Question

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

Answer 1

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

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

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

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

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