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

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

By Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE, this can be expressed as:
 * $\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$

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

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

From Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution:

Note that when $x = 0$, $e^t = x$ has no solution for $t$, and so is excluded from this solution.