0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.