Question

If $y=({\sin }^{-1}x{)}^2$, then show that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$

Answer 1

$y=(si{n}^{-1}x{)}^2$

on differentiating wrt x we get

$\Rightarrow \frac{dy}{dx}=2si{n}^{-1}x\times \frac{1}{\sqrt{1-x^2}}$ _______ (1)

again differentiating wrt x we get

$\Rightarrow \frac{d^{2}y}{d{x}^2}=2\frac{\frac{1}{\sqrt{1-x^2}}\times \sqrt{1-x^2}}{(1-x^2)}-si{n}^{-1}\times \frac{1x-2x}{2\sqrt{1-x^2}}$

$(1-x^2)\frac{d^{2}y}{d{x}^2}=2[1+x\frac{si{n}^{-1}x}{\sqrt{1-x^2}}]$

$(1-x^2)\frac{d^{2}y}{d{x}^2}-2x\frac{si{n}^{-1}x}{\sqrt{1-x^2}}=2$

From equation (1)

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