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:
 * $y = C_1 e^{5 x} + C_2 e^{-x}$

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:
 * $m_1 = 5$
 * $m_2 = -1$

These are real and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
 * $y = C_1 e^{5 x} + C_2 e^{-x}$