Question

If $xy=1+\log y$ and $k.\frac{dy}{dx}+y^2=0$ then k is

• A. $1+xy$
• B. $\frac{1}{xy-1}$
• C. $xy-1$
• D. $1-2xy$

Answer 1

The correct option is C $xy-1$

$xy=1+\log y$

Differentiating both sides, we get

$x\frac{dy}{dx}+y=\frac{1}{y}\times \frac{dy}{dx}$

Multiplying by y throughout, we get

$xy\frac{dy}{dx}+y^2=\frac{dy}{dx}$

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

$\Rightarrow k=xy-1$