Solve the differrential equation $$y'' - 4y' + 13y' = 6e^{2x}\cos(3x)$$ where $y(0)=3$ and $y'(0)=-8$.
I think we start like…
For the homogenous case $$\lambda^2 -4\lambda + 13 = 0 $$ \begin{align} \lambda &= \frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ &= \frac{4 \pm \sqrt{16-52}}{2} \\ &= \frac{4 \pm \sqrt{-36}}{2} \\ &= \frac{4 \pm 6i}{2} \\ &= 2 \pm 3i \end{align} $$\lambda_1 = 2+3i \quad \text{and} \quad \lambda_2 = 2-3i$$
The homogenous equation has a general solution
$$y_{\text{h}} = Ae^{(2+3i)x}\cos(3x) +Be^{(2-3i)x}\sin(3x)$$
Since the R.H.S. is an non-homogenous equation “$6e^{2x}\cos(3x)$” we choose the trial function $$y_{\text{p}} = (\alpha e^{(2+3i)x} \cos(3x)+\beta e^{(2-3i)x}\sin(3x))$$
I’m not sure how to procede from here or whether I am going in the right direction.