dx/dt=x-y dy/dt=x +3y

$\frac{dt}{dx}=x-y$

$\dfrac{dy}{dt}=x+3y$

$\begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix}=\begin{pmatrix} 1 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

Find the eigenvalues 

$\begin{pmatrix} 1-\lambda & -1 \\ 1 & 3-\lambda \end{pmatrix}=0$

$(1-\lambda)(3-\lambda)+1=0$

$\lambda^2-4\lambda+4=0$

$\lambda_1=\lambda_2=2$

 Find the eigenvectors.

$\lambda=2$

$\begin{pmatrix} 1-2 & -1 \\ 1 & 3-2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=0$

The eigen vector is

$v=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

For repeated eigenvalues $\lambda_1=\lambda_2=2$ and the eigenvector $v=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

the general solution takes form

$\begin{pmatrix} x \\ y \end{pmatrix}=c_1e^{2t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}+c_2 te^{2t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}+c_2e^{2t}\begin{pmatrix} u_1 \\ u_2 \end{pmatrix}$

where

$\begin{pmatrix} 1-2 & -1 \\ 1 & 3-2 \end{pmatrix}\begin{pmatrix} u_1 \\ u_2 \end{pmatrix}=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

$\begin{pmatrix} u_1 \\ u_2 \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}$

The solution is

$\begin{pmatrix} x \\ y \end{pmatrix}=c_1e^{2t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}+c_2 te^{2t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}+c_2e^{2t}\begin{pmatrix} 1 \\ 0 \end{pmatrix}$