Question

Let $y=\frac{1}{1+x+\ln x}$, Then

• A. $x\frac{dy}{dx}+y=x$
• B. $x\frac{dy}{dx}=y(y\ln x-1)$
• C. $x^2\frac{dy}{dx}=y^2+1-x^2$
• D. $x{\left[\frac{dy}{dx}\right]}^2=y-x$

Answer 1

The correct option is B $x\frac{dy}{dx}=y(y\ln x-1)$

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