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Mathematics LibreTexts

10.5E: Constant Coefficient Homogeneous Systems II (Exercises)

  • Page ID
    18301
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    \( \newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\\[1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill$\displaystyle#1$\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}} \)

    Q10.5.1

    In Exercises 10.5.1-10.5.12 find the general solution.

    1. \({\bf y}'=\left[\begin{array}{cc}{3}&{4}\\{-1}&{7}\end{array}\right]{\bf y}\)

    2. \({\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\{1}&{-2}\end{array}\right]{\bf y}\)

    3. \({\bf y}'=\left[\begin{array}{cc}{-7}&{4}\\{-1}&{-11}\end{array}\right]{\bf y}\)

    4. \({\bf y}'=\left[\begin{array}{cc}{3}&{1}\\{-1}&{1}\end{array}\right]{\bf y}\)

    5. \({\bf y}'=\left[\begin{array}{cc}{4}&{12}\\{-3}&{-8}\end{array}\right]{\bf y}\)

    6. \({\bf y}'=\left[\begin{array}{cc}{-10}&{9}\\{-4}&{2}\end{array}\right]{\bf y}\)

    7. \({\bf y}'=\left[\begin{array}{cc}{-13}&{16}\\{-9}&{11}\end{array}\right]{\bf y}\)

    8. \({\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{1}\\{-4}&{6}&{1}\\{0}&{4}&{2}\end{array}\right]{\bf y}\)

    9. \({\bf y}'=\frac{1}{3}\left[\begin{array}{ccc}{1}&{1}&{-3}\\{-4}&{-4}&{3}\\{-2}&{1}&{0}\end{array}\right]{\bf y}\)

    10. \({\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{-1}\\{-2}&{0}&{2}\\{-1}&{3}&{-1}\end{array}\right]{\bf y}\)

    11. \({\bf y}'=\left[\begin{array}{ccc}{4}&{-2}&{-2}\\{-2}&{3}&{-1}\\{2}&{-1}&{3}\end{array}\right]{\bf y}\)

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

    Q10.5.2

    In Exercises 10.5.13-10.5.23 solve the initial value problem.

    13. \({\bf y}'=\left[\begin{array}{cc}{-11}&{8}\\{-2}&{-3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\{2}\end{array}\right]\)

    14. \({\bf y}'=\left[\begin{array}{cc}{15}&{-9}\\{16}&{-9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{5}\\{8}\end{array}\right]\)

    15. \({\bf y}'=\left[\begin{array}{cc}{-3}&{-4}\\{1}&{-7}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{2}\\{3}\end{array}\right]\)

    16. \({\bf y}'=\left[\begin{array}{cc}{-7}&{24}\\{-6}&{17}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{3}\\{1}\end{array}\right]\)

    17. \({\bf y}'=\left[\begin{array}{cc}{-7}&{3}\\{-3}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\{2}\end{array}\right]\)

    18. \({\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{0}\\{1}&{-1}&{-2}\\{-1}&{-1}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\{5}\\{-7}\end{array}\right]\)

    19. \({\bf y}'=\left[\begin{array}{ccc}{-2}&{2}&{1}\\{-2}&{2}&{1}\\{-3}&{3}&{2}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\{-2}\\{0}\end{array}\right]\)

    20. \({\bf y}'=\left[\begin{array}{ccc}{-7}&{-4}&{4}\\{-1}&{0}&{1}\\{-9}&{-5}&{6}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\{9}\\{-1}\end{array}\right]\)

    21. \({\bf y}'=\left[\begin{array}{ccc}{-1}&{-4}&{-1}\\{3}&{6}&{1}\\{-3}&{-2}&{3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-2}\\{1}\\{3}\end{array}\right]\)

    22. \({\bf y}'=\left[\begin{array}{ccc}{4}&{-8}&{-4}\\{-3}&{-1}&{-3}\\{1}&{-1}&{9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-4}\\{1}\\{-3}\end{array}\right]\)

    23. \({\bf y}'=\left[\begin{array}{ccc}{-5}&{-1}&{11}\\{-7}&{1}&{13}\\{-4}&{0}&{8}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\{2}\\{2}\end{array}\right]\)

    Q10.5.3

    The coefficient matrices in Exercises 10.5.24-10.5.32 have eigenvalues of multiplicity \(3\). Find the general solution.

    24. \({\bf y}'=\left[\begin{array}{ccc}{5}&{-1}&{1}\\{-1}&{9}&{-3}\\{-2}&{2}&{4}\end{array}\right]{\bf y}\)

    25. \({\bf y}'=\left[\begin{array}{ccc}{1}&{10}&{-12}\\{2}&{2}&{3}\\{2}&{-1}&{6}\end{array}\right]{\bf y}\)

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

    27. \({\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{-2}\\{-1}&{5}&{-3}\\{1}&{1}&{1}\end{array}\right]{\bf y}\)

    28. \({\bf y}'=\left[\begin{array}{ccc}{-2}&{-12}&{10}\\{2}&{-24}&{11}\\{2}&{-24}&{8}\end{array}\right]{\bf y}\)

    29. \({\bf y}'=\left[\begin{array}{ccc}{-1}&{-12}&{8}\\{1}&{-9}&{4}\\{1}&{-6}&{1}\end{array}\right]{\bf y}\)

    30. \({\bf y}'=\left[\begin{array}{ccc}{-4}&{0}&{-1}\\{-1}&{-3}&{-1}\\{1}&{0}&{-2}\end{array}\right]{\bf y}\)

    31. \({\bf y}'=\left[\begin{array}{ccc}{-3}&{-3}&{4}\\{4}&{5}&{-8}\\{2}&{3}&{-5}\end{array}\right]{\bf y}\)

    32. \({\bf y}'=\left[\begin{array}{ccc}{-3}&{-1}&{0}\\{1}&{-1}&{0}\\{-1}&{-1}&{-2}\end{array}\right]{\bf y}\)

    Q10.5.4

    33. Under the assumptions of Theorem 10.5.1, suppose \({\bf u}\) and \(\hat{\bf u}\) are vectors such that

    \[(A-\lambda_1I){\bf u}={\bf x}\quad\mbox{and }\quad (A-\lambda_1I)\hat{\bf u}={\bf x},\nonumber \]

    and let

    \[{\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t} \quad\mbox{and }\quad \hat{\bf y}_2=\hat{\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}.\nonumber \]

    Show that \({\bf y}_2-\hat{\bf y}_2\) is a scalar multiple of \({\bf y}_1={\bf x}e^{\lambda_1t}\).

    34. Under the assumptions of Theorem 10.5.2, let

    \[\begin{aligned} {\bf y}_1 &={\bf x} e^{\lambda_1t},\\ {\bf y}_2&={\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\ {\bf y}_3&={\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}.\end{aligned}\nonumber \]

    Complete the proof of Theorem 10.5.2 by showing that \({\bf y}_3\) is a solution of \({\bf y}'=A{\bf y}\) and that \(\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}\) is linearly independent.

    35. Suppose the matrix \[A=\left[\begin{array}{cc}{a_{11}}&{a_{12}}\\{a_{21}}&{a_{22}}\end{array}\right]\nonumber\] has a repeated eigenvalue \(\lambda_1\) and the associated eigenspace is one-dimensional. Let \({\bf x}\) be a \(\lambda_1\)-eigenvector of \(A\). Show that if \((A-\lambda_1I){\bf u}_1={\bf x}\) and \((A-\lambda_1I){\bf u}_2={\bf x}\), then \({\bf u}_2-{\bf u}_1\) is parallel to \({\bf x}\). Conclude from this that all vectors \({\bf u}\) such that \((A-\lambda_1I){\bf u}={\bf x}\) define the same positive and negative half-planes with respect to the line \(L\) through the origin parallel to \({\bf x}\).

    Q10.5.5

    In Exercises 10.5.36-10.5.45 plot trajectories of the given system.

    36. \({\bf y}'=\left[\begin{array}{cc}{-3}&{-1}\\{4}&{1}\end{array}\right]{\bf y}\)

    37. \({\bf y}'=\left[\begin{array}{cc}{2}&{-1}\\{1}&{0}\end{array}\right]{\bf y}\)

    38. \({\bf y}'=\left[\begin{array}{cc}{-1}&{-3}\\{3}&{5}\end{array}\right]{\bf y}\)

    39. \({\bf y}'=\left[\begin{array}{cc}{-5}&{3}\\{-3}&{1}\end{array}\right]{\bf y}\)

    40. \({\bf y}'=\left[\begin{array}{cc}{-2}&{-3}\\{3}&{4}\end{array}\right]{\bf y}\)

    41. \({\bf y}'=\left[\begin{array}{cc}{-4}&{-3}\\{3}&{2}\end{array}\right]{\bf y}\)

    42. \({\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\{1}&{-2}\end{array}\right]{\bf y}\)

    43. \({\bf y}'=\left[\begin{array}{cc}{0}&{1}\\{-1}&{2}\end{array}\right]{\bf y}\)

    44. \({\bf y}'=\left[\begin{array}{cc}{-2}&{1}\\{-1}&{0}\end{array}\right]{\bf y}\)

    45. \({\bf y}'=\left[\begin{array}{cc}{0}&{-4}\\{1}&{-4}\end{array}\right]{\bf y}\)