Answer :
This looks like a system of differential equations.
[tex]\begin{cases}x' + 2y' + x = 0 \\ x' - y' + y = 0 \\x(0)=0, y(0)=1\end{cases}[/tex]
Eliminating x' gives
[tex](x'+2y'+x)-(x'-y'+y) = 0 - 0 \implies 3y' - y + x = 0[/tex]
and eliminating y' gives
[tex](x'+2y'+x) + 2(x'-y'+y)=0+2\times0 \implies 3x' + x + 2y = 0[/tex]
so that we can rewrite the system as
[tex]\begin{cases}3x' = -x - 2y \\ 3y' = -x + y\end{cases}[/tex]
or equivalently in matrix form as
[tex]\begin{bmatrix}x\\y\end{bmatrix}' = \dfrac13 \begin{bmatrix}-1&-2\\-1&1\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}[/tex]
Compute the eigenvalues for the coefficient matrix:
[tex]\det\begin{bmatrix}-1-\lambda&-2\\-1&1-\lambda\end{bmatrix} = (-1-\lambda)(1-\lambda) - 2 = \lambda^2 -3 = 0 \implies \lambda=\pm\sqrt3[/tex]
Compute the corresponding eigenvectors:
[tex]\lambda=\sqrt3 \implies \begin{bmatrix}-1-\sqrt3&-2\\-1&1-\sqrt3\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \\\\ \implies v_1=(1-\sqrt3)v_2 \implies \vec v = \begin{bmatrix}1-\sqrt3\\1\end{bmatrix}[/tex]
[tex]\lambda=-\sqrt3 \implies \begin{bmatrix}-1+\sqrt3&-2\\-1&1+\sqrt3\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \\\\ \implies v_1 = (1+\sqrt3)v_2 \implies \vec v = \begin{bmatrix}1+\sqrt3\\1\end{bmatrix}[/tex]
We end up multiplying the matrix by 1/3, so the eigenvalues also get scaled by 1/3 and λ = ±1/√3. The eigenvectors stay the same.
Then the characteristic solution to the system is
[tex]\begin{bmatrix}x\\y\end{bmatrix} = C_1 e^{t/\sqrt3} \begin{bmatrix}\frac{1-\sqrt3}3\\\frac13\end{bmatrix} + C_2 e^{-t/\sqrt3} \begin{bmatrix}\frac{1+\sqrt3}3\\\frac13\end{bmatrix}[/tex]
Use the initial conditions to solve for the constants.
[tex]\begin{bmatrix}0\\1\end{bmatrix} = C_1 \begin{bmatrix}\frac{1-\sqrt3}3\\\frac13\end{bmatrix} + C_2 \begin{bmatrix}\frac{1+\sqrt3}3\\\frac13\end{bmatrix}[/tex]
[tex]\implies \begin{cases}(1-\sqrt3) C_1 + (1+\sqrt3) C_2 = 0 \\ C_1 + C_2 = 3\end{cases}[/tex]
[tex]\implies C_1=\dfrac{3+\sqrt3}2, C_2=\dfrac{3-\sqrt3}2[/tex]
Then the particular solution is
[tex]\begin{bmatrix}x\\y\end{bmatrix} = \dfrac{3+\sqrt3}6 e^{t/\sqrt3} \begin{bmatrix}1-\sqrt3\\1\end{bmatrix} + \dfrac{3-\sqrt3}6 e^{-t/\sqrt3} \begin{bmatrix}1+\sqrt3\\1\end{bmatrix}[/tex]
or
[tex]\begin{cases}x(t) = -\dfrac1{\sqrt3} e^{\sqrt3 \, t} + \dfrac1{\sqrt3} e^{-\sqrt3\,t} \\\\ y(t) = \dfrac{3+\sqrt3}6 e^{\sqrt3\,t} + \dfrac{3-\sqrt3}6 e^{-\sqrt3\,t}\end{cases}[/tex]