Linear Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = (x^2 - 1)^2

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

has the general solution:
 * $y = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$

Proof
$(1)$ can be manipulated into the form:
 * $y'' - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = x^2 - 1$

It can be seen that this is a nonhomogeneous linear second order ODE in the form:
 * $y'' + \map P x y' + \map Q x y = \map R x$

where:
 * $\map P x = -\dfrac {2 x} {x^2 - 1}$
 * $\map Q x = \dfrac 2 {x^2 - 1}$
 * $\map R x = x^2 - 1$

First we establish the solution of the corresponding homogeneous linear second order ODE:
 * $\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$

From Second Order ODE: $\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$, this has the general solution:
 * $y_g = C_1 x + C_2 \paren {x^2 + 1}$

It remains to find a particular solution $y_p$ to $(1)$.

Expressing $y_g$ in the form:
 * $y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$

we have:

By the Method of Variation of Parameters, we have that:


 * $y_p = v_1 y_1 + v_2 y_2$

where:

where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.

We have that:

Hence:

It follows that:

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:


 * $y = y_g + y_p = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$

is the general solution to $(1)$.