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# 10.6E: Constant Coefficient Homogeneous Systems III (Exercises)

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ ## Q10.6.1 In Exercises 10.6.1-10.6.16 find the general solution. 1. $${\bf y}'=\left[\begin{array}{cc}{-1}&{2}\\{-5}&{5}\end{array}\right]{\bf y}$$ 2. $${\bf y}'=\left[\begin{array}{cc}{-11}&{4}\\{-26}&{9}\end{array}\right]{\bf y}$$ 3. $${\bf y}'=\left[\begin{array}{cc}{1}&{2}\\{-4}&{5}\end{array}\right]{\bf y}$$ 4. $${\bf y}'=\left[\begin{array}{cc}{5}&{-6}\\{3}&{-1}\end{array}\right]{\bf y}$$ 5. $${\bf y}'=\left[\begin{array}{ccc}{3}&{-3}&{1}\\{0}&{2}&{2}\\{5}&{1}&{1}\end{array}\right]{\bf y}$$ 6. $${\bf y}'=\left[\begin{array}{ccc}{-3}&{3}&{1}\\{1}&{-5}&{-3}\\{-3}&{7}&{3}\end{array}\right]{\bf y}$$ 7. $${\bf y}'=\left[\begin{array}{ccc}{2}&{1}&{-1}\\{0}&{1}&{1}\\{1}&{0}&{1}\end{array}\right]{\bf y}$$ 8. $${\bf y}'=\left[\begin{array}{ccc}{-3}&{1}&{-3}\\{4}&{-1}&{2}\\{4}&{-2}&{3}\end{array}\right]{\bf y}$$ 9. $${\bf y}'=\left[\begin{array}{cc}{5}&{-4}\\{10}&{1}\end{array}\right]{\bf y}$$ 10. $${\bf y}'=\frac{1}{3}\left[\begin{array}{cc}{7}&{-5}\\{2}&{5}\end{array}\right]{\bf y}$$ 11. $${\bf y}'=\left[\begin{array}{cc}{3}&{2}\\{-5}&{1}\end{array}\right]{\bf y}$$ 12. $${\bf y}'=\left[\begin{array}{cc}{34}&{52}\\{-20}&{-30}\end{array}\right]{\bf y}$$ 13. $${\bf y}'=\left[\begin{array}{ccc}{1}&{1}&{2}\\{1}&{0}&{-1}\\{-1}&{-2}&{-1}\end{array}\right]{\bf y}$$ 14. $${\bf y}'=\left[\begin{array}{ccc}{3}&{-4}&{-2}\\{-5}&{7}&{-8}\\{-10}&{13}&{-8}\end{array}\right]{\bf y}$$ 15. $${\bf y}'=\left[\begin{array}{ccc}{6}&{0}&{-3}\\{-3}&{3}&{3}\\{1}&{-2}&{6}\end{array}\right]{\bf y}$$ 16. $${\bf y}'=\left[\begin{array}{ccc}{1}&{2}&{-2}\\{0}&{2}&{-1}\\{1}&{0}&{0}\end{array}\right]{\bf y}$$ ## Q10.6.2 In Exercises 10.6.17-10.6.24 solve the initial value problem. 17. $${\bf y}'=\left[\begin{array}{cc}{4}&{-6}\\{3}&{-2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{5}\\{2}\end{array}\right]$$ 18. $${\bf y}'=\left[\begin{array}{cc}{7}&{15}\\{-3}&{1}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{5}\\{1}\end{array}\right]$$ 19. $${\bf y}'=\left[\begin{array}{cc}{7}&{-15}\\{3}&{-5}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{17}\\{7}\end{array}\right]$$ 20. $${\bf y}'=\frac{1}{6}\left[\begin{array}{cc}{4}&{-2}\\{5}&{2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{1}\\{-1}\end{array}\right]$$ 21. $${\bf y}'=\left[\begin{array}{ccc}{5}&{2}&{-1}\\{-3}&{2}&{2}\\{1}&{3}&{2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{4}\\{0}\\{6}\end{array}\right]$$ 22. $${\bf y}'=\left[\begin{array}{ccc}{4}&{4}&{0}\\{8}&{10}&{-20}\\{2}&{3}&{-2}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{8}\\{6}\\{5}\end{array}\right]$$ 23. $${\bf y}'=\left[\begin{array}{ccc}{1}&{15}&{-15}\\{-6}&{18}&{-22}\\{-3}&{11}&{-15}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{15}\\{17}\\{10}\end{array}\right]$$ 24. $${\bf y}'=\left[\begin{array}{ccc}{4}&{-4}&{4}\\{-10}&{3}&{15}\\{2}&{-3}&{1}\end{array}\right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{c}{16}\\{14}\\{6}\end{array}\right]$$ ## Q10.6.3 25. Suppose an $$n\times n$$ matrix $$A$$ with real entries has a complex eigenvalue $$\lambda=\alpha+i\beta$$ ($$\beta\ne0$$) with associated eigenvector $${\bf x}={\bf u}+i{\bf v}$$, where $${\bf u}$$ and $${\bf v}$$ have real components. Show that $${\bf u}$$ and $${\bf v}$$ are both nonzero. 26. Verify that \[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \quad \text{and}\quad \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),\nonumber$

are the real and imaginary parts of

$e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}).\nonumber$

27. Show that if the vectors $${\bf u}$$ and $${\bf v}$$ are not both $${\bf 0}$$ and $$\beta\ne0$$ then the vector functions

$\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad \mbox{ and }\quad \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)\nonumber$

are linearly independent on every interval.

28. Suppose $${\bf u}=\left[\begin{array}{c}{u_{1}}\\{u_{2}}\end{array}\right]$$ and $${\bf v}=\left[\begin{array}{c}{v_{1}}\\{v_{2}}\end{array}\right]$$ are not orthogonal; that is, $$({\bf u},{\bf v})\ne0$$.

1. Show that the quadratic equation $({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0\nonumber$ has a positive root $$k_1$$ and a negative root $$k_2=-1/k_1$$.
2. Let $${\bf u}_1^{(1)}={\bf u}-k_1{\bf v}$$, $${\bf v}_1^{(1)}={\bf v}+k_1{\bf u}$$, $${\bf u}_1^{(2)}={\bf u}-k_2{\bf v}$$, and $${\bf v}_1^{(2)}={\bf v}+k_2{\bf u}$$, so that $$({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0$$, from the discussion given above. Show that ${\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \quad \text{and} \quad {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.\nonumber$
3. Let $${\bf U}_1$$, $${\bf V}_1$$, $${\bf U}_2$$, and $${\bf V}_2$$ be unit vectors in the directions of $${\bf u}_1^{(1)}$$, $${\bf v}_1^{(1)}$$, $${\bf u}_1^{(2)}$$, and $${\bf v}_1^{(2)}$$, respectively. Conclude from (a) that $${\bf U}_2={\bf V}_1$$ and $${\bf V}_2=-{\bf U}_1$$, and that therefore the counterclockwise angles from $${\bf U}_1$$ to $${\bf V}_1$$ and from $${\bf U}_2$$ to $${\bf V}_2$$ are both $$\pi/2$$ or both $$-\pi/2$$.

## Q10.6.4

In Exercises 10.6.29-10.6.32 find vectors $${\bf U}$$ and $${\bf V}$$ parallel to the axes of symmetry of the trajectories, and plot some typical trajectories.

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

30. $${\bf y}'=\left[\begin{array}{cc}{-15}&{10}\\{-25}&{15}\end{array}\right]{\bf y}$$

31. $${\bf y}'=\left[\begin{array}{cc}{-4}&{8}\\{-4}&{4}\end{array}\right]{\bf y}$$

32. $${\bf y}'=\left[\begin{array}{cc}{-3}&{-15}\\{3}&{3}\end{array}\right]{\bf y}$$

## Q10.6.5

In Exercises 10.6.33-10.6.40 find vectors $${\bf U}$$ and $${\bf V}$$ parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory.

33. $${\bf y}'=\left[\begin{array}{cc}{-5}&{6}\\{-12}&{7}\end{array}\right]{\bf y}$$

34. $${\bf y}'=\left[\begin{array}{cc}{5}&{-12}\\{6}&{-7}\end{array}\right]{\bf y}$$

35. $${\bf y}'=\left[\begin{array}{cc}{4}&{-5}\\{9}&{-2}\end{array}\right]{\bf y}$$

36. $${\bf y}'=\left[\begin{array}{cc}{-4}&{9}\\{-5}&{2}\end{array}\right]{\bf y}$$

37. $${\bf y}'=\left[\begin{array}{cc}{-1}&{10}\\{-10}&{-1}\end{array}\right]{\bf y}$$

38. $${\bf y}'=\left[\begin{array}{cc}{-1}&{-5}\\{20}&{-1}\end{array}\right]{\bf y}$$

39. $${\bf y}'=\left[\begin{array}{cc}{-7}&{10}\\{-10}&{9}\end{array}\right]{\bf y}$$

40. $${\bf y}'=\left[\begin{array}{cc}{-7}&{6}\\{-12}&{5}\end{array}\right]{\bf y}$$