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

10.7E: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises)

  • Page ID
    18303
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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

    \( \newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\\[1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill$\displaystyle#1$\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}} \)

    In Exercises [exer:10.7.1}– [exer:10.7.10} find a particular solution.

    [exer:10.7.1]\(\{{\bf y}'=\twobytwo{-1}{-4}{-1}{-1}{\bf y}+\left[\begin{array}{cc}21e^{4t}\\8e^{-3t}\end{array}\right]}\)&

    [exer:10.7.2]\(\{{\bf y}'={1\over5}\twobytwo{-4}3{-2}{-11}{\bf y}+\left[\begin{array}{cc}50e^{3t}\\10e^{-3t}\end{array}\right]}\)

    [exer:10.7.3]\(\{{\bf y}'=\twobytwo1221{\bf y}+\left[\begin{array}{cc}1\\t\end{array}\right]}\)&

    [exer:10.7.4]\(\{{\bf y}'=\twobytwo{-4}{-3}65{\bf y}+\left[\begin{array}{cc}2\\-2e^t\end{array}\right]}\)

    [exer:10.7.5]\(\{{\bf y}'=\twobytwo{-6}{-3}1{-2}{\bf y}+\left[\begin{array}{cc}4e^{-3t}\\4e^{-5t}\end{array}\right]}\)&

    [exer:10.7.6]\(\{{\bf y}'=\twobytwo01{-1}0{\bf y}+\left[\begin{array}{cc}1\\t\end{array}\right]}\)

    [exer:10.7.7]\(\{{\bf y}'=\threebythree31{-1}351{-6}24{\bf y}+\left[\begin{array}{cc}3\\6\\3\end{array}\right]}\)&

    [exer:10.7.8]\(\{{\bf y}'=\threebythree3{-1}{-1}{-2}324{-1}{-2}{\bf y}+\left[\begin{array}{cc}1\\e^t\\e^t\end{array}\right]}\)

    [exer:10.7.9] \(\{{\bf y}'=\threebythree{-3}222{-3}222{-3}{\bf y}+\left[\begin{array}{cc}e^t\\e^{-5t}\\e^t\end{array}\right]}\)

    [exer:10.7.10]\(\{{\bf y}'={1\over3}\threebythree11{-3}{-4}{-4}3{-2}10{\bf y}+\left[\begin{array}{cc}e^t\\e^t\\e^t\end{array}\right]}\)

    [exer:10.7.11] \(\{{\bf y}'={1\over t}\left[\begin{array}{rc}1&t\\-t&1\end{array}\right]{\bf y} +t\left[\begin{array}{cc} \cos t\\\sin t\end{array}\right]; \quad Y=t\left[\begin{array}{rc}\cos t&\sin t\\-\sin t&\cos t \end{array}\right]}\)

    [exer:10.7.12] \(\{{\bf y}'={1\over t}\left[\begin{array}{cc}1&t\\t&1\end{array}\right]{\bf y} +\left[\begin{array}{c} t\\t^2\end{array}\right]; \quad Y=t\left[\begin{array}{cr} e^t&e^{-t}\\e^t&-e^{-t} \end{array}\right]}\)

    [exer:10.7.13] \(\{{\bf y}'={1\over t^2-1}\left[\begin{array}{rr}t&-1\\-1&t\end{array}\right]{\bf y}+ t\left[\begin{array}{r} 1\\-1\end{array}\right]; \quad Y=\left[\begin{array}{cc} t&1\\1&t \end{array}\right]}\)

    [exer:10.7.14] \(\{{\bf y}'={1\over 3}\left[\begin{array}{cc}1&-2e^{-t}\\2e^t&-1\end{array}\right]{\bf y}+ \left[\begin{array}{c}e^{2t} \\e^{-2t}\end{array}\right]; \quad Y=\left[\begin{array}{cc} 2&e^{-t}\\e^t&2 \end{array}\right]}\)

    [exer:10.7.15] \(\{{\bf y}'={1\over 2t^4}\left[\begin{array}{cc}3t^3&t^6\\1&-3t^3\end{array}\right]{\bf y}+ {1\over t}\left[\begin{array}{c} t^2\\1\end{array}\right]; \quad Y={1\over t^2}\left[\begin{array}{rc} t^3&t^4\\-1&t \end{array}\right]}\)

    [exer:10.7.16] \(\{{\bf y}'= \left[\begin{array}{cc}\{1\over t-1} &-\{e^{-t}\over t-1}\

    \[5pt]\{e^t\over t+1}&\{1\over t+1}\end{array}\right]{\bf y}+\left[\begin{array}{c} t^2-1\\t^2-1\end{array}\right]; \quad Y=\left[\begin{array}{cc} t&e^{-t}\\e^t&t \end{array}\right]}\)

    [exer:10.7.17] \(\{{\bf y}'={1\over t}\threebythree110021{-2}22{\bf y} +\threecol121 \quad Y=\left[\begin{array}{ccr} t^2&t^3&1\\t^2&2t^3&-1\\0&2t^3&2 \end{array}\right]}\)

    [exer:10.7.18] \(\{{\bf y}'= \left[\begin{array}{ccc}3&e^t&e^{2t}\\e^{-t}&2&e^t\\e^{-2t}&e^{-t}&1\end{array}\right]{\bf y} +\left[\begin{array}{c} e^{3t}\\0 \\ 0\end{array}\right]; \quad Y=\left[\begin{array}{crr} e^{5t}&e^{2t}&0\\e^{4t}&0&e^t\\e^{3t}&-1&-1 \end{array}\right]}\)

    [exer:10.7.19] \(\{{\bf y}'={1\over t}\left[\begin{array}{crc}1&t&0\\0&1&t\\0&-t&1\end{array}\right]{\bf y} +\left[\begin{array}{c} t\\t \\t \end{array}\right]; \quad Y=t\left[\begin{array}{crr} 1&\cos t&\sin t \\0&-\sin t&\cos t\\0&-\cos t&-\sin t \end{array}\right]}\)

    [exer:10.7.20] \(\{{\bf y}'=-{1\over t}\left[\begin{array}{crr}e^{-t}&-t&1-e^{-t}\\e^{-t}&1&-t-e^{-t}\\e^{-t}&-t &1-e^{-t}\end{array}\right]{\bf y} +{1\over t}\left[\begin{array}{c} e^t\\0 \\e^t \end{array}\right]; \quad Y={1\over t}\left[\begin{array}{crc} e^t&e^{-t}&t\\e^t&-e^{-t}&e^{-t}\\e^t&e^{-t}&0 \end{array}\right]}\)

    [exer:10.7.21] Prove Theorem [thmtype:10.7.1}.

    [exer:10.7.22]

    Convert the scalar equation

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

    into an equivalent \(n\times n\) system

    \[{\bf y}'=A(t){\bf y}+{\bf f}(t). \eqno{\rm (B)}\]

    Suppose (A) is normal on an interval \((a,b)\) and \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of

    \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=0 \eqno{\rm (C)}\]

    on \((a,b)\). Find a corresponding fundamental matrix \(Y\) for

    \[{\bf y}'=A(t){\bf y} \eqno{\rm (D)}\]

    on \((a,b)\) such that

    \[y=c_1y_1+c_2y_2+\cdots+c_ny_n\]

    is a solution of (C) if and only if \({\bf y}=Y{\bf c}\) with

    \[{\bf c}=\left[\begin{array}{c}c_1\\c_2\\\vdots\\c_n\end{array}\right]\]

    is a solution of (D).

    Let \(y_p=u_1y_1+u_1y_2+\cdots+u_ny_n\) be a particular solution of (A), obtained by the method of variation of parameters for scalar equations as given in Section 9.4, and define

    \[{\bf u}=\left[\begin{array}{c}u_1\\u_2\\\vdots\\u_n\end{array}\right].\]

    Show that \({\bf y}_p=Y{\bf u}\) is a solution of (B).

    Let \({\bf y}_p=Y{\bf u}\) be a particular solution of (B), obtained by the method of variation of parameters for systems as given in this section. Show that \(y_p=u_1y_1+u_1y_2+\cdots+u_ny_n\) is a solution of (A).

    [exer:10.7.23] Suppose the \(n\times n\) matrix function \(A\) and the \(n\)–vector function \({\bf f}\) are continuous on \((a,b)\). Let \(t_0\) be in \((a,b)\), let \({\bf k}\) be an arbitrary constant vector, and let \(Y\) be a fundamental matrix for the homogeneous system \({\bf y}'=A(t){\bf y}\). Use variation of parameters to show that the solution of the initial value problem

    \[{\bf y}'=A(t){\bf y}+{\bf f}(t),\quad {\bf y}(t_0)={\bf k}\]

    is

    \[{\bf y}(t)=Y(t)\left( Y^{-1}(t_0){\bf k}+\int_{t_0}^t Y^{-1}(s){\bf f}(s)\, ds\right).\]