If a curve $y = f(x)$ passes through the point (1, -1) and satisfies the differential equation $y(1+xy) dx=x\, dy, $ then $f(-\frac{1}{2})$ is equal to

$\frac{4}{5}$

The correct answer is option (4) : $\frac{4}{5}$

The differential equation is

$y(1+xy )dx=xdy$

$⇒y\, dx -x\, dy = -xy^2 dx$

$⇒\frac{ydx-xdy}{y^2}=xdx$

$⇒d(\frac{x}{y}) = -xdx$

On integrating, we obtain

$\frac{x}{y}=-\frac{x^2}{2}+C$ ..............(i)

It is given that the curve given by (i) passes through the point (1, -1).

$∴-1=-\frac{1}{2}+C ⇒C=-\frac{1}{2}$

Putting $C=-\frac{1}{2}$ in (i), we obtain

$y(x^2 +1) + 2x=0$ ..............(ii)

Putting $ x=-\frac{1}{2} $ in (ii), we obtain $y =\frac{4}{5}$. Hence, $f(-\frac{1}{2})= \frac{4}{5}$