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10.7.1: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises)

  • Page ID
    30800
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    Q10.7.1

    In Exercises 10.7.1-10.7.10 find a particular solution.

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

    2. \({\bf y}'=\frac{1}{5}\left[\begin{array}{cc}{-4}&{3}\\[4pt]{-2}&{-11}\end{array} \right]{\bf y}+\left[\begin{array}{c}{50e^{3t}}\\[4pt]{10e^{-3t}} \end{array}\right]\)

    3. \({\bf y}'=\left[\begin{array}{cc}{1}&{2}\\[4pt]{2}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\[4pt]{t} \end{array}\right]\)

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

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

    6. \({\bf y}'=\left[\begin{array}{cc}{0}&{1}\\[4pt]{-1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\[4pt]{t} \end{array}\right]\)

    7. \({\bf y}'=\left[\begin{array}{ccc}{3}&{1}&{-1}\\[4pt]{3}&{5}&{1}\\[4pt]{-6}&{2}&{4}\end{array} \right]{\bf y}+\left[\begin{array}{c}{3}\\[4pt]{6}\\[4pt]{3} \end{array}\right]\)

    8. \({\bf y}'=\left[\begin{array}{ccc}{3}&{-1}&{-1}\\[4pt]{-2}&{3}&{2}\\[4pt]{4}&{-1}&{-2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\[4pt]{e^{t}}\\[4pt]{e^{t}} \end{array}\right]\)

    9. \({\bf y}'=\left[\begin{array}{ccc}{-3}&{2}&{2}\\[4pt]{2}&{-3}&{2}\\[4pt]{2}&{2}&{-3}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{t}}\\[4pt]{e^{-5t}}\\[4pt]{e^{t}} \end{array}\right]\)

    10. \({\bf y}'=\frac{1}{3}\left[\begin{array}{ccc}{1}&{1}&{-3}\\[4pt]{-4}&{-4}&{3}\\[4pt]{-2}&{1}&{0}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{t}}\\[4pt]{e^{t}}\\[4pt]{e^{t}} \end{array}\right]\)

    Q10.7.2

    In Exercises 10.7.11-10.7.20 find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system.

    11. \({\bf y}'=\frac{1}{t}\left[\begin{array}{cc}{1}&{t}\\[4pt]{-t}&{1}\end{array} \right]{\bf y}+t\left[\begin{array}{c}{\cos t}\\[4pt]{\sin t}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{\cos t}&{\sin t}\\[4pt]{-\sin t}&{\cos t}\end{array} \right]\)

    12. \({\bf y}'=\frac{1}{t}\left[\begin{array}{cc}{1}&{t}\\[4pt]{t}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t}\\[4pt]{t^{2}}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{e^{t}}&{e^{-t}}\\[4pt]{e^{t}}&{-e^{-t}}\end{array} \right]\)

    13. \({\bf y}'=\frac{1}{t^{2}-1}\left[\begin{array}{cc}{t}&{-1}\\[4pt]{-1}&{t}\end{array} \right]{\bf y}+t\left[\begin{array}{c}{1}\\[4pt]{-1}\end{array} \right];\quad Y=\left[\begin{array}{cc}{t}&{1}\\[4pt]{1}&{t}\end{array} \right]\)

    14. \({\bf y}'=\frac{1}{3}\left[\begin{array}{cc}{1}&{-2e^{-t}}\\[4pt]{2e^{t}}&{-1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{e^{2t}}\\[4pt]{e^{-2t}}\end{array} \right];\quad Y=\left[\begin{array}{cc}{2}&{e^{-t}}\\[4pt]{e^{t}}&{2}\end{array} \right]\)

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

    16. \({\bf y}'=\left[\begin{array}{cc}{\frac{1}{t-1}}&{-\frac{e^{-t}}{t-1}}\\[4pt]{\frac{e^{t}}{t+1}}&{\frac{1}{t+1}}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t^{2}-1}\\[4pt]{t^{2}-1}\end{array} \right];\quad Y=t\left[\begin{array}{cc}{t}&{e^{-t}}\\[4pt]{e^{t}}&{t}\end{array} \right]\)

    17. \({\bf y}' = \frac{1}{t}\left[\begin{array}{ccc}{1}&{1}&{0}\\[4pt]{0}&{2}&{1}\\[4pt]{-2}&{2}&{2}\end{array} \right]{\bf y}+\left[\begin{array}{c}{1}\\[4pt]{2}\\[4pt]{1}\end{array} \right];\quad Y=\left[\begin{array}{ccc}{t^{2}}&{t^{3}}&{1}\\[4pt]{t^{2}}&{2t^{3}}&{-1}\\[4pt]{0}&{2t^{3}}&{2}\end{array} \right]\)

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

    19. \({\bf y}' = \frac{1}{t}\left[\begin{array}{ccc}{1}&{t}&{0}\\[4pt]{0}&{1}&{t}\\[4pt]{0}&{-t}&{1}\end{array} \right]{\bf y}+\left[\begin{array}{c}{t}\\[4pt]{t}\\[4pt]{t}\end{array} \right];\quad Y=t\left[\begin{array}{ccc}{1}&{\cos t}&{\sin t}\\[4pt]{0}&{-\sin t}&{\cos t}\\[4pt]{0}&{-\cos t}&{-\sin t}\end{array} \right]\)

    20. \({\bf y}' = -\frac{1}{t}\left[\begin{array}{ccc}{e^{-t}}&{-t}&{1-e^{-t}}\\[4pt]{e^{-t}}&{1}&{-t-e^{-t}}\\[4pt]{e^{-t}}&{-t}&{1-e^{-t}}\end{array} \right]{\bf y}+\frac{1}{t}\left[\begin{array}{c}{e^{t}}\\[4pt]{0}\\[4pt]{e^{t}}\end{array} \right];\quad Y=\frac{1}{t}\left[\begin{array}{ccc}{e^{t}}&{e^{-t}}&{t}\\[4pt]{e^{t}}&{-e^{-t}}&{e^{-t}}\\[4pt]{e^{t}}&{e^{-t}}&{0}\end{array} \right]\)

    Q10.7.3

    21. Prove Theorem 10.7.1.

    22.

    1. Convert the scalar equation \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t) \tag{A} \] into an equivalent \(n\times n\) system \[{\bf y}'=A(t){\bf y}+{\bf f}(t). \tag{B} \]
    2. Suppose (A) is normal on an interval \((a,b)\) and \(\{y_1,y_2,\dots,y_n\}\) is a fundamental set of solutions of \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=0 \tag{C} \] on \((a,b)\). Find a corresponding fundamental matrix \(Y\) for \[{\bf y}'=A(t){\bf y} \tag{D} \] on \((a,b)\) such that \[y=c_1y_1+c_2y_2+\cdots+c_ny_n\nonumber \] is a solution of (C) if and only if \({\bf y}=Y{\bf c}\) with \[{\bf c}=\left[\begin{array}{c}c_1\\[4pt]c_2\\[4pt]\vdots\\[4pt]c_n\end{array}\right]\nonumber \] is a solution of (D).
    3. 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 \[{\bf u}=\left[\begin{array}{c}u_1\\[4pt]u_2\\[4pt]\vdots\\[4pt]u_n\end{array}\right].\nonumber \] Show that \({\bf y}_p=Y{\bf u}\) is a solution of (B).
    4. 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).

    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

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

    is

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


    This page titled 10.7.1: Variation of Parameters for Nonhomogeneous Linear Systems (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.