Question

If $y=(co{s}^{-1}x{)}^2$, prove that $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}-2=0$. Hence find $y_2$ when $x=0$.

Answer 1

$y=({\cos }^{-1}{)}^2\frac{dy}{dx}=\frac{d}{dx}[{\cos }^{-1}x{]}^2\frac{dy}{dx}=2{\cos }^{-1}x\frac{d}{dx}\left({\cos }^{-1}x\right)\frac{dy}{dx}=2{\cos }^{-1}x\frac{(-1)}{\sqrt{(}1-x^2)}\frac{d^2}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left[2{\cos }^{-1}x\right(-1\left)\right]\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}=-2\frac{d}{dx}\left({\cos }^{-1}x\right)\times \frac{1}{\sqrt{1-x^2}}+2{\cos }^{-1}x\frac{d}{dx}\left[-1\sqrt{1-x^2}\right]\frac{d^{2}y}{d{x}^2}=(-2)\frac{(-1)}{\sqrt{1-x^2}}+2{\cos }^{-1}x\frac{d}{dx}[-\frac{1}{\sqrt{1-x^2}}]\frac{d^{2}y}{d{x}^2}=\frac{(-2)(-1)}{\sqrt{1-x^2}}+x{\cos }^{-1}x\frac{1}{2}(1-x^2{)}^{\frac{3}{2}}.(-2x\left)\right(1-x^2)\frac{d^{2}y}{d{x}^2}=(1-x^2\left)\right[\frac{2}{(1-x^2)}+\frac{(-2x){\cos }^{-1}x}{(1-x^2{)}^{\frac{3}{2}}}]=2+\frac{{\cos }^{-1}x(-2x)}{\sqrt{1-x^2}}+x\frac{dy}{dx}=\frac{2\times {\cos }^{-1}x}{\sqrt{1-x^2}}(1-x^2)\frac{d^2}{d{x}^2}-x\frac{dy}{dx}-x\frac{dy}{dx}-2=\frac{2+{\cos }^{-1}x(-2x)}{\sqrt{1-x^2}}-\frac{2\times {\cos }^{-1}\left(x\right)}{\sqrt{1-x^2}}-2=0$

Hence proved

$y_2(x=0)=(\frac{d^{2}y}{d{x}^2}{)}_{(}x=0)=\frac{2}{(1-0^2)}+\frac{(-2\times 0){\cos }^{-1}\left(10\right)}{(1-0^2{)}^{\frac{3}{2}}}=2\therefore y_2=0whenx=0$