Linear Second Order ODE/2 x^2 y'' + 10 x y' + 8 y = 0

Theorem
The second order ODE:
 * $(1): \quad 2 x^2 y'' + 10 x y' + 8 y = 0$

has the solution:
 * $y = C_1 x^{-2} + C_2 x^{-2} \ln x$

Proof
Let $(1)$ be rewritten as:
 * $x^2 y'' + 5 x y' + 4 y = 0$

It can be seen to be an instance of the Cauchy-Euler Equation:
 * $x^2 y'' + p x y' + q y = 0$

where:
 * $p = 5$
 * $q = 4$

By Conversion of Cauchy-Euler Equation to Linear Equation, this can be expressed as:
 * $\dfrac {\mathrm d^2 y} {\mathrm d t^2} + \left({p - 1}\right) \dfrac {\mathrm d y} {\mathrm d t^2} + q y = 0$

by making the substitution:
 * $x = e^t$

Hence it can be expressed as:
 * $(2): \quad \dfrac {\mathrm d^2 y} {\mathrm d t^2} + 4 \dfrac {\mathrm d y} {\mathrm d t^2} + 4 y = 0$

It can be seen that $(2)$ is a constant coefficient homogeneous linear second order ODE.

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

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

These are real and equal.

So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(2)$ is: