Question

Verify that the given function is a solution of the differential equation, $y=(\sin^{-1}x)^{2} +c; (1-x^{2})\frac{d^{2}y}{dx^{2}}-x \frac{dy}{dx}=2$

Answer 1

To verify that the given function is a solution of the differential equation, $y=(\sin^{-1}x)^{2} +c;\\\\ (1-x^{2})\frac{d^{2}y}{dx^{2}}-x \frac{dy}{dx}=2$

we have to differentiate it twice because differential equation has order 2

$y=(\sin^{-1}x)^{2} +c \\ \\ \frac{dy}{dx} = 2 {sin}^{ - 1} x( \frac{1}{ \sqrt{1 - {x}^{2} } } )...eq1 \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = 2 {sin}^{ - 1} x(0 - \frac{1 \times (1 - {x}^{2})( - 2x) }{2} ) \div (1 - {x}^{2} ) + 2 \frac{1}{ \sqrt{1 - {x}^{2} } } ( \frac{1}{ \sqrt{1 - {x}^{2} } } ) \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = {sin}^{ - 1} x( \frac{2x(1 - {x}^{2}) }{1 - {x}^{2} } ) + \frac{2}{1 - {x}^{2} } \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = 2x \: {sin}^{ - 1} x + \frac{2}{1 - {x}^{2} } ....eq2 \\ \\(1 - {x}^{2} )\frac{ {d}^{2} y}{ {dx}^{2} } = 2x(1 - {x}^{2} ) {sin}^{ - 1} x + 2\\\\(1 - {x}^{2} )\frac{ {d}^{2} y}{ {dx}^{2} } - x(\sqrt{1 - {x}^{2}} )\frac{dy}{dx} = 2$

it can't solve further,their might some typing error.