Question

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

Answer 1

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

$\frac{dy}{dx}=\frac{1}{\sqrt[2]{2\frac{1-x}{1+x}}}\times \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$

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

$\frac{dy}{dx}=\frac{1}{2}\frac{(1+x{)}^{1/2}}{(1-x{)}^{1/2}}\frac{[-1-x-1+x]}{(1+x{)}^2}$

$\frac{dy}{dx}=\frac{1}{2}\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)\left[\frac{-x}{(1-x{)}^{1/2}(1+x{)}^{3/2}}\right]$

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

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