Question

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

Answer 1

given

$y=\sqrt{\frac{1-x}{1+x}}$ rationalizing the numerator by multilying with $1+x$ we get

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

therefore $\frac{y}{\sqrt{1-x^2}}=\frac{1}{1+x}$

dividing with $\sqrt{1-x^2}$ we get

$\frac{y}{1-x^2}=\frac{1}{(1+x)\sqrt{1-x^2}}$ let it be equation 1

now differentiating the equation we get

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

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

[since from equation 1]

therefore $(1-x^2)dy/dx+y=0$