Question

If $y=\sqrt{\frac{1-x}{1+x}},$ prove that $(1-x^2)\frac{dy}{dx}+y=0$

Answer 1

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

From $\left(1\right)$

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