#### Answer

$y(x) = \frac{1}{2}x(5x^{3} - 3)$ satisfies the differential equation $(1-x^{2})y" - 2xy' + 12y = 0$

#### Work Step by Step

The Legendre differential equation of 3rd order is:
$(1-x^{2})y" - 2xy' + 3(4y) = (1-x^{2})y" - 2xy' + 12y = 0$
Given that $y(x) = \frac{1}{2}x(5x^{3} - 3) = \frac{5x^{3}}{2} - \frac{3x}{2}$ is the solution of the equation above, differentiating it should get:
$y'(x) = \frac{3}{2}(5x^{2} - 1)$
$y"(x) = 15x$
Substitute $y(x)$, $y'(x)$ and $y"(x)$ into the above differential equation:
$15x(1 - x^{2}) - 2x(\frac{3}{2}(5x^{2} - 1)) + 12(\frac{5x^{3}}{2} - \frac{3x}{2})$
= $x^{3}( -15 - 15 + 30) + x(15 + 3 - 18)
=0$
As such, $y(x) = \frac{1}{2}x(5x^{3} - 3)$ satisfies the differential equation
$(1-x^{2})y" - 2xy' + 12y = 0$.