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
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:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where:
 * $P \left({x}\right) = \dfrac 1 x$
 * $Q \left({x}\right) = \dfrac 1 {x^2}$

From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:
 * $y_2 \left({x}\right) = v \left({x}\right) y_1 \left({x}\right)$

where:
 * $\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \, \mathrm d x} \, \mathrm d x$

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:


 * $y = C_1 x + k \left({-\dfrac 1 {2 x} }\right)$

where $k$ is arbitrary.

Setting $C_2 = -\dfrac k 2$ yields the result:
 * $y = C_1 x + \dfrac {C_2} x$