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

## 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.