How to solve coupled second order differential equations I have the following coupled differential equations:
$$ 2y''- 3y' + 2z' + 3y + z = e^{2x}$$
$$y''- 3y' + z' + 2y - z = 0 $$
I'm not sure how to solve them as if I try $y = Ae^{\lambda x} $ and $z = Be^{\lambda x}$, I only get one value for $\lambda$ ($\lambda = 1$) so am not sure how to form a complementary function.
Any suggestions?
Thanks.

It is visible after some contemplation that the highest derivatives of $z$ and $y$ occur in the same linear combination in both equations, marking the system as a DAE system of index at least one.
To better handle the structure, define $u=y'+z$ for this combination and eliminate $z$ from the system.
\begin{align}
2u' - 4y' +  3y + u &= e^{2x}\\
u'- 2y'  + 2y - u &= 0
\end{align}
Again the terms in the highest derivatives $y',u'$ occur in the same linear combination. So set $v=u-2y=y'+z-2y$ and eliminate $u$ against $v$ to get an explicit index-1 DAE system.
\begin{align}
2v' +  5y + v &= e^{2x}\\
v'   - v &= 0
\end{align}
This now can be isolated for the highest order derivatives (or non-derivatives) $(y,v')$ to get
\begin{align}
 5y  &= e^{2x}-3v\\
v'   &= v
\end{align}
This system, which resulted just from easily reversible variable substitutions, now indeed has only one integration constant.
So, inserting backwards,
$$
v=Ce^x,\\
y=\frac15(e^{2x}-3Ce^x),\\
u=v+2y=\frac15(2e^{2x}-Ce^x)\\
z=u-y'=\frac15(2e^{2x}-Ce^x)-\frac15(2e^{2x}-3Ce^x)=\frac25Ce^x
$$
