<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}

{\bf Question}

Find the general solution to the differential equation

\[
\dot{\bf x}=M{\bf x}\qquad\hbox{ with }\quad {\bf x}=\pmatrix{x
\cr y \cr}
\]

for each of the following matrices $M$:

\begin{enumerate}

\item
\[
M=\pmatrix{3 &amp;-2 \cr
           2 &amp;-2 \cr} \qquad(*)
\]

\item
\[
M=\pmatrix{0 &amp;1 \cr
           8 &amp;-2 \cr}
\]

\item
\[
M=\pmatrix{3 &amp;-5 \cr
           2 &amp;1 \cr} \qquad(*)
\]

\item
\[
M=\pmatrix{3 &amp; -1 \cr
           1 &amp; 1 \cr} \qquad(*)
\]



\end{enumerate}



\vspace{0.25in}

{\bf Answer}

\begin{enumerate}

\item
\[
\dot{\bf x}= \pmatrix{3 &amp;-2 \cr
           2 &amp;-2 \cr}{\bf x} \qquad \hbox{ so try }\quad {\bf x}={\bf
           x}_0e^{\lambda t}
\]
Hence we need $(M - \lambda I) {\bf x}_0=0$ which implies find
$\lambda$ so that $\left| M-\lambda I \right|=0$ \\(or else ${\bf
x}_0={\bf 0}$).\\ $\ds{\left| \pmatrix{3 &amp;-2 \cr 2 &amp;-2 \cr}-
\lambda \pmatrix{1 &amp; 0 \cr 0 &amp; 1 \cr}\right|=0}$ implies
$\ds{\left|\pmatrix{3-\lambda &amp;-2 \cr 2 &amp;-2-\lambda
\cr}\right|=0}$.\\ $(3-\lambda)(-2-\lambda)+4=0$ so that
$\lambda^2-\lambda -2=0$ hence $\lambda=2,-1$\\ For $\lambda=2$ we
must find ${\bf x_0}$.\\ $(M-\lambda I){\bf x}_0={\bf 0}$ implies
$\ds{\left(\pmatrix{3 &amp;-2 \cr 2 &amp;-2 \cr}- 2 \pmatrix{1 &amp; 0 \cr 0 &amp;
1 \cr}\right)\pmatrix{x_0 \cr y_0 \cr}=0}$\\ $\ds{\pmatrix{1 &amp;-2
\cr 2 &amp;-4 \cr}\pmatrix{x_0 \cr y_0 \cr}={\bf 0}}$ implying $x_0=D$
and $y_0=D/2$.\\ Taking $D=2$, for example gives, $\ds{{\bf
x}_1=\pmatrix{2 \cr 1 \cr}e^{2t}}$.\\ For $\lambda=-1$ we must
find ${\bf x_0}$.\\ $(M-\lambda I){\bf x}_0={\bf 0}$ implies
$\ds{\left(\pmatrix{3 &amp;-2 \cr 2 &amp;-2 \cr}+ \pmatrix{1 &amp; 0 \cr 0 &amp; 1
\cr}\right)\pmatrix{x_0 \cr y_0 \cr}=0}$\\ $\ds{\pmatrix{4 &amp;-2 \cr
2 &amp;-1 \cr}\pmatrix{x_0 \cr y_0 \cr}={\bf 0}}$ implying $x_0=D$ and
$y_0=2D$.\\ Taking $D=1$, for example gives, $\ds{{\bf
x}_2=\pmatrix{1 \cr 2 \cr}e^{-t}}$.\\ The general solution is
therefore given by:\\ $\ds{{\bf x}(t)= A\pmatrix{2 \cr 1
\cr}e^{2t} + s\pmatrix{1 \cr 2 \cr}e^{-t} }$.\\ Or\quad
$x=2Ae^{2t}+Be^{-t}$\quad and \quad$y=Ae^{2t}+2Be^{-t}$.

\item
\[
\dot{\bf x}= \pmatrix{0 &amp;1 \cr
           8 &amp;-2 \cr}{\bf x} \qquad \hbox{ so try }\quad {\bf x}={\bf
           x}_0e^{\lambda t}
\]
Hence we need $(M - \lambda I) {\bf x}_0=0$ which implies find
$\lambda$ so that $\left| M-\lambda I \right|=0$ \\ $\ds{\left|
\pmatrix{0 &amp; 1 \cr 8 &amp;-2 \cr}- \lambda \pmatrix{1 &amp; 0 \cr 0 &amp; 1
\cr}\right|=0}$ implies $\ds{\left|\pmatrix{-\lambda &amp; 1 \cr 8
&amp;-2-\lambda \cr}\right|=0}$.\\ $\lambda(2+\lambda)-8=0$ so that
$\lambda^2+2\lambda -8=0$ hence $\lambda=2,-4$\\ For $\lambda=2$
we must find ${\bf x_0}$.\\ $(M-\lambda I){\bf x}_0={\bf 0}$
implies $\ds{\left(\pmatrix{0 &amp; 1 \cr 8 &amp;-2 \cr}- 2 \pmatrix{1 &amp; 0
\cr 0 &amp; 1 \cr}\right)\pmatrix{x_0 \cr y_0 \cr}=0}$\\
$\ds{\pmatrix{-2 &amp; 1 \cr 8 &amp;-4 \cr}\pmatrix{x_0 \cr y_0 \cr}={\bf
0}}$ implying $x_0=D$ and $y_0=2D$.\\ Taking $D=1$, for example
gives, $\ds{{\bf x}_1=\pmatrix{1 \cr 2 \cr}e^{2t}}$.\\ For
$\lambda=-4$ we must find ${\bf x_0}$.\\ $(M-\lambda I){\bf
x}_0={\bf 0}$ implies $\ds{\left(\pmatrix{0 &amp; 1 \cr 8 &amp; -2 \cr}+
\pmatrix{1 &amp; 0 \cr 0 &amp; 1 \cr}\right)\pmatrix{x_0 \cr y_0
\cr}=0}$\\ $\ds{\pmatrix{4 &amp; 1 \cr 8 &amp; 2 \cr}\pmatrix{x_0 \cr y_0
\cr}={\bf 0}}$ implying $x_0=D$ and $y_0=-4D$.\\ Taking $D=1$, for
example gives, $\ds{{\bf x}_2=\pmatrix{1 \cr -4 \cr}e^{-4t}}$.\\
The general solution is therefore given by:\\ $\ds{{\bf x}(t)=
A\pmatrix{1 \cr 2 \cr}e^{2t} + B\pmatrix{1 \cr -4 \cr}e^{-4t}
}$.\\ Or\quad $x=Ae^{2t}+Be^{-4t}$\quad and
\quad$y=2Ae^{2t}-4Be^{-4t}$.

\item
\[
\dot{\bf x}= \pmatrix{3 &amp; -5 \cr
           2 &amp; 1 \cr}{\bf x} \qquad \hbox{ so try }\quad {\bf x}={\bf
           x}_0e^{\lambda t}
\]
Hence we need $(M - \lambda I) {\bf x}_0=0$ which implies find
$\lambda$ so that $\left| M-\lambda I \right|=0$ \\ $\ds{\left|
\pmatrix{3 &amp; -5 \cr 2 &amp; 1 \cr}- \lambda \pmatrix{1 &amp; 0 \cr 0 &amp; 1
\cr}\right|=0}$ implies $\ds{\left|\pmatrix{3-\lambda &amp; -5 \cr 2 &amp;
1-\lambda \cr}\right|=0}$.\\ $(3-\lambda)(1-\lambda)+10=0$ so that
$\lambda^2-4\lambda +13=0$ hence $\lambda=2+3i,2-3i$\\ For
$\lambda=2+3i$ we must find ${\bf x_0}$.\\ $(M-\lambda I){\bf
x}_0={\bf 0}$ implies $\ds{\left(\pmatrix{3 &amp; -5 \cr 2 &amp; 1 \cr}-
(2+3i) \pmatrix{1 &amp; 0 \cr 0 &amp; 1 \cr}\right)\pmatrix{x_0 \cr y_0
\cr}=0}$\\ $\ds{\pmatrix{1-3i &amp; -5 \cr 2 &amp; -1-3i \cr}\pmatrix{x_0
\cr y_0 \cr}={\bf 0}}$ implying $x_0=D$ and $y_0=(1-3i)D/5$.\\
Taking $D=5$, for example gives, $\ds{{\bf x}_1=\pmatrix{5 \cr
1-3i \cr}e^{(2+3i)t}}$.\\ For $\lambda=2-3i$ we must find ${\bf
x_0}$.\\ $(M-\lambda I){\bf x}_0={\bf 0}$ implies
$\ds{\left(\pmatrix{3 &amp; -5 \cr 2 &amp; 1 \cr} -(2-3i) \pmatrix{1 &amp; 0
\cr 0 &amp; 1 \cr}\right)\pmatrix{x_0 \cr y_0 \cr}=0}$\\
$\ds{\pmatrix{1+3i &amp; -5 \cr 2 &amp; -1+3i \cr}\pmatrix{x_0 \cr y_0
\cr}={\bf 0}}$ implying $x_0=D$ and $y_0=(1+3i)D/5$.\\ Taking
$D=5$, for example gives, $\ds{{\bf x}_2=\pmatrix{5 \cr 1+3i
\cr}e^{(2-3i)t}}$.\\ The general solution is therefore given by:\\
$\ds{{\bf x}(t)= A\pmatrix{5 \cr 1-3i \cr}e^{(2+3i)t} +
D\pmatrix{5 \cr 1+3i \cr}e^{(2-3i)t} }$.\\ To get this into the
form of a real solution take $A=a_1+a_2i$ and $B=a_1-a_2i$ where
both $a_1$ and $a_2$ are real numbers so that B is the complex
conjugate of A.\\ In addition use the fact that
$e^{(2-3i)t}=e^{2t}(\cos 3t - i \sin 3t)$\\ Thus $\ds{{\bf x}(t)=
a_1 e^{2t} \pmatrix{10\cos 3t \cr 2 \cos 3t + 6 \sin 3t \cr} a_2
e^{2t} \pmatrix{10\sin 3t \cr 6 \cos 3t - 2 \sin 3t \cr}
 }$\\
Or\quad $x=(10a_1\cos 3t +10 a_2 \sin 3t)e^{2t}$\\ and \quad
$y=((2a_1+6a_2)\cos 3t + (6a_1-2a_2) \sin 3t)e^{2t}$\\


\item
\[
\dot{\bf x}= \pmatrix{3 &amp; -1 \cr
           1 &amp; 1 \cr}{\bf x} \qquad \hbox{ so try }\quad {\bf x}={\bf
           x}_0e^{\lambda t}
\]
Hence we need $(M - \lambda I) {\bf x}_0=0$ which implies find
$\lambda$ so that $\left| M-\lambda I \right|=0$ \\ $\ds{\left|
\pmatrix{3 &amp; -1 \cr 1 &amp; 1 \cr}- \lambda \pmatrix{1 &amp; 0 \cr 0 &amp; 1
\cr}\right|=0}$ \\ $(3-\lambda_(1-\lambda)+1=0$ so that
$\lambda^2-4\lambda +4=0$ hence $\lambda=2,2 (repeated root)$\\
For $\lambda=2$ we must find ${\bf x_0}$.\\ $(M-\lambda I){\bf
x}_0={\bf 0}$ implies $\ds{\left(\pmatrix{3 &amp; -1 \cr 1 &amp; 1 \cr}- 2
\pmatrix{1 &amp; 0 \cr 0 &amp; 1 \cr}\right)\pmatrix{x_0 \cr y_0
\cr}=0}$\\ $\ds{\pmatrix{1 &amp; -1 \cr 1 &amp; -1 \cr}\pmatrix{x_0 \cr
y_0 \cr}={\bf 0}}$ implying $x_0=D$ and $y_0=D$.\\ Taking $D=1$,
for example gives, $\ds{{\bf x}_1=\pmatrix{1 \cr 1 \cr}e^{2t}}$.\\
Now try $\ds{{\bf x}_2=({\bf v}_0 + {\bf x}_1 t) e^{2t}}$ so
that\\ $\ds{2({\bf v}_0+{\bf x}_1 t)+{\bf x}_0= \pmatrix{3 &amp; -1
\cr 1 &amp; 1}({\bf v}_0 + {\bf x}_1 t) }$\\ $\ds{ 2{\bf v}_0 +{\bf
x}_1-\pmatrix{3 &amp; -1 \cr 1 &amp; 1 \cr} {\bf v}_0 =\left(\pmatrix{3 &amp;
-1 \cr 1 &amp; 1 \cr} - 2I \right){\bf x}_1 t }$\\ And since ${\bf
x}_1$ satifies the equation given earlier it follows that:\\
$\ds{\left(\pmatrix{3 &amp; -1 \cr 1 &amp; 1 \cr} - 2\pmatrix{1 &amp; 0 \cr 0
&amp; 1 \cr}\right) {\bf v}_0 = {\bf x}_1 }$\\ Hence $\ds{\pmatrix{1 &amp;
-1 \cr 1 &amp; -1 \cr}\pmatrix{v_0 \cr v_1 \cr}= \pmatrix{1 \cr 1
\cr}}$ which gives$\ds{ {\bf v}_0=\pmatrix{1/2 \cr -1/2 \cr}}$.\\
Hence $\ds{{\bf x}_2= \left(\pmatrix{1/2 \cr -1/2 \cr} + t
\pmatrix{1 \cr 1 \cr}\right)e^{2t} }$\\ The general solution is
therefore given by:\\ $\ds{{\bf x}(t)= A \left(\pmatrix{1/2 \cr
-1/2 \cr} + t \pmatrix{1 \cr 1 \cr}\right)e^{2t} +B \pmatrix{1 \cr
1 \cr}e^2t }$.\\ Or\quad $x=(A(2t+1)+B)e^{2t}$\quad and
\quad$y=(A(2t-1)+B)e^{2t}$.



\end{enumerate}


\end{document}
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