Question

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

Answer 1

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

$\frac{dy}{dx}=2si{n}^{-1}x\frac{1}{\sqrt{1-x^2}}$

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

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

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

$=2$

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