\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION


\begin{description}

\item[(a)]
Set up the following system of linear differential equations as a
matrix proble:

$$\frac{dx}{dt}-x-3y=0,\ \frac{dy}{dt}-3x-y-4z=0,\
\frac{dz}{dt}-4y-z=0.$$

\item[(b)]
Write down the general form of the solution to the problem.

\item[(c)]
Find the particular solution subject to the initial conditions
$\frac{dx}{dt}=6,\ \frac{dy}{dt}=-10,\ \frac{dz}{dt}=8$ when
$t=0$.

\end{description}



ANSWER


\begin{description}

\item[(a)]

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{ccc}1&3&0\\3&1&4\\0&4&1\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)$$

\item[(b)]
General solution
$\mathbf{v}_1e^{\lambda_1t}+\mathbf{v}_2e^{\lambda_2
t}+\mathbf{v}_3e^{\lambda_3 t}$ where $\mathbf{\lambda}_i$ are
vectors corresponding to $\lambda_i$.

\begin{eqnarray*}
\left|\begin{array}{ccc}1-\lambda&3&0\\3&1-\lambda&4
\\0&4&1-\lambda\end{array}\right|&=&
(1-\lambda)\left[(1-\lambda)^2-4^2\right]-3\left[3
(1-\lambda)\right]\\
&=&(1-\lambda)\left[(1-\lambda)^2-5^2\right]\\
&=&(1-\lambda)(-4-\lambda)(6-\lambda)
\end{eqnarray*}

$\lambda=-4,1,6$

Eigenvectors:

$\lambda=-4$

$$\left(\begin{array}{ccc}5&3&0\\3&5&4\\0&4&5\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=\mathbf{0}\Rightarrow
k_1\left(\begin{array}{c}-3\\5\\-4\end{array}\right)$$

$\lambda=1$

$$\left(\begin{array}{ccc}0&3&0\\3&0&4\\0&4&0\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\mathbf{0}\Rightarrow
k_2\left(\begin{array}{c}4\\0\\-3\end{array}\right)$$

$\lambda=6$

$$\left(\begin{array}{ccc}-5&3&0\\3&-5&4\\0&4&-5\end{array}
\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\mathbf{0}\Rightarrow
k_3\left(\begin{array}{c}3\\5\\4\end{array}\right)$$

General solution

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=k_1\left(
\begin{array}{c}-3\\5\\-4\end{array}\right)e^{-4t}+k_2\left(
\begin{array}{c}4\\0\\-3\end{array}\right)e^t+k_3\left(
\begin{array}{c}3\\5\\4\end{array}\right)e^{6t}$$

\item[(c)]
For the particular solution solve

$$-4k_1\left(\begin{array}{c}-3\\5\\-4\end{array}\right)+k_2\left(
\begin{array}{c}4\\0\\-3\end{array}\right)+6k_3\left(\begin{array}{c}
3\\5\\4\end{array}\right)=\left(\begin{array}{c}6\\-10\\8\end{array}\right)$$

to get

\begin{eqnarray*}
k_1&=&-\frac{1}{2}\\ k_2&=&0\\ k_3&=&0
\end{eqnarray*}

\end{description}




\end{document}