# Solving Differential Equation: Adding Solutions

$$y''+2y'+2y=e^{-3x}$$

Apparently one can solve this differential equation by first solving for the general solution of the left-hand side (= 0):

$$y''+2y'+2y= 0$$

yields

$$y = e^{-x}(c_{1}sin(x)+c_2cos(x))$$

Here's what I don't quite understand: The next step is to find an equation in the form of
$$y = se^{-3x}$$ which is a solution to the original $$y''+2y'+2y=e^{-3x}$$.

This gives $$y = \frac{1}{5}e^{-3x}$$. (s = 1/5)

$$y = e^{-x}(c_{1}sin(x)+c_2cos(x)) + \frac{1}{5}e^{-3x}$$ is the general solution to $$y''+2y'+2y=e^{-3x}$$.
Can someone explain how one would reach this solution? How would one know that one of the solutions is in the form of $$y = se^{-3x}$$? Is it just a matter of guesswork? Why would you add these particular two solutions together?
• It looks like the RHS of your differential equation should be $e^{-3x}$ instead of $e^{3x}$. – D.B. Dec 26 '18 at 3:16