# Solving a second order homogenous ODE with non-real complex roots

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?

• $r=\pm i\sqrt{48}$ or $y=C_1\cos \sqrt{48}x+C_2\sin \sqrt{48}x$. – njguliyev Oct 5 '13 at 21:35
• 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. – Bob Shannon Oct 5 '13 at 21:37
• Do you know that $e^{ix}=\cos x + i \sin x$? – 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:

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

• 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. – Bob Shannon Oct 5 '13 at 21:59
• 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. – Amzoti Oct 5 '13 at 21:59
• 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! – Bob Shannon Oct 5 '13 at 22:04
• Well, we have $x$ as the independent variable. Regards – Amzoti Oct 5 '13 at 22:06
• Needs another TU! +1 – Namaste Oct 6 '13 at 0:52