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QUESTION


\begin{description}

\item[(i)]
Set up the following system of linear diffenrential equations as a
matrix problem:

$$\frac{dx}{dt}+y+z=3x\ \frac{dy}{dt}+6y+z=x\
\frac{dz}{dt}+y=2z.$$

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

\item[(iii)]
Find the particular solution subject to the initial conditions
$\frac{dx}{dt}=-218,\ \frac{dy}{dt}=-78,\ \frac{dz}{dt}=52$ when
$t=0$.

\end{description}




ANSWER


\begin{description}

\item[(i)]
$$D\left(\begin{array}{ccc}3&amp;-1&amp;-1\\1&amp;-6&amp;-1\\0&amp;-1&amp;2\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)$$

\item[(ii)]
The general solution has form

$$k_1\mathbf{v}_1e^{\lambda_1t}+k_2\mathbf{v}_2e^{
\lambda_2t}+k_3\mathbf{v}_3e^{\lambda_3t}$$

where $\{\mathbf{v}_i\}=$ eigenvectors and $\{\lambda_i\}=$
eigenvalues of $A$

Solve $|A-\lambda I|=0$ for $\lambda$

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

Eigenvalues $\lambda=-6,2,3$.

$\lambda_1=-6$ solve
$\left(\begin{array}{ccc}9&amp;-1&amp;-1\\1&amp;0&amp;-1\\0&amp;-1&amp;8
\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)
=\left(\begin{array}{c}0\\0\\0\end{array}\right)$

$8x=y=8z,\
\mathbf{v}_1=\left(\begin{array}{c}1\\8\\1\end{array}\right)$

$\lambda_2=2$ solve
$\left(\begin{array}{ccc}-1&amp;-1&amp;-1\\1&amp;-8&amp;-1\\0&amp;-1&amp;0
\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)
=\left(\begin{array}{c}0\\0\\0\end{array}\right)$

$y=0,\ x=z,\
\mathbf{v}_2=\left(\begin{array}{c}1\\0\\1\end{array}\right)$

$\lambda_3=3$ solve
$\left(\begin{array}{ccc}0&amp;-1&amp;-1\\1&amp;-9&amp;-1\\0&amp;-1&amp;-1
\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)
=\left(\begin{array}{c}0\\0\\0\end{array}\right)$

$-y=z,\ x=8y,\
\mathbf{v}_3=\left(\begin{array}{c}8\\1\\-1\end{array}\right)$

Solution is

\begin{eqnarray*}
x&amp;=&amp;k_1e^{-6t}+k_2e^{2t}+8k_3e^{3t}\\ y&amp;=&amp;8k_1e^{-6t}+k_3e^{3t}\\
z&amp;=&amp;k_1e^{-6t}+k_2e^{2t}-k_3e^{3t}
\end{eqnarray*}

\item[(iii)]

When $t=0$,

$$D\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{ccc}-6k_1&amp;+2k_2&amp;+24k_3\\-48k_1&amp;&amp;+3k_3\\-6k_1&amp;+2k_2&amp;-3k_3
\end{array}\right)$$

Solve
$\left(\begin{array}{ccc}-6&amp;2&amp;24\\-48&amp;0&amp;3\\-6&amp;2&amp;-3\end{array}\right)
\left(\begin{array}{c}k_1\\k_2\\k_3
\end{array}\right)=\left(\begin{array}{c}-218\\-78\\52\end{array}\right)$,
the initial conditions.

\begin{tabular}{clc}
row(1)-row(3)&amp;$27k_3=-270$&amp;$k_3=-10$\\
row(2)&amp;$-48k_1-30=-78$&amp;$k_1=1$\\ row(3)&amp;$-6+2k_2+30=52$&amp;$k_2=14$
\end{tabular}

Therefore the particular solution is

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(
\begin{array}{c}e^{-6t}+14e^{2t}-80e^{3t}\\ 8e^{-6t}-10e^{3t}\\
e^{-6t}+14e^{2t}+10e^{3t}\end{array}\right)$$

\end{description}




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