Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/Proof 1

Proof
Consider the functions:


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

We have that:

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

Hence it can be seen that:

are particular solutions to $(1)$.

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

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 $0$.