Question

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

Answer 1

$y=\frac{\sqrt{1-x}}{\sqrt{1+x}}$
Differentiating with respect to $x$, we get
$\frac{dy}{dx}=\frac{1}{2}{\left(\frac{1-x}{1+x}\right)}^{1/2-1}\times \frac{d}{dx}\frac{1-x}{1+x}$
$=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\times \frac{(1+x)(-1)-(1-x)\left(1\right)}{(1+x{)}^2}$
$=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\times \frac{-1-x-1+x}{(1+x{)}^2}$
$\frac{dy}{dx}=-\sqrt{\frac{1+x}{1-x}}\times \frac{1}{(1+x{)}^2}$
Multiplying both sides by $(1-x^2)$
$(1-x^2)\frac{dy}{dx}=-\sqrt{\frac{1+x}{1-x}}\times \frac{1}{(1+x{)}^2}(1-x^2)$
$(1-x^2)\frac{dy}{dx}=-\sqrt{\frac{1-x}{1+x}}$
$\Rightarrow (1-x^2)\frac{dy}{dx}=-y$
$\therefore (1-x^2)\frac{dy}{dx}+y=0$ (proved)