Question

If $y={\left({\sin }^{-1}x\right)}^2+c$ then prove that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$

Answer 1

$y=(si{n}^{-1}x{)}^2\therefore (\frac{dy}{dx})=2si{n}^{-1}x(\frac{1}{\sqrt{1-x^2}})\therefore (\frac{d^{2}y}{d{x}^2})=2\left[\right(\frac{\left(\frac{1}{\sqrt{1-x^2}}\right)\times \sqrt{1-x^2}-si{n}^{-1}x\left(\frac{1}{2\sqrt{1-x^2}}\right)\times -2x}{(1-x^2)}\left)\right]=(\frac{2}{1-x^2})(1+(\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}\left)\right)\therefore LHS=(1-x^2\left)\right(\frac{d^{2}y}{d{x}^2})-x(\frac{dy}{dx})=2(1+(\frac{xsi{n}^{-1}x}{\sqrt{1-x^2}}\left)\right)-x2si{n}^{-1}x(\frac{1}{\sqrt{1-x^2}})=2$