# How to solve $\ddot{y} + a_{1}\dot{y} + (a_{2}+a_{3}e^{-2a_{4}t})y = a_{5}e^{-a_{4}t}$

I have been in trouble finding a general solution for the following differential equation. $$\begin{equation} \ddot{y} + a_{1}\dot{y} + (a_{2}+a_{3}e^{-2a_{4}t})y = a_{5}e^{-a_{4}t} \end{equation}$$ with the following initial conditions: $$\begin{equation} y(0) = \dot{y}(0)=0 \end{equation}$$ I have tried a few methods to solve this equation but none of them seems promising.

1. Variation of parameters: I need to know at least one particular solution to apply this method but I could not find one. Is there any method to find one? (Reference: Chapter$$3.9.2.1$$ of https://www.jirka.org/diffyqs/html/nonhomogsys_section.html)

2. Laplace transform: Let Y be the Laplace transform of y, then the Laplace transform of this equation is: $$\begin{equation} (s^{2}+a_{1}s+a_{2})Y(s) + a_{3}Y(s+2a_{4}) = \frac{a_{5}}{s+a_{4}} \end{equation}$$ but I have no idea how to deal with $$Y(s+2a_{4})$$ term. Moreover, comparison of the first term and the RHS implies that $$Y(s)$$ has an order of $$s^{-3}$$ but comparison of the second term and the RHS implies that $$Y(s+2a_{4})$$ has an order of $$s^{-1}$$, which seems to be impossible...

I would appreciate if you could give me any piece of advice. Thank you.