Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0

Theorem
The second order ODE:
 * $(1): \quad x^2 y'' + x y' - 4 y = 0$

has the solution:
 * $y = C_1 x^2 + \dfrac {C_2} {x^2}$

Proof
Note that:

and so by inspection:
 * $y_1 = x^2$

is a particular solution of $(1)$.

$(1)$ can be expressed as:
 * $(2): \quad y'' + \dfrac 1 x y' - \dfrac 4 {x^2} y = 0$

which is in the form:
 * $y'' + P \left({x}\right) y' + Q \left({x}\right) y = 0$

where:
 * $P \left({x}\right) = \dfrac 1 x$
 * $Q \left({x}\right) = - \dfrac 4 {x^2}$

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:


 * $y = C_1 x^2 + k \left({- \dfrac 1 {4 x^2}}\right)$

where $k$ is arbitrary.

Setting $C_2 = - \dfrac k 4$ yields the result:
 * $y = C_1 x^2 + \dfrac {C_2} {x^2}$