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

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

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

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

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

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

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

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

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

Solution is

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

\item[(iii)]

When $t=0$,

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

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

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

Therefore the particular solution is

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

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