Question

dx/dt=x-y

dy/dt=x+3y

Answer 1

We can express the system in the form $X'=AX$ , where:

$$X=\begin{bmatrix} x \\ y \end{bmatrix},\quad X'=\begin{bmatrix} x' \\ y' \end{bmatrix},\quad A=\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}$$

We found a normal-shaped Jordan matrix similar to the original:

$$|A-\lambda I|=\begin{vmatrix} 1-\lambda & -1 \\ 1 & 3-\lambda \end{vmatrix}=(\lambda -2)^2$$

Therefore, its eigenvalue is $\lambda=2$ with algebraic multiplicity $2$.

We find the eigenvectors of eigenvalue $\lambda=2$:

$$v=(-1,1)$$

We know that for repeated real eigenvalues $\lambda$ with multiplicity $2$ and eigenvector $v$ , the general solution takes the form:

$$X=c_1e^{\lambda t}v+c_2e^{\lambda t}(tv+u)$$

where $u$ is a solution to $(A-\lambda I)u=v$. We can see that:

$$u=(1,0)$$

Finally, the solution is:

$$X=c_1e^{2t}\begin{bmatrix} -1 \\ 1 \end{bmatrix}+c_2e^{2t}\left( t+\begin{bmatrix} -1 \\ 1 \end{bmatrix}+\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right).$$