Linear Second Order ODE/2 y'' - 4 y' + 8 y = 0

Theorem
The second order ODE:
 * $(1): \quad 2 y'' - 4 y + 8 y = 0$

has the general solution:
 * $y = C_1 e^{2 x} + C_2 e^{-3 x}$

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

Let $(1)$ be written in the form:
 * $y'' - 2 y + 4 y = 0$

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

From Solution to Quadratic Equation: Real Coefficients, the roots of $(2)$ are:
 * $m_1 = 1 + \sqrt 3 i$
 * $m_2 = 1 - \sqrt 3 i$

These are complex and unequal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
 * $y = e^x \left({C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}\right)$