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QUESTION

Solve the following system of linear differential equations
subject to the initial conditions $\frac{dx}{dt}=4,\
\frac{dy}{dt}=20,\ \frac{dz}{dt}=-48$ when %t=0$:

$$\left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\\
\frac{dz}{dt}\end{array}\right)=\left(\begin{array}{ccc}4&0&-1\\-1&3&1\\-1&-1&-5
\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)$$



ANSWER


The general solution is

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=k_1\mathbf{v}_1e^{\lambda_1t}+k_2\mathbf{v}_2e^{\lambda_2t}+k_3\mathbf{v}_3e^{\lambda_3^t}$$

where $k_1,\ k_2,\ k_3$ are arbitrary real constants,
$\mathbf{v}_1,\ \mathbf{v}_2,\ \mathbf{v}_3$ are the eigenvectors
of $A$ and $\lambda_1,\ \lambda_2,\ \lambda_3$ are the
corresponding eigenvalues.

When
$A=\left(\begin{array}{ccc}4&0&-1\\-1&3&1\\-1&-1&-5\end{array}\right)$
the eigenvalues $\lambda_i$ satisfy
$(4-\lambda)((3-\lambda)(-5-\lambda)$

$+1)-\left(1+(3-\lambda)\right)$ or
$(4-\lambda)\left((3-\lambda)(-5-\lambda)\right)$ so
$\lambda_1=4,\ \lambda_2=3,\ \lambda_3=-5$.

Eigenvectors are:

$\lambda_1=4:$

$$\left(\begin{array}{ccc}0&0&-1\\-1&-1&1\\-1&-1&-9
\end{array}\right)\mathbf{v}_1=\mathbf{0}\Rightarrow
\mathbf{v}_1=\left(\begin{array}{c}1\\-1\\0\end{array}\right)$$

$\lambda_2=3:$

$$\left(\begin{array}{ccc}1&0&-1\\-1&0&1\\-1&-1&-8
\end{array}\right)\mathbf{v}_2=\mathbf{0}\Rightarrow
\mathbf{v}_2=\left(\begin{array}{c}1\\-9\\1\end{array}\right)$$

$\lambda_3=-5:$

$$\left(\begin{array}{ccc}9&0&-1\\-1&8&1\\-1&-1&0
\end{array}\right)\mathbf{v}_3=\mathbf{0}\Rightarrow
\mathbf{v}_3=\left(\begin{array}{c}1\\-1\\9\end{array}\right)$$

so the general solution is

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=
k_1\left(\begin{array}{c}1\\-1\\0\end{array}\right)e^{4t}
+k_2\left(\begin{array}{c}1\\-9\\1\end{array}\right)e^{3t}
+k_3\left(\begin{array}{c}1\\-1\\9\end{array}\right)e^{-5t}$$

$$\left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\\
\frac{dz}{dt}\end{array}\right)=
4k_1\left(\begin{array}{c}1\\-1\\0\end{array}\right)e^{4t}
+3k_2\left(\begin{array}{c}1\\-9\\1\end{array}\right)e^{3t}
-5k_3\left(\begin{array}{c}1\\-1\\9\end{array}\right)e^{-5t}$$

so initial conditions give

$$\left(\begin{array}{c}4\\20\\-48\end{array}\right)=
\left(\begin{array}{ccc}4&3&-5\\-4&-27&5\\0&3&-45\end{array}\right)
\left(\begin{array}{c}k_1\\k_2\\k_3\end{array}\right)$$

adding rows 1 and 2 gives

$$\left(\begin{array}{c}4\\24\\-48\end{array}\right)=
\left(\begin{array}{ccc}4&3&-5\\0&-23&0\\0&3&-45\end{array}\right)
\left(\begin{array}{c}k_1\\k_2\\k_3\end{array}\right)\Rightarrow
k_2=-1$$

so $-3-45k_3=-48\Rightarrow k_3=1\Rightarrow 4k_1-3-5=4\Rightarrow
k_1=3$ to give the solution

$$\left(\begin{array}{c}x\\y\\z\end{array}\right)=
\left(\begin{array}{c}3\\-3\\0\end{array}\right)e^{4t}
-\left(\begin{array}{c}1\\-9\\1\end{array}\right)e^{3t}
+\left(\begin{array}{c}1\\-1\\9\end{array}\right)e^{-5t}$$




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