Question

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

Answer 1

We have,

$y={\left({\sin }^{-1}x\right)}^2+c$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=2\left({\sin }^{-1}x\right)\times \frac{1}{\sqrt{1-x^2}}+0$

$\frac{dy}{dx}=\frac{2\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}$

Again differentiating both sides w.r.t $x$, we get

$\frac{d^{2}y}{d{x}^2}=\frac{\left(\sqrt{1-x^2}\right)\times \frac{2}{\sqrt{1-x^2}}-2\left({\sin }^{-1}x\right)\times \frac{1}{2\sqrt{1-x^2}}\times (0-2x)}{{\left(\sqrt{1-x^2}\right)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{2+2\left({\sin }^{-1}x\right)\times \frac{x}{\sqrt{1-x^2}}}{\sqrt{1-x^2}}$

$\frac{d^{2}y}{d{x}^2}=\frac{2+\frac{2x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}}{(1-x^2)}$

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

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

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

Hence, proved.