Form the differential equation
having $y = {\left( {{{\sin }^{ - 1}}x} \right)^2} + A{\cos ^{ - 1}}x + B$, where $A$ and $B$
are arbitrary constants, as its general solution.

Given that, $y = {\left( {{{\sin }^{ - 1}}x} \right)^2} + A{\cos ^{ - 1}}x + B$

On differentiating w.r.t. $x$, we get
$\frac{{dy}}{{dx}} = \frac{{2{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }} + \frac{{( - A)}}{{\sqrt {1 - {x^2}} }}$

$\Rightarrow$ $\sqrt {1 - {x^2}} \frac{{dy}}{{dx}} = 2{\sin ^{ - 1}}x - A$

Again, differentiating w.r.t. $x$,

we get $\sqrt {1 - {x^2}} \frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} \cdot \frac{{ - 2x}}{{2\sqrt {1 + {x^2}} }} = \frac{2}{{\sqrt {1 - {x^2}} }}$

$\Rightarrow$ $\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - \frac{x}{{\sqrt {1 - {x^2}} }} \cdot \sqrt {1 - {x^2}} \frac{{dy}}{{dx}} = 2$

$\Rightarrow$ $\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} = 2$

$\Rightarrow$ $\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} - 2 = 0$

which is the required differential equation.