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Mathematics LibreTexts

4.7E: Exercises

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    18489
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    In Exercises \((4.7E.1)\) to \((4.7E.10)\), find a particular solution.

    Exercise \(\PageIndex{1}\)

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

    Answer

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    Exercise \(\PageIndex{2}\)

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

    Answer

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    Exercise \(\PageIndex{3}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array}{cc}1\\t\end{array}\right]}\)

    Answer

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    Exercise \(\PageIndex{4}\)

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

    Answer

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    Exercise \(\PageIndex{5}\)

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

    Answer

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    Exercise \(\PageIndex{6}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 0 & 1 \\ {-1} & 0 \end{array} \right] {\bf y} + \left[\begin{array}{cc}1\\t\end{array}\right]}\)

    Answer

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    Exercise \(\PageIndex{7}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & 1 & {-1} \\ 3 & 5 & 1 \\ {-6} & 2 & 4 \end{array} \right] {\bf y} + \left[ \begin{array}{cc}3\\6\\3\end{array}\right]}\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ 3 & {-1} & {-1} \\ {-2} & 3 & 2 \\ 4 & {-1} & {-2} \end{array} \right] {\bf y} + \left[ \begin{array}{cc}1\\e^t\\e^t\end{array}\right]}\)

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    Exercise \(\PageIndex{9}\)

    \(\displaystyle{{\bf y}' = \left[ \begin{array} \\ {-3} & 2 & 2 \\ 2 & {-3} & 2 \\ 2 & 2 & {-3} \end{array} \right] {\bf y} + \left[ \begin{array}{cc}e^t\\e^{-5t}\\e^t\end{array}\right]}\)

    Answer

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    Exercise \(\PageIndex{10}\)

    \(\displaystyle{{\bf y}' = {1\over3} \left[ \begin{array} \\ 1 & 1 & {-3} \\ {-4} & {-4} & 3 \\ {-2} & 1 & 0 \end{array} \right]{\bf y} + \left[ \begin{array}{cc}e^t\\e^t\\e^t\end{array}\right]}\)

    Answer

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    In Exercises \((4.7E.11)\) to \((4.7E.20)\), find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system.

    Exercise \(\PageIndex{11}\)

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

    Answer

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    Exercise \(\PageIndex{12}\)

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

    Answer

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    Exercise \(\PageIndex{13}\)

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

    Answer

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    Exercise \(\PageIndex{14}\)

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

    Answer

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    Exercise \(\PageIndex{15}\)

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

    Answer

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    Exercise \(\PageIndex{16}\)

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

    Answer

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    Exercise \(\PageIndex{17}\)

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

    Answer

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    Exercise \(\PageIndex{18}\)

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

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    Exercise \(\PageIndex{19}\)

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

    Answer

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    Exercise \(\PageIndex{20}\)

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

    Answer

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    Exercise \(\PageIndex{21}\)

    Prove Theorem \((4.7.1)\).

    Answer

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    Exercise \(\PageIndex{22}\)

    (a) Convert the scalar equation

    \begin{equation} \label{eq:4.7E.1}
    P_0(t) y^{(n)} + P_1(t) y^{(n-1)} + \cdots + P_n(t) y = F(t)
    \end{equation}

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

    \begin{equation} \label{eq:4.7E.2}
    {\bf y}' = A(t) {\bf y} + {\bf f} (t).
    \end{equation}

    (b) Suppose \eqref{eq:4.7E.1} is normal on an interval \((a,b)\) and \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of

    \begin{equation} \label{eq:4.7E.3}
    P_0(t) y^{(n)} + P_1(t) y^{(n-1)} + \cdots + P_n(t) y = 0
    \end{equation}

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

    \begin{equation} \label{eq:4.7E.4}
    {\bf y}' = A(t) {\bf y}
    \end{equation}

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

    \begin{eqnarray*}
    y = c_1 y_1 + c_2 y_2 + \cdots + c_n y_n
    \end{eqnarray*}

    is a solution of \eqref{eq:4.7E.3} if and only if \({\bf y}=Y{\bf c}\) with

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

    is a solution of \eqref{eq:4.7E.4}.

    (c) Let \(y_p=u_1y_1+u_1y_2+\cdots+u_ny_n\) be a particular solution of \eqref{eq:4.7E.1}, obtained by the method of variation of parameters for scalar equations as given in Section \((3.4)\), and define

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

    Show that \({\bf y}_p=Y{\bf u}\) is a solution of \eqref{eq:4.7E.2}.

    (d) Let \({\bf y}_p=Y{\bf u}\) be a particular solution of \eqref{eq:4.7E.2}, 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 \eqref{eq:4.7E.1}.

    Answer

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    Exercise \(\PageIndex{23}\)

    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

    \begin{eqnarray*}
    {\bf y}' = A(t) {\bf y} + {\bf f} (t), \quad {\bf y} (t_0) = {\bf k}
    \end{eqnarray*}

    is

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

    Answer

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