Question

Form the differential equation corresponding to $y=({\sin }^{-1}x{)}^2+A{\cos }^{-1}x+B$ where A, B are arbitrary constants.

Answer 1

Let $y=({\sin }^{-1}{)}^2+A{\cos }^{-1}x+B$
so,
$\frac{dy}{dx}=2{\sin }^{-1}x-\frac{1}{\sqrt{1-x^2}}+A\times \frac{-1}{\sqrt{1-x^2}}+0$
$\frac{dy}{dx}=\frac{2\sin -1x}{\sqrt{1-x^2}}-\frac{A}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=2{\sin }^{-1}x-A$
Again differenttating we get-
$\sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-\frac{1}{2\sqrt{1-x^2}}-2x=\frac{2}{\sqrt{1-x^2}}$
$\Rightarrow \sqrt{1-x^2}\frac{d^{2}y}{d{x}^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$
$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=2$