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.

Define $\mathbb{Y}(s)=\mathcal{L}(y(t))$ and $\mathbb{Z}(s)=\mathcal{L}(z(t))$ where $\mathcal{L}$ denotes the laplace transform.
Put $a=y(0),b=y'(0),c=z(0)$.
With a little algebra your DE becomes $$(2s^2-3s+3)\mathbb{Y}(s)+(2s+1)\mathbb{Z}(s)=-as+\frac{1}{s-2}+2b+2c$$ $$(s^2-3s+2)\mathbb{Y}(s)+(s-1)\mathbb{Z}(s)=as-3a+b+c$$ This can be expressed as a linear system $$\begin{pmatrix}2s^2-3s+3&2s+1\\ s^2-3s+2&s-1\end{pmatrix}\begin{pmatrix}\mathbb{Y}(s)\\ \mathbb{Z}(s)\end{pmatrix}=\begin{pmatrix}-as+\frac{1}{s-2}+2b+2c\\ as-3a+b+c\end{pmatrix}$$ Can you finish? Keep in mind, once you solve this system via the inverse laplace transform $\mathcal{L}^{-1},$ the constants $a,b,c,d$ will be your arbitrary constants in your general solution.
