\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION


Solve the following system of linear differential equations,
subject to the initial conditions $\frac{dx}{dt}=\frac{dz}{dt}=8$
and $\frac{dy}{dt}=52$ when $t=0$:

$$\left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\\
\frac{dz}{dt}\end{array}\right)=\left(\begin{array}{ccc}3&1&0\\-1&-6&-1
\\-1&-1&2\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right).$$



ANSWER


Solution to $\mathbf{x}'=A\mathbf{x}$ where $A$ is an $n\times n$
matrix with eigenvalues $\lambda_1,\ldots,\lambda_n$ and the
eigenvectors $\mathbf{x}_1,\ldots\mathbf{x}_n$ (each with an
arbitrary constant) is

$$\mathbf{x}=\mathbf{x}_1e^{\lambda_1t}+
\mathbf{x}_2e^{\lambda_2t}+\ldots+\mathbf{x}_ne^{\lambda_nt}.$$

Eigenvalues of $A$:

\begin{eqnarray*}
\left|\begin{array}{ccc}3-\lambda&1&0\\-1&-6-\lambda&-1\\-1&-1&2-\lambda\end{array}\right|&=&0\\
(3-\lambda)\left[(2-\lambda)(-6-\lambda)-1\right]-[-(2-\lambda)-1]+0&=&0\\
(3-\lambda)[\lambda^2+4\lambda-13]+(3-\lambda)&=&0\\
(3-\lambda)(\lambda^2+4\lambda-12)&=&0\\
(3-\lambda)(\lambda+6)(\lambda-2)&=&0
\end{eqnarray*}

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

$\lambda=2$

$$\left(\begin{array}{ccc}1&1&0\\-1&-8&-1\\-1&-1&0\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}x\\y\\z\end{array}\right)\ \
\left.\begin{array}{r}x+y=0\\-x-8y-z=0\\-x-y=0\end{array}\right\}
\begin{array}{c}y=-x\\z=7x\end{array}$$

Let $x=\alpha, y=-\alpha, z=7\alpha$, a suitable eigenvector is
$\mathbf{x}_1=\left(\begin{array}{c}\alpha\\-\alpha\\7\alpha\end{array}\right)$.


$\lambda=3$

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

Let $x=\beta, y=0, z=-\beta$, a suitable eigenvector is
$\mathbf{x}_2=\left(\begin{array}{c}\beta\\0\\-\beta\end{array}\right)$.


$\lambda=-6$

$$\left(\begin{array}{ccc}9&1&0\\-1&0&-1\\-1&-1&8\end{array}\right)
\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}x\\y\\z\end{array}\right)\ \
\left.\begin{array}{r}9x+y=0\\-x-z=0\\-x-y+8z=0\end{array}\right\}
\begin{array}{c}y=-9x\\z=-x\end{array}$$

Let $ x=\gamma, y=-9\gamma, z=-\gamma $, a suitable eigenvector is
$\mathbf{x}_1=\left(\begin{array}{c}\gamma\\-9\gamma\\-\gamma
\end{array}\right)$.

Hence the general solution:

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\alpha\left(\begin{array}{c}1\\-1\\7\end{array}\right)e^{2t}+
\beta\left(\begin{array}{c}1\\0\\-1\end{array}\right)e^{3t}+
\gamma\left(\begin{array}{c}1\\-9\\-1\end{array}\right)e^{-6t}$$

\begin{tabular}{c|c}
$x=\alpha e^{2t}+\beta e^{3t}+\gamma
e^{-6t}$&$\frac{dx}{dt}=2\alpha e^{2t}+3\beta e^{3t}-6\alpha
e^{-6t}$\\ $y=-\alpha e^{2t}-9\gamma
e^{-6t}$&$\frac{dy}{dt}=-2\alpha e^{2t}+54\gamma e^{-6t}$\\
$z=6\alpha e^{2t}-\beta e^{3t}-\gamma
e^{-6t}$&$\frac{dz}{dt}=14\alpha e^{2t}-3\beta e^{3t}+6\gamma
e^{-6t}$ \end{tabular}

When $t=0,\ \frac{dx}{dt}=\frac{dz}{dt}=8$ and $\frac{dy}{dt}=52$:

\begin{eqnarray}
8&=&2\alpha+3\beta-6\alpha\\ 52&=&-2\alpha+54\gamma\\
8&=&14\alpha-3\beta+6\gamma
\end{eqnarray}

(1)+(3)$\Rightarrow 16=16\alpha\Rightarrow \alpha=1$

substitute $\alpha=1$ into (2) :$52-2+54\gamma\Rightarrow\alpha=1$

substitute $\alpha=\gamma=1$ into (1):
$\beta=\frac{1}{3}(8+6-2)=4$

Particular solution:

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}1\\-1\\7\end{array}\right)e^{2t}+4\left(\begin{array}{c}1\\0\\-1\end{array}\right)e^{3t}+\left(\begin{array}{c}1\\-9\\-1\end{array}\right)e^{-6t}.$$




\end{document}