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?

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