Linear Second Order ODE/x^2 y'' + x y' - y = 0
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 1
The particular solution:
- $y_1 = x$
can be found by inspection.
Let $(1)$ be written as:
- $(2): \quad y + \dfrac {y'} x - \dfrac y {x^2} = 0$
which is in the form:
- $y + \map P x y' + \map Q x y = 0$
where:
- $\map P x = \dfrac 1 x$
- $\map Q x = \dfrac 1 {x^2}$
From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
- $\map {y_2} x = \map v x \, \map {y_1} x$
where:
- $\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
is also a particular solution of $(1)$.
We have that:
\(\ds \int P \rd x\) | \(=\) | \(\ds \int \frac 1 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{-\int P \rd x}\) | \(=\) | \(\ds e^{-\ln x}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 x\) |
Hence:
\(\ds v\) | \(=\) | \(\ds \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x\) | Definition of $v$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac 1 {x^2} \frac 1 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac {\d x} {x^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {2 x^2}\) |
and so:
\(\ds y_2\) | \(=\) | \(\ds v y_1\) | Definition of $y_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \paren {-\dfrac 1 {2 x^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {2 x}\) |
From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $y = C_1 x + k \paren {-\dfrac 1 {2 x} }$
where $k$ is arbitrary.
Setting $C_2 = -\dfrac k 2$ yields the result:
- $y = C_1 x + \dfrac {C_2} x$
$\blacksquare$
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:
- $p = 1$
- $q = -1$
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 $x = e^t$ | |||||||||||
\(\ds \) | \(=\) | \(\ds C_1 x + \frac {C_2} x\) |
$\blacksquare$