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# 10.2E: Linear Systems of Differential Equations (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ [exer:10.2.1] Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$ and $$c_2$$. $$\begin{array}{ccl}y'_1&=&2y_1 + 4y_2\\ y_2'&=&4y_1+2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{6t}+c_2\twocol1{-1}e^{-2t}$$ $$\begin{array}{ccl}y'_1&=&-2y_1 - 2y_2\\ y_2'&=&-5y_1 + \phantom{2}y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{-4t}+c_2\twocol{-2}5e^{3t}$$ $$\begin{array}{ccr}y'_1&=&-4y_1 -10y_2\\ y_2'&=&3y_1 + \phantom{1}7y_2;\end{array} \quad {\bf y}=c_1\twocol{-5}3e^{2t}+c_2\twocol2{-1}e^t$$ $$\begin{array}{ccl}y'_1&=&2y_1 +\phantom{2}y_2 \\ y_2'&=&\phantom{2}y_1 + 2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{3t}+c_2\twocol1{-1}e^t$$ [exer:10.2.2] Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$, $$c_2$$, and $$c_3$$. $$\begin{array}{ccr}y'_1&=&- y_1+2y_2 + 3y_3 \\ y_2'&=&y_2 + 6y_3\\y_3'&=&- 2y_3;\end{array}$$ $${\bf y}=c_1\threecol110e^t+c_2\threecol100e^{-t}+c_3\threecol1{-2}1e^{-2t}$$ $$\begin{array}{ccc}y'_1&=&\phantom{2y_1+}2y_2 + 2y_3 \\ y_2'&=&2y_1\phantom{+2y_2} + 2y_3\\y_3'&=&2y_1 + 2y_2;\phantom{+2y_3}\end{array}$$ $${\bf y}=c_1\threecol{-1}01e^{-2t}+c_2\threecol0{-1}1e^{-2t}+c_3\threecol111e^{4t}$$ $$\begin{array}{ccr}y'_1&=&-y_1 +2y_2 + 2y_3\\ y_2'&=&2y_1 -\phantom{2}y_2 +2y_3\\y_3'&=&2y_1 + 2y_2 -\phantom{2}y_3;\end{array}$$ $${\bf y}=c_1\threecol{-1}01e^{-3t}+c_2\threecol0{-1}1e^{-3t}+c_3\threecol111e^{3t}$$ $$\begin{array}{ccr}y'_1&=&3y_1 - \phantom{2}y_2 -\phantom{2}y_3 \\ y_2'&=&-2y_1 + 3y_2 + 2y_3\\y_3'&=&\phantom{-}4y_1 -\phantom{3}y_2 - 2y_3;\end{array}$$ $${\bf y}=c_1\threecol101e^{2t}+c_2\threecol1{-1}1e^{3t}+c_3\threecol1{-3}7e^{-t}$$ [exer:10.2.3] Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. $$\begin{array}{ccl}y'_1 &=&\phantom{-2}y_1+\phantom{4}y_2\\ y_2'&=&-2y_1 + 4y_2,\end{array} \begin{array}{ccr}y_1(0)&=&1\\y_2(0)&=&0;\end{array}$$ $${\bf y}=2\twocol11e^{2t}-\twocol12e^{3t}$$ $$\begin{array}{ccl}y'_1 &=&5y_1 + 3y_2 \\ y_2'&=&- y_1 + y_2,\end{array} \begin{array}{ccr}y_1(0)&=&12\\y_2(0)&=&-6;\end{array}$$ $${\bf y}=3\twocol1{-1}e^{2t}+3\twocol3{-1}e^{4t}$$ [exer:10.2.4] Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. $$\begin{array}{ccr}y'_1&=&6y_1 + 4y_2 + 4y_3 \\ y_2'&=&-7y_1 -2y_2 - y_3,\\y_3'&=&7y_1 + 4y_2 + 3y_3\end{array},\; \begin{array}{ccr}y_1(0)&=&3\\ y_2(0)&=&-6\\ y_3(0)&=&4\end{array}$$ $${\bf y}=\threecol1{-1}1e^{6t}+2\threecol1{-2}1e^{2t}+\threecol0{-1}1e^{-t}$$ $$\begin{array}{ccr}y'_1&=& \phantom{-}8y_1 + 7y_2 +\phantom{1}7y_3 \\ y_2'&=&-5y_1 -6y_2 -\phantom{1}9y_3,\\y_3'&=& \phantom{-}5y_1 + 7y_2 +10y_3,\end{array}\ \begin{array}{ccr}y_1(0)&=&2\\ y_2(0)&=&-4\\ y_3(0)&=&3\end{array}$$ $${\bf y}=\threecol1{-1}1e^{8t}+\threecol0{-1}1e^{3t}+\threecol1{-2}1e^t$$ [exer:10.2.5] Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$ and $$c_2$$. $$\begin{array}{ccc}y'_1&=&-3y_1+2y_2+3-2t \\ y_2'&=&-5y_1+3y_2+6-3t\end{array}$$ $${\bf y}=c_1\left[\begin{array}{c}2\cos t\\3\cos t-\sin t\end{array}\right]+c_2\left[\begin{array}{c}2\sin t\\3\sin t+\cos t \end{array}\right]+\twocol1t$$ $$\begin{array}{ccc}y'_1&=&3y_1+y_2-5e^t \\ y_2'&=&-y_1+y_2+e^t\end{array}$$ $${\bf y}=c_1\twocol{-1}1e^{2t}+c_2\left[\begin{array}{c}1+t\\-t\end{array} \right]e^{2t}+\twocol13e^t$$ $$\begin{array}{ccl}y'_1&=&-y_1-4y_2+4e^t+8te^t \\ y_2'&=&-y_1-\phantom{4}y_2+e^{3t}+(4t+2)e^t\end{array}$$ $${\bf y}=c_1\twocol21e^{-3t}+c_2\twocol{-2}1e^t+\left[\begin{array}{c} e^{3t}\\2te^t\end{array}\right]$$ $$\begin{array}{ccc}y'_1&=&-6y_1-3y_2+14e^{2t}+12e^t \\ y_2'&=&\phantom{6}y_1-2y_2+7e^{2t}-12e^t\end{array}$$ $${\bf y}=c_1\twocol{-3}1e^{-5t}+c_2\twocol{-1}1e^{-3t}+ \left[\begin{array}{c}e^{2t}+3e^t\\2e^{2t}-3e^t\end{array}\right]$$ [exer:10.2.6] Convert the linear scalar equation \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y(t)=F(t) \eqno{\rm (A)}$

into an equivalent $$n\times n$$ system

${\bf y'}=A(t){\bf y}+{\bf f}(t),$

and show that $$A$$ and $${\bf f}$$ are continuous on an interval $$(a,b)$$ if and only if (A) is normal on $$(a,b)$$.

[exer:10.2.7] A matrix function

$Q(t)=\matfunc qrs$

is said to be differentiable if its entries $$\{q_{ij}\}$$ are differentiable. Then the derivative $$Q'$$ is defined by

$Q'(t)=\matfunc {q'}rs.$

Prove: If $$P$$ and $$Q$$ are differentiable matrices such that $$P+Q$$ is defined and if $$c_1$$ and $$c_2$$ are constants, then

$(c_1P+c_2Q)'=c_1P'+c_2Q'.$

Prove: If $$P$$ and $$Q$$ are differentiable matrices such that $$PQ$$ is defined, then

$(PQ)'=P'Q+PQ'.$

[exer:10.2.8] Verify that $$Y' = AY$$.

$$\{Y = \twobytwo {e^{6t}}{e^{-2t}} {e^{6t}}{-e^{-2t}}, \quad A = \twobytwo 2 4 4 2}$$

$$\{Y = \twobytwo {e^{-4t}} {-2e^{3t}} {e^{-4t}} {5e^{3t}}, \quad A = \twobytwo {-2} {-2} {-5} {1}}$$

$$\{Y = \twobytwo {-5e^{2t}} {2e^t} {3e^{2t}} {-e^t}, \quad A = \twobytwo {-4} {-10} 3 7}$$

$$\{Y = \twobytwo {e^{3t}} {e^t} {e^{3t}} {-e^t}, \quad A = \twobytwo 2 1 1 2}$$

$$Y = \left[\begin{array}{crr} e^t&e^{-t}& e^{-2t}\\ e^t&0&-2e^{-2t}\\ 0&0&e^{-2t}\end{array}\right], \quad A = \threebythree {-1} 2 {3} {0} 1 6 0 0 {-2}$$

$$\{Y = \cthreebythree {-e^{-2t}} {-e^{-2t}} {e^{4t}} 0 {\phantom{-} e^{-2t}} {e^{4t}} {e^{-2t}} 0 {e^{4t}}, \quad A = \threebythree 0 2 2 2 0 2 2 2 0}$$

$$\{Y = \cthreebythree {e^{3t}} {e^{-3t}} 0 {e^{3t}} 0 {-e^{-3t}} {e^{3t}} {e^{-3t}} {\phantom{-}e^{-3t}}, \quad A = \threebythree {-9}66{-6}36{-6}63}$$

$$Y = \left[\begin{array}{crr} e^{2t}&e^{3t}& e^{-t}\\ 0&-e^{3t}&-3e^{-t}\\ e^{2t}&e^{3t}&7e^{-t}\end{array}\right] , \quad A = \threebythree 3 {-1} {-1}{-2} 3 2 4 {-1} {-2}$$

[exer:10.2.9] Suppose

${\bf y}_1=\twocol{y_{11}}{y_{21}}\mbox{\quad and \quad}{\bf y}_2=\twocol{y_{12}}{y_{22}}$

are solutions of the homogeneous system

${\bf y}'=A(t){\bf y}, \eqno{\rm (A)}$

and define

$Y= \twobytwo{y_{11}}{y_{12}}{y_{21}}{y_{22}}.$

Show that $$Y'=AY$$.

Show that if $${\bf c}$$ is a constant vector then $${\bf y}= Y{\bf c}$$ is a solution of (A).

State generalizations of

and

## b

for $$n\times n$$ systems.

[exer:10.2.10] Suppose $$Y$$ is a differentiable square matrix.

Find a formula for the derivative of $$Y^2$$.

Find a formula for the derivative of $$Y^n$$, where $$n$$ is any positive integer.

State how the results obtained in

and

## b

are analogous to results from calculus concerning scalar functions.

[exer:10.2.11] It can be shown that if $$Y$$ is a differentiable and invertible square matrix function, then $$Y^{-1}$$ is differentiable.

Show that ($$Y^{-1})'= -Y^{-1}Y'Y^{-1}$$. (Hint: Differentiate the identity $$Y^{-1}Y=I$$.)

Find the derivative of $$Y^{-n}=\left(Y^{-1}\right)^n$$, where $$n$$ is a positive integer.

State how the results obtained in

and

## b

are analogous to results from calculus concerning scalar functions.

[exer:10.2.12] Show that Theorem [thmtype:10.2.1} implies Theorem [thmtype:9.1.1}.

[exer:10.2.13] Suppose $${\bf y}$$ is a solution of the $$n\times n$$ system $${\bf y}'=A(t){\bf y}$$ on $$(a,b)$$, and that the $$n\times n$$ matrix $$P$$ is invertible and differentiable on $$(a,b)$$. Find a matrix $$B$$ such that the function $${\bf x}=P{\bf y}$$ is a solution of $${\bf x}'=B{\bf x}$$ on $$(a,b)$$.