Solve differential equation y'' + 4y' + 5 =0 How to go about solving this:
$$y'' + 4y' + 5 = 0$$
$$y = Ae^{px} + Be^{qx}$$
I know the following:
p and q are solutions to the characteristic equation 
$am^2 + bm + c = 0$
So in this case
$m^2 + 4m + 5 = 0$
However I do not know what to do after this.
 A: If you want all solution real: 
$m^2+4m+5=0$ $\implies$ $m=-2\pm i$.
Then $e^{(-2+i)t}=e^{-2t}(\cos{t}+i\sin{t}) $ and $e^{(-2-i)t}=e^{-2t}(\cos{t}-i\sin{t})$ are two solutions.
Clearly linear combinations these solutions still are solutions. Then
$\phi(t)=\dfrac{e^{(-2+i)t}+e^{(-2-i)t}}{2}=\dfrac{e^{-2t}(\cos{t}+i\sin{t})+e^{-2t}(\cos{t}-i\sin{t})}{2}=e^{-2t}\cos{t}$
and 
$\xi(t)==\dfrac{e^{(-2+i)t}-e^{(-2-i)t}}{2i}=\dfrac{e^{-2t}(\cos{t}+i\sin{t})-e^{-2t}(\cos{t}-i\sin{t})}{2i}=e^{-2t}\sin{t}$
We have $\phi$ and $\xi$ are two solutions linearly independent. Therefore
all real solution are $C_1e^{-2t}\cos{t}+C_2e^{-2t}\sin{t}$ for $C_1,C_2\in \mathbb R$  
A: So, $$m=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot5}}2=-2\pm i$$
Then as the roots are unequal, $$y=Ae^{(-2+i)x}+Be^{(-2-i)x}$$ where $A,B$ are arbitrary constants.
$$y=e^{-2x}(Ae^{ix}+Be^{-ix})$$
$$=e^{-2x}\{(A+B)\cos x+i(A-B)\sin x\}$$ using Euler's identity.
or, $$y=e^{-2x}(c_1\cos x+c_2\sin x)$$ where $c_1=A+B,c_2=i(A-B)$ are arbitrary constants.
