Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

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

  • Page ID
    18315
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    In Exercises [exer:9.4.1}– [exer:9.4.21} find a particular solution, given the fundamental set of solutions of the complementary equation.

    [exer:9.4.1] \(x^3y'''-x^2(x+3)y''+2x(x+3)y'-2(x+3)y=-4x^4\); \(\{x,\,x^2,\,xe^x\}\)

    [exer:9.4.2] \(y'''+6xy''+(6+12x^2)y'+(12x+8x^3)y=x^{1/2}e^{-x^2}\); \(\{e^{-x^2},\,xe^{-x^2},\,x^2e^{-x^2}\}\)

    [exer:9.4.3] \(x^3y'''-3x^2y''+6xy'-6y=2x\); \(\{x,x^2,x^3\}\)

    [exer:9.4.4] \(x^2y'''+2xy''-(x^2+2)y'=2x^2\);\(\{1,\,e^x/x,\,e^{-x}/x\}\)

    [exer:9.4.5] \(x^3y'''-3x^2(x+1)y''+3x(x^2+2x+2)y'-(x^3+3x^2+6x+6)y=x^4e^{-3x}\);\(\{xe^x,\,x^2e^x,\,x^3e^x\}\)

    [exer:9.4.6] \(x(x^2-2)y'''+(x^2-6)y''+x(2-x^2)y'+(6-x^2)y=2(x^2-2)^2\); \(\{e^x,\,e^{-x},\,1/x\}\)

    [exer:9.4.7] \(xy'''-(x-3)y''-(x+2)y'+(x-1)y=-4e^{-x}\); \(\{e^x,\,e^x/x,\,e^{-x}/x\}\)

    [exer:9.4.8] \(4x^3y'''+4x^2y''-5xy'+2y=30x^2\); \(\{\sqrt x,\,1/\sqrt x,\,x^2\}\)

    [exer:9.4.9] \(x(x^2-1)y'''+(5x^2+1)y''+2xy'-2y=12x^2\); \(\{x,\,1/(x-1),\,1/(x+1)\}\)

    [exer:9.4.10] \(x(1-x)y'''+(x^2-3x+3)y''+xy'-y=2(x-1)^2\); \(\{x,\,1/x,e^x/x\}\)

    [exer:9.4.11] \(x^3y'''+x^2y''-2xy'+2y=x^2\); \(\{x,\,x^2,\,1/x\}\)

    [exer:9.4.12] \(xy'''-y''-xy'+y=x^2\); \(\{x,\,e^x,\,e^{-x}\}\)

    [exer:9.4.13] \(xy^{(4)}+4y'''=6 \ln |x|\); \(\{1,\,x,\,x^2,\,1/x\}\)

    [exer:9.4.14] \(16x^4y^{(4)}+96x^3y'''+72x^2y''-24xy'+9y=96x^{5/2}\); \(\{\sqrt x,\,1/\sqrt x,\,x^{3/2},\,x^{-3/2}\}\)

    [exer:9.4.15] \(x(x^2-6)y^{(4)}+2(x^2-12)y'''+x(6-x^2)y''+2(12-x^2)y'=2(x^2-6)^2\);\(\{1,\,1/x,\,e^x,\,e^{-x}\}\)

    [exer:9.4.16] \(x^4y^{(4)}-4x^3y'''+12x^2y''-24xy'+24y=x^4\); \(\{x,\,x^2,\,x^3,\,x^4\}\)

    [exer:9.4.17] \(x^4y^{(4)}-4x^3y'''+2x^2(6-x^2)y''+4x(x^2-6)y'+(x^4-4x^2+24)y=4x^5e^x\);\(\{xe^x,\,x^2e^x,\,xe^{-x},\,x^2e^{-x}\}\)

    [exer:9.4.18] \(x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=12x^2\); \(\{x,x^2,1/x,1/x^2\}\)

    [exer:9.4.19] \(xy^{(4)}+4y'''-2xy''-4y'+xy=4e^x\); \(\{e^x,\,e^{-x},\,e^x/x,\,e^{-x}/x\}\)

    [exer:9.4.20] \(xy^{(4)}+(4-6x)y'''+(13x-18)y''+(26-12x)y'+(4x-12)y=3e^x\); \(\{e^x,\,e^{2x},\,e^x/x,\,e^{2x}/x\}\)

    [exer:9.4.21] \(x^4y^{(4)}-4x^3y'''+x^2(12-x^2)y''+2x(x^2-12)y'+2(12-x^2)y=2x^5\); \(\{x,\,x^2,\,xe^x,\,xe^{-x}\}\)

    [exer:9.4.22] \(x^3y'''-2x^2y''+3xy'-3y=4x, \quad y(1)=4,\quad y'(1)=4, \quad y''(1)=2\); \(\{x,\,x^3,\,x \ln x\}\)

    [exer:9.4.23] \(x^3y'''-5x^2y''+14xy'-18y=x^3, \quad y(1)=0,\quad y'(1)=1,\quad y''(1)=7\); \(\{x^2,\, x^3,\,x^3 \ln x\}\)

    [exer:9.4.24] \(\{(5-6x)y'''+(12x-4)y''+(6x-23)y'+(22-12x)y=-(6x-5)^2e^x}\)\(\{y(0)=-4, \quad y'(0)=-{3\over2},\quad y''(0)=-19}\); \(\{e^x,\,e^{2x},\,xe^{-x} \}\)

    [exer:9.4.25] \(x^3y'''-6x^2y''+16xy'-16y=9x^4, \quad y(1)=2,\quad y'(1)=1,\quad y''(1)=5\);\(\{x,\,x^4,\,x^4 \ln |x|\}\)

    [exer:9.4.26] \((x^2-2x+2)y'''-x^2y''+2xy'-2y=(x^2-2x+2)^2, \quad y(0)=0,\quad y'(0)=5\),\(y''(0)=0\);\(\{x,\,x^2,\,e^x\}\)

    [exer:9.4.27] \(x^3y'''+x^2y''-2xy'+2y=x(x+1), \quad y(-1)=-6,\quad y'(-1)=\{43\over6},\quad y''(-1)= -\{5\over2}\);\(\{x,\,x^2,\,1/x\}\)

    [exer:9.4.28] \((3x-1)y'''-(12x-1)y''+9(x+1)y'-9y=2e^x(3x-1)^2, \quad y(0)=\{3\over4}\),\(y'(0)=\{5\over4}, \quad y''(0)=\{1\over4}\); \(\{x+1,\,e^x,\,e^{3x}\}\)

    [exer:9.4.29] \((x^2-2)y'''-2xy''+(2-x^2)y'+2xy=2(x^2-2)^2, \quad y(0)=1,\quad y'(0)=-5\),\(y''(0)=5\);\(\{x^2,\,e^x,\,e^{-x}\}\)

    [exer:9.4.30] \(x^4y^{(4)}+3x^3y'''-x^2y''+2xy'-2y=9x^2, \quad y(1)=-7,\quad y'(1)= -11,\quad y''(1)=-5\), \(y'''(1)=6; \quad \{x,\,x^2,\,1/x,\,x\ln x\}\)

    [exer:9.4.31] \((2x-1)y^{(4)}-4xy'''+(5-2x)y''+4xy'-4y=6(2x-1)^2, \quad y(0)=\{55\over4}, \quad y'(0)=0\),\(y''(0)=13, \quad y'''(0)=1\);\(\{x,\,e^x,\,e^{-x},\,e^{2x}\}\)

    [exer:9.4.32] \(4x^4y^{(4)}+24x^3y'''+23x^2y''-xy'+y=6x\),\(y(1)=2,\quad y'(1)=0,\quad y''(1)=4,\quad y'''(1)=-\{37\over4}\); \(\{x,\sqrt x,1/x,1/\sqrt x\}\)

    [exer:9.4.33] \(x^4y^4+5x^3y'''-3x^2y''-6xy'+6y=40x^3, \quad y(-1)=-1, \; y'(-1)=-7\),

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

    [exer:9.4.34] Suppose the equation

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

    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

    \[y_p^{(n)}=u_1y_1^{(n)}+u_2y_2^{(n)}+\cdots+u_ny_n^{(n)}+{F\over P_0}.\nonumber \]

    Show that \(y_p\) is the solution of the initial value problem

    \[\begin{array}{r} P_0(x)y^{(n)}+P_1(x)y^{(n-1)}+\cdots+P_n(x)y=F(x),\\[4pt] \hspace*{20pt} y(x_0)=0,\; y'(x_0)=0,\dots,\quad y^{(n-1)}(x_0)=0. \end{array}\nonumber \]

    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

    \[\left.{\partial^{j}G(x,t)\over\partial x^j}\right|_{x=t}= \left\{\begin{array}{cl} 0,&1\le j\le n-2,\\[4pt] \{1\over P_0(t)},&j=n-1. \end{array}\right.\nonumber \]

    Show that

    \[y_p^{(j)}(x)=\left\{\begin{array}{cl}\{\int_{x_0}^x {\partial^{j}G(x,t)\over\partial x^j}F(t)\,dt},&1\le j\le n-1,\\[10pt] \{{F(x)\over P_0(x)}+ \int_{x_0}^x{\partial^{(n)}G(x,t)\over\partial x^n}F(t)\,dt},&j=n. \end{array}\right.\nonumber \]

    [exer:9.4.35] \(y'''+2y'-y'-2y=F(x); \quad \{e^x,\,e^{-x},e^{-2x}\}\)

    [exer:9.4.36] \(x^3y'''+x^2y''-2xy'+2y=F(x); \quad \{x,\,x^2,\,1/x\}\)

    [exer:9.4.37] \(x^3y'''-x^2(x+3)y''+2x(x+3)y'-2(x+3)y=F(x); \{x,x^2,xe^x\}\)

    [exer:9.4.38] \(x(1-x)y'''+(x^2-3x+3)y''+xy'-y=F(x); \quad \{x,\,1/x,\,e^x/x\}\)

    [exer:9.4.39] \(y^{(4)}-5y''+4y=F(x); \quad \{e^x,\,e^{-x},\,e^{2x},\,e^{-2x}\}\)

    [exer:9.4.40] \(xy^{(4)}+4y'''=F(x); \quad \{1,\,x,\,x^2,\,1/x\}\)

    [exer:9.4.41] \(x^4y^{(4)}+6x^3y'''+2x^2y''-4xy'+4y=F(x)\); \(\{x,x^2,1/x,1/x^2\}\)

    [exer:9.4.42] \(xy^{(4)}-y'''-4xy'+4y'=F(x); \quad \{1,\,x^2,\,e^{2x}, e^{-2x}\}\)