Find the solution of $y''+4y'+5y=0$ that satisfies the initial condition $ϕ(0)=1$ and $ϕ'(0)=0$. Find the solution of the following differential equation that satisfies the initial condition $ϕ(0)=1$ and $ϕ'(0)=0$:
$$y''+4y'+5y=0$$
After solving the problem I found that the solution to be $$ϕ(x) = c_1e^{-2x}(\cos(2x) - i\sin(2x)) + c_2 e^{-2x}(\cos(2x) + i\sin(2x))$$
Moving on from this step to find the solution satisfying the given initial condition is a nightmare because I have to deal with $i$. Anyone know an easier way to solve this problem?
 A: Using the solution of form $y = e^{\lambda x}$, I got
$y(x) = c_1 e^{-2x} \sin(x) + c_2 e^{-2x} \cos(x)$ instead since you can rearrange the terms and absorb the $i$ into $c$.
If you rearrange your solution,
$c_1 e^{-2x}(\cos(2x) - i\sin(2x)) + c_2 e^{-2x}(\cos(2x) + i \sin(2x)) = e^{-2x}(c_1 \cos(2x) + c_2 \cos(2x)) + e^{-2x}((-i c_1 sin(2x) + ic_2 \sin(2x))  = e^{-2x}(c_1 + c_2) \cos(2x) + e^{-2x}(-i c_1 + ic_2) \sin(2x) $
Now you can rename $c_1 := (c_1 + c_2)$ and $c_2 := (-i c_1 + ic_2)$.
Solving the characteristic equation, we obtain $\lambda = -2 \pm i$. Thus
$$y(x) = c_1 e^{(-2+i)x} + c_2 e^{-(2+i)x}$$
Using Euler formular $e^{a + ib} = e^a \cos(b) + ie^a \sin(b)$ to obtain the reduced form.
A: $$ϕ(0)=1\Rightarrow  c_1 + c_2 =1$$
$$ϕ'(x)=-2c_1e^{-2x}(\cos(2x) - i\sin(2x)) -2 c_2 e^{-2x}(\cos(2x) + i\sin(2x))+ c_1e^{-2x}(-2\sin(2x)-2i\cos(2x)) + c_2 e^{-2x}(-2\sin(2x) + 2i\cos(2x))$$
$$ϕ'(0)=0\Rightarrow -2c_1-2 c_2 -2ic_1+2ic_2=0 \Rightarrow -c_1- c_2 -ic_1+ic_2=0\Rightarrow -ic_1+ic_2=1\Rightarrow c_1-c_2=i$$
Finally :
$c_1=\frac{1+i}2$ and $c_2=\frac{1-i}2$
