Linear 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:
- $\map {y_1} x = x$
- $\map {y_2} x = x^2$
We have that:
\(\ds \frac \d {\d x} \, x\) | \(=\) | \(\ds 1\) | Power Rule for Derivatives | |||||||||||
\(\ds \frac \d {\d x} \, x^2\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d^2} {\d x^2} \, x\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \frac {\d^2} {\d x^2} \, x^2\) | \(=\) | \(\ds 2\) |
Putting $x$ and $x^2$ into $(1)$ in turn:
\(\ds x^2 \cdot 0 - 2 x \cdot 1 + 2 x\) | \(=\) | \(\ds 2 x - 2 x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
\(\ds x^2 \cdot 2 - 2 x \cdot 2 x + 2 x^2\) | \(=\) | \(\ds 2 x^2 - 4 x^2 + 2 x^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
Hence it can be seen that:
\(\ds \map {y_1} x\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \map {y_2} x\) | \(=\) | \(\ds x^2\) |
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
\(\ds \map W {y_1, y_2}\) | \(=\) | \(\ds \begin{vmatrix} x & x^2 \\ 1 & 2 x \end{vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 x^2 - x^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2\) |
So the Wronskian of $y_1$ and $y_2$ is zero only when $x = 0$.
Let $\Bbb I = \closedint a b$ 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:
- $y'' + \map P x y' + \map Q x y = 0$
where $\map P x = \dfrac {-2 x} {x^2}$ and $\map Q x = \dfrac 2 {x^2}$.
So by Real Rational Function is Continuous: $P$ and $Q$ are continuous on $\closedint a b$
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 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 Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution:
\(\ds y\) | \(=\) | \(\ds C_1 e^t + C_2 e^{2 t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C_1 x + C_2 x^2\) | 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.