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

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ In Exercises [exer:10.6.1}– [exer:10.6.16} find the general solution. [exer:10.6.1] $$\{\bf y'=\twobytwo{-1}2{-5}5\bf y }$$ [exer:10.6.2] $$\{\bf y'=\twobytwo{-11}4{-26}9\bf y }$$ [exer:10.6.3] $$\{\bf y'=\twobytwo12{-4}5\bf y}$$ [exer:10.6.4] $$\{\bf y'=\twobytwo5{-6}3{-1}\bf y}$$ [exer:10.6.5] $$\{\bf y'=\threebythree3{-3}1022511\bf y}$$ [exer:10.6.6] $$\{\bf y'=\threebythree{-3}311{-5}{-3}{-3}73\bf y}$$ [exer:10.6.7] $$\{\bf y'=\threebythree21{-1}011101\bf y}$$ [exer:10.6.8] $$\{\bf y'=\threebythree{-3}1{-3}4{-1}24{-2}3\bf y}$$ [exer:10.6.9] $$\{\bf y'=\twobytwo5{-4}{10}1\bf y}$$ [exer:10.6.10] $$\{\bf y'={1\over3}\twobytwo7{-5}25\bf y}$$ [exer:10.6.11] $$\{\bf y'=\twobytwo32{-5}1\bf y}$$ [exer:10.6.12] $$\{\bf y'=\twobytwo{34}{52}{-20}{-30}\bf y}$$ [exer:10.6.13] $$\{\bf y' =\threebythree11210{-1}{-1}{-2}{-1}\bf y}$$ [exer:10.6.14] $$\{\bf y' =\threebythree3{-4}{-2}{-5}7{-8}{-10}{13}{-8}\bf y}$$ [exer:10.6.15] $$\{\bf y'=\threebythree60{-3}{-3}331{-2}6\bf y'}$$ [exer:10.6.16] $$\{\bf y'=\threebythree12{-2}02{-1}100\bf y'}$$ [exer:10.6.17] $$\{\bf y'=\twobytwo4{-6}3{-2}\bf y,\quad \bf y(0)=\twocol52}$$ [exer:10.6.18] $$\{\bf y'=\twobytwo7{15}{-3}1\bf y,\quad \bf y(0)=\twocol51}$$ [exer:10.6.19] $$\{\bf y'=\twobytwo7{-15}3{-5}\bf y,\quad \bf y(0)=\twocol{17}7}$$ [exer:10.6.20] $$\{\bf y'={1\over6}\twobytwo4{-2}52\bf y,\quad \bf y(0)=\twocol1{-1}}$$ [exer:10.6.21] $$\{\bf y'=\threebythree52{-1}{-3}22132\bf y,\quad \bf y(0)=\threecol406}$$ [exer:10.6.22] $$\{\bf y'=\threebythree4408{10}{-20}23{-2}\bf y,\quad \bf y(0)=\threecol865}$$ [exer:10.6.23] $$\{\bf y'=\threebythree1{15}{-15}{-6}{18}{-22}{-3}{11}{-15}\bf y,\quad \bf y(0)=\threecol{15}{17}{10}}$$ [exer:10.6.24] $$\{\bf y'=\threebythree4{-4}4{-10}3{15}2{-3}1\bf y,\quad \bf y(0)=\threecol{16}{14}6}$$ [exer:10.6.25] 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. [exer:10.6.26] Verify that \[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \mbox{\quad and\quad} \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),$

are the real and imaginary parts of

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

[exer:10.6.27] Show that if the vectors $${\bf u}$$ and $${\bf v}$$ are not both $${\bf 0}$$ and $$\beta\ne0$$ then the vector functions

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

are linearly independent on every interval.

[exer:10.6.28] Suppose $${\bf u}=\{\twocol{u_1}{u_2}}$$ and $${\bf v}=\{\twocol{v_1}{v_2}}$$ are not orthogonal; that is, $$({\bf u},{\bf v})\ne0$$.

$({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0$

has a positive root $$k_1$$ and a negative root $$k_2=-1/k_1$$.

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

${\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \mbox{\quad and \quad} {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.$

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 $$-\pi/2$$.

 [exer:10.6.29]   $$\{\bf y'=\twobytwo3{-5}5{-3}\bf y}$$ [exer:10.6.30]   $$\{\bf y'=\twobytwo{-15}{10}{-25}{15}\bf y}$$
 [exer:10.6.31]   $$\{\bf y'=\twobytwo{-4}8{-4}4\bf y}$$ [exer:10.6.32]   $$\{\bf y'=\twobytwo{-3}{-15}33\bf y}$$
 [exer:10.6.33]   $$\{\bf y'=\twobytwo{-5}6{-12}7\bf y}$$ [exer:10.6.34]   $$\{\bf y'=\twobytwo5{-12}6{-7}\bf y}$$
 [exer:10.6.35]   $$\{\bf y'=\twobytwo4{-5}9{-2}\bf y}$$ [exer:10.6.36]   $$\{\bf y'=\twobytwo{-4}9{-5}2\bf y}$$
 [exer:10.6.37]   $$\{\bf y'=\twobytwo{-1}{10}{-10}{-1}\bf y}$$ [exer:10.6.38]   $$\{\bf y'=\twobytwo{-1}{-5}{20}{-1}\bf y}$$
 [exer:10.6.39]   $$\{\bf y'=\twobytwo{-7}{10}{-10}9\bf y}$$ [exer:10.6.40]   $$\{\bf y'=\twobytwo{-7}6{-12}5\bf y}$$