So, I'm trying to prove the solution of $$ \ddot{x} + b\dot{x} + cx = 0 $$ I have successfully done it for when the root $b^2 - 4c $ is positive or 0, but am now struggling to get it for negative roots. So far, I have $$ \lambda = \frac{-b}{2} \pm \frac{\sqrt{b^2-4c}}{2} = \frac{-b}{2} \pm i*\frac{\sqrt{4c-b^2}}{2} $$ I now set variables: $$ \delta = \frac{-b}{2} $$ $$ w = \frac{\sqrt{4c-b^2}}{2} $$
And set those in for a solution: $$ x(t) = c_1*e^{\delta t + wit} + c_2 * e^{\delta t - wit} $$ We can factorize out $e^{\delta t} $, and we get: $$ x(t) = e^{\delta t} (c_1*e^{wit} - c_2*e^{wit}) $$ Now, the solution is supposed to be $$ x(t) = e^{\delta t}(c_1 * cos(wt) + c_2 * sin(wt)) $$ but I'm struggling converting it into that. I have tried using Eulers formula, but so far just gotten a mess.
You are almost there. So far you have
$$ x(t) = e^{\delta t} (c_1 e^{i \omega t} + c_2 e^{-i \omega t}) $$
now use Euler's identity
\begin{eqnarray} x(t) &=& e^{\delta t}\left(c_1(\cos \omega t + i\sin \omega t) + c_2(\cos \omega t - i \sin \omega t)\right) \\ &=& e^{\delta t}\left(\underbrace{(c_1 + c_2)}_{a_1}\cos \omega t + \underbrace{i (c_1 - c_2)}_{a_2}\sin \omega t\right) \\ &=& e^{\delta t}(a_1 \cos \omega t + a_2 \sin \omega t) \end{eqnarray}