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

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


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

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)

Then the final correct answer is to add these two together:

$$y = e^{-x}(c_{1}sin(x)+c_2cos(x)) + \frac{1}{5}e^{-3x}$$ 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?

I know that linear combinations of (different) solutions form the general solution...is that related to the solution of this problem in any way?

  • $\begingroup$ It looks like the RHS of your differential equation should be $e^{-3x}$ instead of $e^{3x}$. $\endgroup$ – D.B. Dec 26 '18 at 3:16
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    $\begingroup$ There is some theory that you can use to find that form for the answer, but I prefer to leverage the existence and uniqueness theorems in order to sweep all of that under the rug. Once you find the answer, it no longer matters how you found it. As for adding solutions together, that is justified by the fact that this is a linear differential equation. These are very special indeed. $\endgroup$ – ImNotTheGuy Dec 26 '18 at 3:16
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    $\begingroup$ This might help: stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/… $\endgroup$ – shree.pat18 Dec 26 '18 at 7:15
  • $\begingroup$ en.wikipedia.org/wiki/Method_of_undetermined_coefficients $\endgroup$ – Hans Lundmark Dec 26 '18 at 9:29

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