Find Solutions For Second Order Differential Equation. 
Find General Solution for
  $$y''+4y'+5y=0$$$$b^2-4c< 0$$
  $$y_1=e^\frac{-bx}{2}\cos \frac{\sqrt{4c-b^2}}{2}x$$
  $$y_2=e^\frac{-bx}{2}\sin \frac{\sqrt{4c-b^2}}{2}x$$
  Given following, find solutions for which
  $$y(0)=1,y'(0)=0$$
  $$y(0)=12,y'(0)=-5$$
  $$y_1=e^\frac{-4x}{2}\cos \frac{\sqrt{20-16}}{2}x$$
  $$y_2=e^\frac{-4x}{2}\sin \frac{\sqrt{20-16}}{2}x$$
  $$y_1=e^{-2x}\cos x$$
  $$y_2=e^{-2x}\sin x$$
  But where to from here?

 A: $$y′′+4y′+5y=0$$
Take auxiliary equation,
let 
$$y=e^{mx}$$
$$\frac{dy}{dx}=me^{mx}$$
$$\frac{d^2y}{dx^2}=m^2e^{mx}$$
Substituting into the differential equation we obtain
$$m^2e^{mx}+4me^{mx}+5e^{mx}=0$$ provided $$e^{mx} \neq0$$
$$m^2+4m+5=0$$
Solving it using quadratic formula we have
$$m=-2\pm i$$
The solution is then 
$$y=e^{-2x}[c_1cosx+c_2sinx]$$
Given $$y(0)=1$$
$$c_1=1$$
Also,
$$y'=-2e^{-2x}[c_1cosx+c_2sinx]+e^{-2x}[-c_1sinx+c_2cosx]$$
$$y'(0)=0$$
$$c_2-2c_1=0$$
$$c_2=2$$
The particular solution is then 
$$y=e^{-2x}[cosx+2sinx]$$
Addendum
Clarification of complex root
We know that complex number takes the form of 
$$z=a+bi$$
Taking exponential on both sides we have
$$e^z=e^{a+bi}$$
Recall that
$$e^{a+b}=e^a.e^b$$
Taking out a we have
$$e^{zx}=e^{ax}.e^{bxi}$$
Euler's equation of complex number is 
$$e^{i \theta}=cos\theta+isin\theta$$
Replacing $\theta$ with $bx$
So we have,
$$e^{ax}.e^{bxi}=e^{ax}cosbx+isinbx$$
Recall the the solution for complex roots also contain its conjugate that is 
$$z_1=a+bi$$
$$z_2=a-bi$$
So the general solution is 
$$y=k_1e^{(a+bi)x}+k_2e^{(a-bi)x}$$
Factoring out $e^{ax}$
$$y=e^{ax}[k_1e^{bix}+k_2e^{-bix}]$$
$$y=e^{ax}[k_1{cosx+sinx}+k_2{cosx-sinx}]$$
$$y=e^{ax}[k_1cosx+k_2cosx+k-1sinx-k_2sinx]$$
let $k_1+k_2=c_1$ and $k_1-k_2=c_2$
$$y=e^{ax}[c_1cosx+c_2sinx]$$
A: The general solution of $y''+4y'+5y=0$ is given by
$y(x)=c_1y_1(x)+c_2y_2(x)$,
($c_1,c_2 \in \mathbb R$), with the functions $y_1,y_2$ from above.
Now determine $c_1$ and $c_2$ such that $y(0)=1,y'(0)=0$
