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

Proof
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:
 * $\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.

We have that:

Hence:

and so:

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:
 * $y = C_1 x + \dfrac {C_2} x$