9.4E: Variation of Parameters for Higher Order Equations (Exercises)

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$trench-1995.jpg$
Contributed by William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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Q9.4.1

In Exercises 9.4.1-9.4.21 find a particular solution, given the fundamental set of solutions of the complementary equation.

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

2. $y'''+6xy''+(6+12x^2)y'+(12x+8x^3)y=x^{1/2}e^{-x^2}$; $\{e^{-x^2},\,xe^{-x^2},\,x^2e^{-x^2}\}$

3. $x^3y'''-3x^2y''+6xy'-6y=2x$; $\{x,x^2,x^3\}$

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

5. $x^3y'''-3x^2(x+1)y''+3x(x^2+2x+2)y'-(x^3+3x^2+6x+6)y=x^4e^{-3x}$;$\{xe^x,\,x^2e^x,\,x^3e^x\}$

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

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

8. $4x^3y'''+4x^2y''-5xy'+2y=30x^2$; $\{\sqrt x,\,1/\sqrt x,\,x^2\}$

9. $x(x^2-1)y'''+(5x^2+1)y''+2xy'-2y=12x^2$; $\{x,\,1/(x-1),\,1/(x+1)\}$

10. $x(1-x)y'''+(x^2-3x+3)y''+xy'-y=2(x-1)^2$; $\{x,\,1/x,e^x/x\}$

11. $x^3y'''+x^2y''-2xy'+2y=x^2$; $\{x,\,x^2,\,1/x\}$

12. $xy'''-y''-xy'+y=x^2$; $\{x,\,e^x,\,e^{-x}\}$

13. $xy^{(4)}+4y'''=6 \ln |x|$; $\{1,\,x,\,x^2,\,1/x\}$

14. $16x^4y^{(4)}+96x^3y'''+72x^2y''-24xy'+9y=96x^{5/2}$; $\{\sqrt x,\,1/\sqrt x,\,x^{3/2},\,x^{-3/2}\}$

15. $x(x^2-6)y^{(4)}+2(x^2-12)y'''+x(6-x^2)y''+2(12-x^2)y'=2(x^2-6)^2$;$\{1,\,1/x,\,e^x,\,e^{-x}\}$

16. $x^4y^{(4)}-4x^3y'''+12x^2y''-24xy'+24y=x^4$; $\{x,\,x^2,\,x^3,\,x^4\}$

17. $x^4y^{(4)}-4x^3y'''+2x^2(6-x^2)y''+4x(x^2-6)y'+(x^4-4x^2+24)y=4x^5e^x$;$\{xe^x,\,x^2e^x,\,xe^{-x},\,x^2e^{-x}\}$

18. $x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=12x^2$; $\{x,x^2,1/x,1/x^2\}$

19. $xy^{(4)}+4y'''-2xy''-4y'+xy=4e^x$; $\{e^x,\,e^{-x},\,e^x/x,\,e^{-x}/x\}$

20. $xy^{(4)}+(4-6x)y'''+(13x-18)y''+(26-12x)y'+(4x-12)y=3e^x$; $\{e^x,\,e^{2x},\,e^x/x,\,e^{2x}/x\}$

21. $x^4y^{(4)}-4x^3y'''+x^2(12-x^2)y''+2x(x^2-12)y'+2(12-x^2)y=2x^5$; $\{x,\,x^2,\,xe^x,\,xe^{-x}\}$

Q9.4.2

In Exercises 9.4.22-9.4.33 solve the initial value problem, given the fundamental set of solutions of the complementary equation. Graph the solution for Exercises 9.4.22, 9.4.26, 9.4.29, and 9.4.30.

22. $x^3y'''-2x^2y''+3xy'-3y=4x, \quad y(1)=4,\quad y'(1)=4, \quad y''(1)=2$; $\{x,\,x^3,\,x \ln x\}$

23. $x^3y'''-5x^2y''+14xy'-18y=x^3, \quad y(1)=0,\quad y'(1)=1,\quad y''(1)=7$; $\{x^2,\, x^3,\,x^3 \ln x\}$

24. $(5-6x)y'''+(12x-4)y''+(6x-23)y'+(22-12x)y=-(6x-5)^2e^x \quad \{y(0)=-4, \quad y'(0)=-{3\over2},\quad y''(0)=-19; \{e^x,\,e^{2x},\,xe^{-x} \}$

25. $x^3y'''-6x^2y''+16xy'-16y=9x^4, \quad y(1)=2,\quad y'(1)=1,\quad y''(1)=5$;$\{x,\,x^4,\,x^4 \ln |x|\}$

26. $(x^2-2x+2)y'''-x^2y''+2xy'-2y=(x^2-2x+2)^2, \quad y(0)=0,\quad y'(0)=5$,$y''(0)=0$;$\{x,\,x^2,\,e^x\}$

27. $x^{3}y'''+x^{2}y''-2xy'+2y=x(x+1), \quad y(-1)=-6,\quad y'(-1)=\frac{43}{6},\quad y''(-1)= -\frac{5}{2};\{x,\,x^2,\,1/x\}$

28. $(3x-1)y'''-(12x-1)y''+9(x+1)y'-9y=2e^x(3x-1)^2, \quad y(0)=\frac{3}{4},\quad y'(0)=\frac{5}{4}, \quad y''(0)=\frac{1}{4}; \{x+1,\,e^x,\,e^{3x}\}$

29. $(x^2-2)y'''-2xy''+(2-x^2)y'+2xy=2(x^2-2)^2, \quad y(0)=1,\quad y'(0)=-5$,$y''(0)=5$;$\{x^2,\,e^x,\,e^{-x}\}$

30. $x^4y^{(4)}+3x^3y'''-x^2y''+2xy'-2y=9x^2, \quad y(1)=-7,\quad y'(1)= -11,\quad y''(1)=-5,\quad y'''(1)=6; \quad \{x,\,x^2,\,1/x,\,x\ln x\}$

31. $(2x-1)y^{(4)}-4xy'''+(5-2x)y''+4xy'-4y=6(2x-1)^2, \quad y(0)=\frac{55}{4}, \quad y'(0)=0,\quad y''(0)=13, \quad y'''(0)=1; \{x,\,e^x,\,e^{-x},\,e^{2x}\}$

32. $4x^4y^{(4)}+24x^3y'''+23x^2y''-xy'+y=6x,\quad y(1)=2,\quad y'(1)=0,\quad y''(1)=4,\quad y'''(1)=-\frac{37}{4}; \{x,\sqrt x,1/x,1/\sqrt x\}$

33. $x^4y^4+5x^3y'''-3x^2y''-6xy'+6y=40x^3, \quad y(-1)=-1, \; y'(-1)=-7$,

$y''(-1)=-1,\quad y'''(-1)=-31$; $\{x,\, x^3,\,1/x,\,1/x^2\}$

Q9.4.3

34. Suppose the equation

\[P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x) \tag{A}\]

is normal on an interval $(a,b)$. Let $\{y_1,y_2,\dots,y_n\}$ be a fundamental set of solutions of its complementary equation on $(a,b)$, let $W$ be the Wronskian of $\{y_1,y_2,\dots,y_n\}$, and let $W_j$ be the determinant obtained by deleting the last row and the $j$-th column of $W$. Suppose $x_0$ is in $(a,b)$, let

\[u_j(x)=(-1)^{(n-j)}\int_{x_0}^x{F(t)W_j(t)\over P_0(t)W(t)}\,dt, \quad 1\le j\le n,\nonumber \]

and define

\[y_p=u_1y_1+u_2y_2+\cdots+u_ny_n.\nonumber \]

Show that $y_p$ is a solution of (A) and that \[y_p^{(r)}=u_1y^{(r)}_1+u_2y_2^{(r)}\cdots+u_ny^{(r)}_n,\quad 1 \le r \le n-1,\nonumber \] and \[y_p^{(n)}=u_1y_1^{(n)}+u_2y_2^{(n)}+\cdots+u_ny_n^{(n)}+{F\over P_0}.\nonumber \] HINT: See the derivation of the method of variation of parameters at the beginning of the section.
Show that $y_p$ is the solution of the initial value problem \[\begin{array}{r} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),\\ y(x_0)=0,\; y'(x_0)=0,\dots,\quad y^{(n-1)}(x_0)=0. \end{array}\nonumber \]
Show that $y_p$ can be written as \[y_p(x)=\int_{x_0}^x G(x,t)F(t)\,dt,\nonumber \] where \[G(x,t)={1\over P_0(t)W(t)}\left|\begin{array}{cccc} y_1(t)&y_2(t)&\cdots&y_n(t)\\[4pt] y_1'(t)&y_2'(t)&\cdots&y_n'(t)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_1^{(n-2)}(t)&y_2^{(n-2)}(t)&\cdots&y_n^{(n-2)}(t)\\[4pt] y_1(x)&y_2(x)&\cdots&y_n(x)\end{array}\right|,\nonumber \] which is called the Green’s function for (A).
Show that \[{\partial^{j}G(x,t)\over\partial x^j} ={1\over P_0(t)W(t)}\left|\begin{array}{cccc} y_1(t)&y_2(t)&\cdots&y_n(t)\\[4pt] y_1'(t)&y_2'(t)&\cdots&y_n'(t)\\[4pt] \vdots&\vdots&\ddots&\vdots\\[4pt] y_1^{(n-2)}(t)&y_2^{(n-2)}(t)&\cdots&y_n^{(n-2)}(t)\\[4pt] y_1^{(j)}(x)&y_2^{(j)}(x)&\cdots&y_n^{(j)}(x)\end{array}\right|,\quad 0\le j\le n.\nonumber \]
Show that if $a<t<b$ then \[\left. \frac{\partial ^{j}G(x,t)}{\partial x^{j}} \right| _{x=t} = \left\{\begin{array}{cl}{0,}&{1\leq j\leq n-2}\\{\frac{1}{P_{0}(t)},}&{j=n-1} \end{array} \right. \nonumber\]
Show that \[y_{p}^{(j)}(x) = \left\{\begin{array}{cl}{\int_{x_{0}}^{x}\frac{\partial ^{j}G(x,t)}{\partial x^{j}}F(t)dt, }&{1\leq j\leq n-1,}\\{\frac{F(x)}{P_{0}(x)}+\int_{x_{0}}^{x}\frac{\partial ^{(n)}G(x,t)}{\partial x^{n}}F(t)dt, }&{j=n.} \end{array} \right. \nonumber\]

Q9.4.4

In Exercises 9.4.35-9.4.42 use the method suggested by Exercise 9.4.34 to find a particular solution in the form $y_{p}=\int_{x_{0}}^{x}G(x,t)F(t)dt$, given the indicated fundamental set of solutions. Assume that $x$ and $x_{0}$ are in an interval on which the equation is normal.

35. $y'''+2y'-y'-2y=F(x); \quad \{e^x,\,e^{-x},e^{-2x}\}$

36. $x^3y'''+x^2y''-2xy'+2y=F(x); \quad \{x,\,x^2,\,1/x\}$

37. $x^3y'''-x^2(x+3)y''+2x(x+3)y'-2(x+3)y=F(x); \{x,x^2,xe^x\}$

38. $x(1-x)y'''+(x^2-3x+3)y''+xy'-y=F(x); \quad \{x,\,1/x,\,e^x/x\}$

39. $y^{(4)}-5y''+4y=F(x); \quad \{e^x,\,e^{-x},\,e^{2x},\,e^{-2x}\}$

40. $xy^{(4)}+4y'''=F(x); \quad \{1,\,x,\,x^2,\,1/x\}$

41. $x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=F(x)$; $\{x,x^2,1/x,1/x^2\}$

42. $xy^{(4)}-y'''-4xy'+4y'=F(x); \quad \{1,\,x^2,\,e^{2x}, e^{-2x}\}$