Consider $y^{''}+y^{'}-2y=0$. Find the general solution to this differential equation. Convert the second order equation to a system of linear equations corresponds to the general solution of the second order equation.
Solution so far
Substitution $y_1=y$, $y_2 =\dot{y}$ yields an equivalent 1st order system
\begin{cases} \dot{y_1}=y_2, \\ \dot{y_2}=2y_1 -y_2 \end{cases} i.e $\dot{y}=Ay$, where $A=\begin{bmatrix} 0& 1 \\ 2 & -1 \end{bmatrix}$
Eigenvalues $\lambda_{1}=1$ , $\lambda_{2}=-2$ eigenvectors $v_{1}=[1,1]^{T}$, $v_{2}=[1,-2]^{T}$ . Hence $P=\begin{bmatrix} 1& 1 \\ 1 & -2 \end{bmatrix}$ with $P^{-1}=\begin{bmatrix} \frac{2}{3}& \frac{1}{3} \\ \frac{1}{3} & \frac{-1}{3} \end{bmatrix}$
Thus the solution of the IVP is given by $y(t)= P \begin{bmatrix} e^t& 0 \\ 0 & e^{-2t} \end{bmatrix} P^{-1} y(0)$
Is what I have correct?