I'm working on solving this problem:

$$ y''+ 48y = 0 $$

For a typical homogenous ODE with real roots we let $y=e^{rx}$ and solve for the roots $r_1$ and $r_2$:

$$ y=e^{rx}$$ $$ y'=re^{rx} $$ $$ y''=r^2e^{rx} $$ $$ r^2e^{rx} + 48e^{rx} = 0 $$ $$ e^{rx} (r^2+48) = 0 $$

Since $e^{rx}$ cannot equal zero we set $r^2+48$ equal to zero then solve. However, this results in a imaginary number. What's a young math student to do in this situation?

  • $\begingroup$ $r=\pm i\sqrt{48}$ or $y=C_1\cos \sqrt{48}x+C_2\sin \sqrt{48}x$. $\endgroup$ – njguliyev Oct 5 '13 at 21:35
  • $\begingroup$ Hrm, I'm very interested in the conversion from a+bi form to trig functions. But I'm not quite sure how that is done. $\endgroup$ – Bob Shannon Oct 5 '13 at 21:37
  • 1
    $\begingroup$ Do you know that $e^{ix}=\cos x + i \sin x$? $\endgroup$ – njguliyev Oct 5 '13 at 21:44

You can also just write off the top:

$$r^2 + 48 = 0 \rightarrow r_{1,2} = \pm i \sqrt{48} = \pm i 4 \sqrt{3}$$

So, we have:

  • $y_1 = e ^{+4\sqrt{3} i~x}$
  • $y_2 = e ^{-4\sqrt{3} i~x}$

We know: $ e^{it} = \cos t + i \sin t$.

So, we get:

$$y(x) = y_1(x) + y_2(x) = c_1 \cos( 4 \sqrt{3}x) + c_2 \sin( 4 \sqrt{3}x)$$

  • $\begingroup$ Wow, that makes sense. I did not know that identity. Oddly though, this answer is being marked as incorrect by the class's web assignment software. I even re-worked the whole problem using the new knowledge you've provided and got the same answer. $\endgroup$ – Bob Shannon Oct 5 '13 at 21:59
  • $\begingroup$ I wonder if they leave it as $\sqrt{48}$? Also, how does it know what variable names you choose for the constants? I also wonder if they leave with the imaginaries, so you might have to try all of those. The more I hear about this SW, the more I dislike it! Also, that is Euler's Formula. $\endgroup$ – Amzoti Oct 5 '13 at 21:59
  • $\begingroup$ It tells you to use x as the independent variable. It's not all too uncommon for bugs to prevent the right answer from being marked (it's written in Perl), so I've e-mailed the professor. I have more confidence in this explanation then some program telling me the answer is wrong. Thanks! $\endgroup$ – Bob Shannon Oct 5 '13 at 22:04
  • 1
    $\begingroup$ Well, we have $x$ as the independent variable. Regards $\endgroup$ – Amzoti Oct 5 '13 at 22:06
  • $\begingroup$ Needs another TU! +1 $\endgroup$ – Namaste Oct 6 '13 at 0:52

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.