Second Order Differential Equation (Can you check my work?)

Your answer is right but i'm not sure what you are doing with the imaginary parts.

First hint, if you have an eqn of the form $f(y,y'')$ then observe that $\frac{d^2y}{dx^2} sin(x) = -sin(x)$ , so try a soln. y = sin(ax), then $y'' = -a^2sin(x)$ so in this casae you have $-a^2sin(x) + 12sin(ax) = 0$ or solve for $-a^2 + 12 = 0$ to get $a = \sqrt{12}$ .

Reverse signs and changing sin(ax) to cos(ax) will also give the soln. so you have a gen soln of type

$y = asin(\sqrt{12}) - b (\sqrt{12})$ .

Now we could do this your method as well, but ur method is cumbersome, and usually should be used for eqn of the form f(y'',y',y) = 0.

To see why we do that weird thing to get rid of the imaginary parts. If we solve the auxillary eqn to get $e^{i\sqrt{12}}$ and $e^{-i\sqrt{12}}$ .

$e^{a + ib} = e^{a}(cos(a) + isin(b)$ Since sin(-x) = -sin(x). Note that $e^{a - ib} = e^{a}(cos(a) - isin(b)$ so since these two are linearly independent solutions, adding them together, $e^{a + ib} + e^{a - ib} = e^{a}cos(a) + e^{a}cos(a) = 2e^{a}cos(a)$

Now as i said above, if cos(a) solves a linear equation, then -sin(a) also solves it, (or vice versa). To show that $2e^{a}cos(a)$ and $-2e^{a}sin(a)$ are linearly iandependent, i can take the wronskian determinant of them. (If you haven't heard of this, go to Wronskian - Wikipedia, the free encyclopedia) $\begin{bmatrix} 2e^{a}cos{a} & -2e^{a}sin(a) \\ -2e^{a}sin(a) & -2e^{a}cos(a) \end{bmatrix}$ taking the determinant of this matrix, we get $-4e^{2a}(cos^{2}(a) + sin^{2}(a) = -4e^{2a}$ which does not equal 0 for any a. So these two are linearly independent and thus,
our general solution is of the form y = acos(a) - bsin(a).