Linear Second Order ODE/x^2 y'' + 2 x y' - 12 y = 0

Theorem
The second order ODE:
 * $(1): \quad x^2 y'' + 2 x y' - 12 y = 0$

has the solution:
 * $y = C_1 x^3 + C_2 x^{-4}$

Proof
It can be seen that $(1)$ is an instance of the Cauchy-Euler Equation:
 * $x^2 y'' + p x y' + q y = 0$

where:
 * $p = 2$
 * $q = -12$

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:
 * $(2): \quad \dfrac {\mathrm d^2 y} {\mathrm d t^2} + \dfrac {\mathrm d y} {\mathrm d t^2} - 12 y = 0$

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

Its auxiliary equation is:
 * $(2): \quad: m^2 + m - 12 = 0$

From Solution to Quadratic Equation: Real Coefficients, the roots of $(2)$ are:
 * $m_1 = 3$
 * $m_2 = -4$

These are real and unequal.

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