Question

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

Answer 1

$\sqrt{x^2+1}=log\{\sqrt{x^2+1}-x\}$

Differentiating with reset to be x we get

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

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

$\Rightarrow \sqrt{x^2+1}\frac{dy}{dx}+\frac{xy}{\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}-x}\left\{\frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}}\right\}$

$\Rightarrow \sqrt{x^2+1}\frac{dy}{dx}+\frac{xy}{\sqrt{x^2+1}}=\frac{-1}{\sqrt{x^2+1}}$

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

$\Rightarrow (x^2+1)\frac{dy}{dx}+xy+1=0.$