I am to find the general solution of the differential equation $$y'' + y' + 4y = 4 \sinh(t) = 2e^t - 2e^{-t}$$

Now, using the method of undetermined coefficients, it is simple to arrive at the particular solution $Y(t) = \frac{1}{3}e^t - \frac{1}{2}e^{-t}$. The text, however, provides the more bizarre general solution $$c_1\cos(\sqrt{15}t/2)+c_2e^{-t/2}\sin(\sqrt{15}t/2) + Y(t)$$

Now, seeing as my $Y(t)$ is apparently correct, I know that I am on the right track, but deriving the general solution from $Y(t)$ confuses me greatly, since the text provides no good example for this.

  • $\begingroup$ What's the solution to the homogeneous equation? (You're missing a factor of $e^{-t/2}$ in the first term of your second displayed expression.) $\endgroup$ – David Mitra Feb 24 '14 at 10:18
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    $\begingroup$ There's a standard procedure to find the general solution to the equation $y'' + y' + 4y = 0$. Did you learn that yet? $\endgroup$ – user99914 Feb 24 '14 at 10:21
  • $\begingroup$ @DavidMitra. Sorry, I did not see your comment. I delete my answer. Cheers. $\endgroup$ – Claude Leibovici Feb 24 '14 at 10:27
  • $\begingroup$ Yes, @John, I did. I was unfamiliar with the idea of solving the homogeneous the find the complementary solution. $\endgroup$ – Andrew Thompson Feb 24 '14 at 10:34

You have found the particular solution, but you need to also find the complementary solution $Y_{c}(t)$, which solves the homogeneous equation (which is $y''+y'+4y=0$). The general solution is then $y=Y_{c}(t)+Y(t)$.

To find $Y_{c}(t)$, guess a solution of $e^{\lambda t}$ to the equation $y''+y'+4y=0$, and find $\lambda$ such that you get two linearly independent solutions.


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