2nd order linear ODE with constant driving term

Question

I have this DE$$y''+ay'=c $$ I tried with undetermined coefficient method, but the ansatz of particular solution being a zeroth order polynomial obviously doesn't work(LHS is 0). How do I solve this problem? do I have to use variation of parameters or the ansatz is not right?

Answer 1

Integrating factor method works fine $$y''+ay'=c$$ use $\mu(x)=e^{ax}$ as integrating factor $$(y'e^{ax})'=ce^{ax}$$ Integrate $$y'e^{ax}=c\int e^{ax}dx =\frac ca {e^{ax}}+K$$ $$y'=e^{-ax} \left (\frac ca {e^{ax}}+K \right )$$ $$y'=\frac ca+Ke^{-ax}$$ Integrate again $$\boxed{y=\frac cax+K_1e^{-ax}+K_2}$$

Answer 2

Integrating once

$$ y'+ay=c x+C_1 $$

and now the first order linear differential equation with solution

$$ y = \frac{c (a x-1)+a C_1}{a^2}+C_2 e^{-a x} $$