Linear Second Order ODE/y'' + 4 y' + 5 y = 0

Theorem
The second order ODE:
 * $(1): \quad y'' + 4 y' + 5 y = 0$

has the general solution:
 * $y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$

Proof
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.

Its auxiliary equation is:
 * $(2): \quad m^2 + 4 m + 5 = 0$

From Solution to Quadratic Equation: Real Coefficients, the roots of $(2)$ are:
 * $m_1 = -2 + i$
 * $m_2 = -2 - i$

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
 * $y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$