Linear Second Order ODE/x^2 y'' + x y' - y = 0/Proof 2
Jump to navigation
Jump to search
Theorem
The second order ODE:
- $(1): \quad x^2 y + x y' - y = 0$
has the general solution:
- $y = C_1 x + \dfrac {C_2} x$
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:
- $p = 1$
- $q = -1$
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:
- $(2): \quad \dfrac {\d^2 y} {\d t^2} - y = 0$
From Linear Second Order ODE: $y - y = 0$, this has the general solution:
\(\ds y\) | \(=\) | \(\ds C_1 e^t + C_2 e^{-t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds C_1 x + C_2 x^{-1}\) | substituting $x = e^t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds C_1 x + \frac {C_2} x\) |
$\blacksquare$