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

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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 1

Consider the functions:

$y_1 \left({x}\right) = x$
$y_2 \left({x}\right) = x^2$


We have that:

\(\displaystyle \frac {\mathrm d} {\mathrm d x} \, x\) \(=\) \(\displaystyle 1\) Power Rule for Derivatives
\(\displaystyle \frac {\mathrm d} {\mathrm d x} \, x^2\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\mathrm d^2} {\mathrm d x^2} \, x\) \(=\) \(\displaystyle 0\)
\(\displaystyle \frac {\mathrm d^2} {\mathrm d x^2} \, x^2\) \(=\) \(\displaystyle 2\)


Putting $x$ and $x^2$ into $(1)$ in turn:

\(\displaystyle x^2 \cdot 0 - 2 x \cdot 1 + 2 x\) \(=\) \(\displaystyle 2 x - 2 x\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


\(\displaystyle x^2 \cdot 2 - 2 x \cdot 2 x + 2 x^2\) \(=\) \(\displaystyle 2 x^2 - 4 x^2 + 2 x^2\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


Hence it can be seen that:

\(\displaystyle y_1 \left({x}\right)\) \(=\) \(\displaystyle x\)
\(\displaystyle y_2 \left({x}\right)\) \(=\) \(\displaystyle x^2\)

are particular solutions to $(1)$.


Calculating the Wronskian of $y_1$ and $y_2$:

\(\displaystyle W \left({y_1, y_2}\right)\) \(=\) \(\displaystyle \begin{vmatrix} x & x^2 \\ 1 & 2 x \end{vmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 x^2 - x^2\)
\(\displaystyle \) \(=\) \(\displaystyle x^2\)

So the Wronskian of $y_1$ and $y_2$ is zero only when $x = 0$.


Let $\Bbb I = \left[{a \,.\,.\, b}\right]$ be a closed real interval such that $0 \notin \Bbb I$.

Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:

$y_1$ and $y_2$ are linearly independent on $\Bbb I$.


We can manipulate $(1)$ is a homogeneous linear second order ODE in the form:

$y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where $P \left({x}\right) = \dfrac {-2 x} {x^2}$ and $Q \left({x}\right) = \dfrac 2 {x^2}$.

So by Real Rational Function is Continuous: $P$ and $Q$ are continuous on $\left[{a \,.\,.\, b}\right]$

Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$(1)$ has the general solution:
$y = C_1 x + C_2 x^2$
on any closed real interval $\Bbb I$ which does not include $0$.

$\blacksquare$


Proof 2

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.