# 10.5.1: Constant Coefficient Homogeneous Systems II (Exercises)

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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. $${\bf y}'=\left[\begin{array}{cc}{3}&{4}\\[4pt]{-1}&{7}\end{array}\right]{\bf y}$$ 2. $${\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\[4pt]{1}&{-2}\end{array}\right]{\bf y}$$ 3. $${\bf y}'=\left[\begin{array}{cc}{-7}&{4}\\[4pt]{-1}&{-11}\end{array}\right]{\bf y}$$ 4. $${\bf y}'=\left[\begin{array}{cc}{3}&{1}\\[4pt]{-1}&{1}\end{array}\right]{\bf y}$$ 5. $${\bf y}'=\left[\begin{array}{cc}{4}&{12}\\[4pt]{-3}&{-8}\end{array}\right]{\bf y}$$ 6. $${\bf y}'=\left[\begin{array}{cc}{-10}&{9}\\[4pt]{-4}&{2}\end{array}\right]{\bf y}$$ 7. $${\bf y}'=\left[\begin{array}{cc}{-13}&{16}\\[4pt]{-9}&{11}\end{array}\right]{\bf y}$$ 8. $${\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{1}\\[4pt]{-4}&{6}&{1}\\[4pt]{0}&{4}&{2}\end{array}\right]{\bf y}$$ 9. $${\bf y}'=\frac{1}{3}\left[\begin{array}{ccc}{1}&{1}&{-3}\\[4pt]{-4}&{-4}&{3}\\[4pt]{-2}&{1}&{0}\end{array}\right]{\bf y}$$ 10. $${\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{-1}\\[4pt]{-2}&{0}&{2}\\[4pt]{-1}&{3}&{-1}\end{array}\right]{\bf y}$$ 11. $${\bf y}'=\left[\begin{array}{ccc}{4}&{-2}&{-2}\\[4pt]{-2}&{3}&{-1}\\[4pt]{2}&{-1}&{3}\end{array}\right]{\bf y}$$ 12. $${\bf y}'=\left[\begin{array}{ccc}{6}&{-5}&{3}\\[4pt]{2}&{-1}&{3}\\[4pt]{2}&{1}&{1}\end{array}\right]{\bf y}$$ ## Q10.5.2 In Exercises 10.5.13-10.5.23 solve the initial value problem. 13. $${\bf y}'=\left[\begin{array}{cc}{-11}&{8}\\[4pt]{-2}&{-3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\[4pt]{2}\end{array}\right]$$ 14. $${\bf y}'=\left[\begin{array}{cc}{15}&{-9}\\[4pt]{16}&{-9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{5}\\[4pt]{8}\end{array}\right]$$ 15. $${\bf y}'=\left[\begin{array}{cc}{-3}&{-4}\\[4pt]{1}&{-7}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{2}\\[4pt]{3}\end{array}\right]$$ 16. $${\bf y}'=\left[\begin{array}{cc}{-7}&{24}\\[4pt]{-6}&{17}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{3}\\[4pt]{1}\end{array}\right]$$ 17. $${\bf y}'=\left[\begin{array}{cc}{-7}&{3}\\[4pt]{-3}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\[4pt]{2}\end{array}\right]$$ 18. $${\bf y}'=\left[\begin{array}{ccc}{-1}&{1}&{0}\\[4pt]{1}&{-1}&{-2}\\[4pt]{-1}&{-1}&{-1}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{6}\\[4pt]{5}\\[4pt]{-7}\end{array}\right]$$ 19. $${\bf y}'=\left[\begin{array}{ccc}{-2}&{2}&{1}\\[4pt]{-2}&{2}&{1}\\[4pt]{-3}&{3}&{2}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\[4pt]{-2}\\[4pt]{0}\end{array}\right]$$ 20. $${\bf y}'=\left[\begin{array}{ccc}{-7}&{-4}&{4}\\[4pt]{-1}&{0}&{1}\\[4pt]{-9}&{-5}&{6}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-6}\\[4pt]{9}\\[4pt]{-1}\end{array}\right]$$ 21. $${\bf y}'=\left[\begin{array}{ccc}{-1}&{-4}&{-1}\\[4pt]{3}&{6}&{1}\\[4pt]{-3}&{-2}&{3}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-2}\\[4pt]{1}\\[4pt]{3}\end{array}\right]$$ 22. $${\bf y}'=\left[\begin{array}{ccc}{4}&{-8}&{-4}\\[4pt]{-3}&{-1}&{-3}\\[4pt]{1}&{-1}&{9}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{-4}\\[4pt]{1}\\[4pt]{-3}\end{array}\right]$$ 23. $${\bf y}'=\left[\begin{array}{ccc}{-5}&{-1}&{11}\\[4pt]{-7}&{1}&{13}\\[4pt]{-4}&{0}&{8}\end{array}\right]{\bf y},\quad {\bf y}(0)=\left[\begin{array}{c}{0}\\[4pt]{2}\\[4pt]{2}\end{array}\right]$$ ## Q10.5.3 The coefficient matrices in Exercises 10.5.24-10.5.32 have eigenvalues of multiplicity $$3$$. Find the general solution. 24. $${\bf y}'=\left[\begin{array}{ccc}{5}&{-1}&{1}\\[4pt]{-1}&{9}&{-3}\\[4pt]{-2}&{2}&{4}\end{array}\right]{\bf y}$$ 25. $${\bf y}'=\left[\begin{array}{ccc}{1}&{10}&{-12}\\[4pt]{2}&{2}&{3}\\[4pt]{2}&{-1}&{6}\end{array}\right]{\bf y}$$ 26. $${\bf y}'=\left[\begin{array}{ccc}{-6}&{-4}&{-4}\\[4pt]{2}&{-1}&{1}\\[4pt]{2}&{3}&{1}\end{array}\right]{\bf y}$$ 27. $${\bf y}'=\left[\begin{array}{ccc}{0}&{2}&{-2}\\[4pt]{-1}&{5}&{-3}\\[4pt]{1}&{1}&{1}\end{array}\right]{\bf y}$$ 28. $${\bf y}'=\left[\begin{array}{ccc}{-2}&{-12}&{10}\\[4pt]{2}&{-24}&{11}\\[4pt]{2}&{-24}&{8}\end{array}\right]{\bf y}$$ 29. $${\bf y}'=\left[\begin{array}{ccc}{-1}&{-12}&{8}\\[4pt]{1}&{-9}&{4}\\[4pt]{1}&{-6}&{1}\end{array}\right]{\bf y}$$ 30. $${\bf y}'=\left[\begin{array}{ccc}{-4}&{0}&{-1}\\[4pt]{-1}&{-3}&{-1}\\[4pt]{1}&{0}&{-2}\end{array}\right]{\bf y}$$ 31. $${\bf y}'=\left[\begin{array}{ccc}{-3}&{-3}&{4}\\[4pt]{4}&{5}&{-8}\\[4pt]{2}&{3}&{-5}\end{array}\right]{\bf y}$$ 32. $${\bf y}'=\left[\begin{array}{ccc}{-3}&{-1}&{0}\\[4pt]{1}&{-1}&{0}\\[4pt]{-1}&{-1}&{-2}\end{array}\right]{\bf y}$$ ## 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},\\[4pt] {\bf y}_2&={\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\[4pt] {\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 $A=\left[\begin{array}{cc}{a_{11}}&{a_{12}}\\[4pt]{a_{21}}&{a_{22}}\end{array}\right]\nonumber$ 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. $${\bf y}'=\left[\begin{array}{cc}{-3}&{-1}\\[4pt]{4}&{1}\end{array}\right]{\bf y}$$

37. $${\bf y}'=\left[\begin{array}{cc}{2}&{-1}\\[4pt]{1}&{0}\end{array}\right]{\bf y}$$

38. $${\bf y}'=\left[\begin{array}{cc}{-1}&{-3}\\[4pt]{3}&{5}\end{array}\right]{\bf y}$$

39. $${\bf y}'=\left[\begin{array}{cc}{-5}&{3}\\[4pt]{-3}&{1}\end{array}\right]{\bf y}$$

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

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

42. $${\bf y}'=\left[\begin{array}{cc}{0}&{-1}\\[4pt]{1}&{-2}\end{array}\right]{\bf y}$$

43. $${\bf y}'=\left[\begin{array}{cc}{0}&{1}\\[4pt]{-1}&{2}\end{array}\right]{\bf y}$$

44. $${\bf y}'=\left[\begin{array}{cc}{-2}&{1}\\[4pt]{-1}&{0}\end{array}\right]{\bf y}$$

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

This page titled 10.5.1: Constant Coefficient Homogeneous Systems II (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.