Find the solution of the sistem $x''=2x+y$ and $y''=x+2y$

I have to find the solution of the sistem $x''=2x+y$ and $y''=x+2y$ to which it applies $x(0)=0$, $x'(0)=2$, $y(0)=0$ and $y'(0)=0$.

First I wrote this two formulas in matrix like this $$\begin{bmatrix} x'' \\ y'' \end{bmatrix}=\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$

Then I calculate eigenvalues of the matrix $\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ where I get $\lambda_{1}=1$ and $\lambda_{2}=3$

For each eigenvalues we got eigenvectors $v_{1}=\begin{bmatrix} 1\\ -1 \end{bmatrix}$ and $v_{2}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

For that we get the solution $$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} e^{t} & e^{3t} \\ -e^{t} & e^{3t} \end{bmatrix} \begin{bmatrix} C_{1} \\ C_{2} \end{bmatrix}$$

We use $x'(0)=2$ and $y'(0)=0$ and we get $C_{1}=C_{2}=1$

Now I have to find solution for $$\begin{bmatrix} x'\\ y' \end{bmatrix}=\begin{bmatrix} e^{t} & e^{3t} \\ -e^{t} & e^{3t} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$

I tried to find eigenvalues for that matrix but I can not find them.

Any help?

Hint: Notice that \begin{align} (x-y)'' = x-y. \end{align} Set $u = x-y$ then you have the following IVP \begin{align} u''= u, \quad u(0) = 0, u'(0) = 2. \end{align}

Hint 2: After you solve for $u$ then you see that $y= x-u$ which means $x''= 2x+x-u =3x -u$ where $u$ is known.

You mentioned in the comment that you need to solve it using the eigenvalue method. There are two ways to do this:

1. Let $x_1 = x$, $x_2 = x'$, $x_3 = y$, $x_4 = y'$. You can check that the given second-order system is equivalent to the following first-order system: \begin{align*} x_1' & = x_2 \\ x_2' & = 2x_1 + x_3 \\ x_3' & = x_4 \\ x_4' & = x_1 + 2x_3. \end{align*} Define the vector function $\mathbf{x} = (x_1, x_2, x_3, x_4)^T$. We have that $\mathbf{x}' = A\mathbf{x}$, where the coefficient matrix is $$ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 2 & 0 \end{bmatrix}. $$ WolframAlpha gives the eigenvalues of $A$ as $\lambda = \pm 1, \pm\sqrt{3}$ which are distinct and so the general solution is $$\mathbf{x}(t) = c_1e^{t}\mathbf{v}_1 + c_2e^{-t}\mathbf{v}_2 + c_1e^{\sqrt{3}t}\mathbf{v}_3 + c_4e^{-\sqrt{3}t}\mathbf{v}_4. $$
2. One can also solve the second-order linear system using the eigenvalue method without rewriting it as an equivalent first-order system. Let $\mathbf{z} = (x, y)^T$. You can check that $\mathbf{z} = e^{\alpha t}\mathbf{v}$ is a solution to $\mathbf{z}'' = B\mathbf{z}$ with $\alpha^2 = \lambda$ and $(\lambda, \mathbf{v})$ an eigenpair of the matrix $B$, i.e., $B\mathbf{v} = \lambda\mathbf{v}$. In this case, $B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ with distinct eigenvalues $\lambda = 1, 3$ and so the general solution is $$\mathbf{z}(t) = c_1e^{t}\mathbf{w}_1 + c_2e^{-t}\mathbf{w}_1 + c_3e^{\sqrt{3}t}\mathbf{w}_2 + c_4e^{-\sqrt{3}t}\mathbf{w}_2.$$