# Solve the given system of differential equations. Q.x′_1 =x_1−3x_2, x′_2 =3x_1+x_2.

Solve the given system of differential equations.
$Q.x{\prime }_{1}={x}_{1}-3{x}_{2},x{\prime }_{2}=3{x}_{1}+{x}_{2}$.
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Step 1
Provided a system of differential equations.
Arrange the system as shown below,
x'=Ax
$x\left(t\right)=\left[\begin{array}{c}{x}_{1}\left(t\right)\\ {x}_{2}\left(t\right)\end{array}\right]$
$A=\left[\begin{array}{cc}1& -3\\ 3& 1\end{array}\right]$
Eigen values of A is given by $|A-\lambda I|=0$
${\left(1-\lambda \right)}^{2}+9=0$
${\left(1-\lambda \right)}^{2}=-9$
$\left(1-\lambda \right)=±3i$
$\lambda =±3i+1$
Take $\lambda =3i+1$
$A-\lambda I\left[\begin{array}{cc}1-3i-1& -3\\ 3& 1-3i-1\end{array}\right]$
$A-\lambda I=\left[\begin{array}{cc}-3i& -3\\ 3& -3i\end{array}\right]$
Step 2
Now
$-3i{x}_{1}-3{x}_{2}=0$
$-3i{x}_{1}=3{x}_{2}$
$-i{x}_{1}={x}_{2}$
Take Eigen vector ${v}_{1}=\left[\begin{array}{c}1\\ -j\end{array}\right]$
Then $x\left(t\right)={e}^{\lambda t}{v}_{1}={e}^{3i+1}{v}_{1}$
$x\left(t\right)={e}^{3i+1}\left[\begin{array}{c}1\\ -j\end{array}\right]$
${x}_{1}\left(t\right)=\left[\begin{array}{c}{e}^{3i+1}\\ -i{e}^{3i+1}\end{array}\right]$
As both the Eigen values are conjugate hence,
${x}_{2}\left(t\right)=\left[\begin{array}{c}{e}^{-3i+1}\\ i{e}^{-3i+1}\end{array}\right]$