Question

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

Answer 1

$y=({\sin }^{-1}x{)}^2$

Differentiating both sides, we get

$y^{\prime }=2{\sin }^{-1}x\cdot \frac{1}{\sqrt{1-x^2}}$

Differentiating both sides once more, we get

$y^{\prime \prime }=2\cdot \frac{1}{\sqrt{1-x^2}}\cdot \frac{1}{\sqrt{1-x^2}}+2{\sin }^{-1}x\cdot (-\frac{1}{2})\cdot \frac{1}{(1-x^2)\sqrt{1-x^2}}\cdot (-2x)$

$\Rightarrow y^{\prime \prime }=\frac{2}{1-x^2}+\frac{2x{\sin }^{-1}x}{(1-x^2)\sqrt{1-x^2}}$

$\Rightarrow (1-x^2)y^{\prime \prime }=2+\frac{2x{\sin }^{-1}x}{\sqrt{1-x^2}}$

$\Rightarrow (1-x^2)y^{\prime \prime }=2+x{y}^{\prime }$

$\Rightarrow (1-x^2)y^{\prime \prime }-x{y}^{\prime }-2=0$

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