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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?

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  • $\begingroup$ $r=\pm i\sqrt{48}$ 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
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    $\begingroup$ Do you know that $e^{ix}=\cos x + i \sin x$? $\endgroup$ – njguliyev Oct 5 '13 at 21:44
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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:

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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