Solving ordinary differential equations of order 2 I'm struggling solving this linear ODE:

$$y''(x) + 2y'(x) = -4$$

I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;
But I'm struggling with the non-homogeneous part:
If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.
What am I doing wrong! Thanks a lot!

Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=x\cdot C$$
where $C$ is a constant to be determined.
