Legendre's Differential Equation/(1 - x^2) y'' - 2 x y' + 2 y = 0

Theorem
The special case of Legendre's differential equation:
 * $(1): \quad \left({1 - x^2}\right) y'' - 2 x y' + 2 y = 0$

has the solution:
 * $y = C_1 x + C_2 \left({\dfrac x 2 \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1}\right)$

Proof
Note that:

and so by inspection:
 * $y_1 = x$

is a particular solution of $(1)$.

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

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

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

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:


 * $y = C_1 x + C_2 \left({\dfrac x 2 \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1}\right)$