Solve $$y'' - 4y' + 5y = 0 $$
Where $y(0) = 0 \ , \ y'(0) = 2$.
So I solve this as a second degree polynomial (no idea why)
$$\frac{4 \pm \sqrt{16-20}}{2} = 2 \pm 2i$$
So the CASE III solution as my book calls it is: $$Ae^{kt} \cos(wt) + B e^{kt} \sin(wt)$$
Where $k = Re$ and $w = Im$.
So anyhow, $$y(0) = A \cos(0) + B \sin (0) \Rightarrow A = 0$$
$$y'(0) = -A \sin(0) + B \cos (0) = 2 \Rightarrow B = 2$$
So the solution is thus $$y(t) = 2 \cos(t)$$
Am I even doing this right? I have no idea what I am doing and it seems that differential equations are just taught this way. Plug this and that into these magical formulas.
We have:
$$m^2-4m+5 = 0 \implies m_1 = 2+i, m_2 = 2 - i$$
This means we have a solution:
$$y(x) = e^{2t}(c_1 \cos t + c_2 \sin t)$$
Now, substitute in the ICs and find:
$$c_1 = 0, c_2 = 2$$
This makes the final solution:
$$y(t) = 2 e^{2t} \sin t$$
Do you see the slight issue when you found the $\sqrt{-4}$ term in your answer? You also found the correct constants, but used the wrong trig term in your solution.
As an alternative to prove this to yourself, let $y(t) = e^{\lambda t}$. Substitute that back into the ODE and see that it gives you the same result.
Also, here are some notes on Complex Roots if you want to understand the theory a bit more.