{2x}, \quad y(0)=1,\; y'(0)=2$; $y_1=e^{2x},\; y_2=xe^{-x}$

31. $(x-1)^2y''-2(x-1)y'+2y=(x-1)^2, \quad y(0)=3,\quad y'(0)=-6$;

$y_1=x-1$, $y_2=x^2-1$

32. $(x-1)^2y''-(x^2-1)y'+(x+1)y=(x-1)^3e^x, \quad y(0)=4,\quad y'(0)=-6$;

$y_1=(x-1)e^x,\quad y_2=x-1$

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 $y_{1}$, $y_{2}$ are solutions of the complementary equation.

33. ${(x^2-1)y''+4xy'+2y=2x, \quad y(0)=0,\; y'(0) =-2; \quad y_1={1\over x-1},\; y_2={1\over x+1}}$

34. ${x^2y''+2xy'-2y=-2x^2, \quad y(1)=1,\; y'(1)= -1; \quad y_1=x,\; y_2={1\over x^2}}$

35. $(x+1)(2x+3)y''+2(x+2)y'-2y=(2x+3)^2, \quad y(0)=0,\quad y'(0)=0$; $y_1=x+2,\quad y_2={1\over x+1}$

Q5.7.5

36. Suppose

$$y_p=\overline y+a_1y_1+a_2y_2 \nonumber$$

is a particular solution of

$$P_0(x)y''+P_1(x)y'+P_2(x)y=F(x), \tag{A}$$

where $y_1$ and $y_2$ are solutions of the complementary equation

$$P_0(x)y''+P_1(x)y'+P_2(x)y=0. \nonumber$$

Show that $\overline y$ is also a solution of (A).

37. Suppose $p$, $q$, and $f$ are continuous on $(a,b)$ and let $x_0$ be in $(a,b)$. Let $y_1$ and $y_2$ be the solutions of

$$y''+p(x)y'+q(x)y=0 \nonumber$$

such that

$$y_1(x_0)=1, \quad y_1'(x_0)=0, \quad y_2(x_0)=0, \quad y_2'(x_0)=1. \nonumber$$

Use variation of parameters to show that the solution of the initial value problem

$$y''+p(x)y'+q(x)y=f(x), \quad y(x_0)=k_0,\; y'(x_0)=k_1, \nonumber$$

is

$$\begin{array}{rcl} y(x) &= k_0y_1(x)+k_1y_2(x) \\[4pt] & +\int^x_{x_0}\left(y_1(t)y_2(x)- y_1(x)y_2(t)\right) f(t)\exp\left(\int^t_{x_0}p(s)\,ds\right)\,dt. \end{array} \nonumber$$

HINT: Use Abel's formula for the Wronskian of $\{ y_{1}, y_{2}\}$, and integrate $u_{1}'$ and $u_{2}'$ from $x_{0}$ to $x$.

Show also that

$$\begin{array}{rcl} y'(x) &= k_0y_1'(x)+k_1y_2'(x) \\[4pt] & +\int^x_{x_0}\left(y_1(t)y_2'(x)-y_1'(x)y_2(t) \right)f(t)\exp\left(\int^t_{x_0}p(s)\,ds\right)\,dt. \end{array} \nonumber$$

38. Suppose $f$ is continuous on an open interval that contains $x_0=0$. Use variation of parameters to find a formula for the solution of the initial value problem

$$y''-y=f(x), \quad y(0)=k_0,\quad y'(0)=k_1. \nonumber$$

39. Suppose $f$ is continuous on $(a,\infty)$, where $a<0$, so $x_0=0$ is in $(a,\infty)$.

1. Use variation of parameters to find a formula for the solution of the initial value problem $$y''+y=f(x), \quad y(0)=k_0,\quad y'(0)=k_1. \nonumber$$ HINT: You will need the addition formulas for the sine and cosine. $$\begin{aligned} \sin (A+B)&=\sin A\cos B +\cos A\sin B \\[4pt] \cos (A+B) &=\cos A\cos B - \sin A\sin B \end{aligned} \nonumber$$ For the rest of this exercise assume that the improper integral $\int_{0}^\infty f(t)\,dt$ is absolutely convergent.
2. Show that if $y$ is a solution of $$y''+y=f(x) \tag{A}$$ on $(a,\infty)$, then $$\lim_{x \to \infty}\left(y(x)-A_0\cos x-A_1\sin x\right)=0 \tag{B}$$ and $$\lim_{x\to\infty}\left(y'(x)+A_0\sin x-A_1\cos x\right)=0, \tag{C}$$ where $$A_0=k_0-\int_0^\infty f(t)\sin t\,dt \quad \text{and} \quad A_1=k_1+\int_0^\infty f(t)\cos t\,dt. \nonumber$$ HINT: Recall from calculus that if $\int _{0}^{\infty} f(t)dt$ converges absolutely, then $\lim_{x\to ∞}\int_{x}^{\infty}|f(t)|dt=0$.
3. Show that if $A_0$ and $A_1$ are arbitrary constants, then there’s a unique solution of $y''+y=f(x)$ on $(a,\infty)$ that satisfies (B) and (C).