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

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Theorem

The second order ODE:

$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$

has the general solution:

$y = C_1 x + C_2 x^2$

on any closed real interval which does not contain $0$.


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 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:

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


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

\(\displaystyle y\) \(=\) \(\displaystyle C_1 e^t + C_2 e^{2 t}\)
\(\displaystyle \) \(=\) \(\displaystyle C_1 x + C_2 x^2\) substituting $x = e^t$

$\blacksquare$

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