How to solve $2y''+3y'-2y=e^{-2x}+1$ How can we find the general solution of the equation $2y''+3y'-2y=e^{-2x}+1$. I know the general solution is of the form $y(x)=y_c(x)+y_p(x)$ and $y_c(x)=c_1e^{x/2}+c_2e^{-2x}$ but I couldn't find $y_p(x)$. Thanks!
 A: If you want a slightly more 'systematic' approach beyond 1234's good Ansatz, you can differentiate the equation again to obtain
$$2y''' + 3y'' - 2y' = -2e^{-2x}$$
Now you can apply standard approaches, as the RHS is an exponential.

Write u = y'. Then the ODE above is 
$$2u'' + 3u' - 2u = -2e^{-2x}$$
The characteristic polynominal $p(D) = 2D^2 + 3D- 2$ has roots $-2, 1/2$. As $-2$ is also the exponent of the RHS, the particular solution of the equation in $u$ is
$$y' = u = f'_p(x) = \frac{-2xe^{-2x}}{p'(2)} = \frac{ -2xe^{-2x}}{-5} = \frac{2}{5}xe^{-2x}$$
Integrating, the particular solution of the original equation is $f_p(x) = -\frac{1}{5} xe^{-2x} + C$. Substituting back into the ODE to solve for $C$,
$$f_p(x) =  -\frac{1}{5}xe^{-2x} +\frac{1}{2}$$
This last step can be made easy by observing that the derivatives of $y_p$ kill the constant, so it must be that $-2y = -2C = 1$.
A: General solutions are $e^{-2x}$ and $e^{\frac{x}{2}}$.
$\frac{-x}{5}e^{-2x}$ is a particular solution of $2y''+3y'-2y=e^{-2x}$ and $\frac{-1}{2}$ is a particular solution of $2y''+3y'-2y=1$. So the solution of the differential equation is
$$Ae^{-2x}+Be^{\frac{x}{2}}-\frac{x}{5}e^{-2x}-\frac{1}{2}.$$
