Solving the system: $$x''+y=1$$ $$y''+x=-1$$ with $x(0)=1,x'(0)=1,y(0)=1,y'(0)=-1$
I tried doing $x_1=x,x_2=x',x_3=y,x_4=y'$ and then solving the system:
$ \begin{pmatrix} x' \\ x'' \\ y'\\ y'' \end{pmatrix}=$ $ \begin{pmatrix} x_2 \\ 1-x_3 \\ x_4\\ 1-x_1 \end{pmatrix}=$ $\begin{pmatrix} 0&1&0&0 \\ 0&0&-1&0 \\0&0&0&1\\ -1&0&0&0 \end{pmatrix}*\begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 0\\ -1\end{pmatrix}$ and then using
$X(t)=e^{(t-t_o)A}X_o+ \int_{t_o}^{t}e^{(t-\tau)A}*b(\tau)*d\tau $
But the process was very complicated I ended up having a wrong solution. Is there a way to simplify things?