Reducing system of differential equations to first order. I have to solve following system of equations:
$$
x''=-x-z+e^{-t},z'=-2x-2z+3e^{-t}
$$
I would like to have formula for $x'$ in order to find fundamental matrix and so on. So from above quations I obtain $z'-2x''=e^{-t}$, integrate sides and have $2x'=z+e^{-t}$.
So I have system of two first-order differential equations $$2x'=z+e^{-t}, z'=-2x-2z+3e^{-t}$$
Is it correct?
 A: you can do this 
$$\begin{cases}x''=-x-z+e^{-t}\\z'=-2x-2z+3e^{-t} \end{cases}$$
 from the first equation find $z$
$$z=-x''-x+e^{-t}$$ and put into second 
$$z'=-2x-2z+3e^{-t}\\
(\color{red} {-x''-x+e^{-t}})'=-2x-2(\color{red} {-x''-x+e^{-t}})+3e^{-t}$$now simplify 
$$-x'''-x'-e^{-t}=-2x+2x''+2x-2e^{-t}+3e^{-t}\\x'''+2x''=-2e^{-t}$$
$$r^3+2r^2=0 \to r=0,0,-2\\x=ae^{0t}+bte^{0t}+ce^{-2t}\\x=a+bt+ce^{-2t}$$then for particular solution you can add $ke^{-t}$ and find $k$
A: Differentiate the first equation & substitute for $z'$
\begin{eqnarray*}
x^{'''}=-x^{'}-z^{'}-e^{-t} \\
=-x^{'}-(-2x-2z+3e^{-t})-e^{-t}
\end{eqnarray*}
Now use the first equation to substitute for $z=-x-x^{''}+e^{-t}$ ... neaten it up a bit ... we have
\begin{eqnarray*}
x^{'''}+2x^{''}+x^{'}=-2e^{-t}
\end{eqnarray*}
So the auxillary equation is $\lambda^{3}+2 \lambda^{2}+\lambda=0$ and the general solution is
\begin{eqnarray*}
x=A +Be^{-t}+Ct e^{-t}.
\end{eqnarray*}
Now to obtain the particular solution try $x=\alpha t^2 e^{-t}$ ... $ \alpha =1$. 
