Linear 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 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.