If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation $y(1+x y) d x=x d y$, then $f\left(-\frac{1}{2}\right)$ is equal to
Answer & explanation
Correct answer: option 4
The differential equation is
$y(1+x y) d x=x d y$
$\Rightarrow y d x-x d y=-x y^2 d x $
$\Rightarrow \frac{y d x-x d y}{y^2}=-x d x $
$\Rightarrow d\left(\frac{x}{y}\right)=-x d x$
On integrating, we obtain
$\frac{x}{y}=-\frac{x^2}{2}+C$ ......(i)
It is given that the curve given by (i) passes thought the point $(1,-1)$.
∴ $-1=-\frac{1}{2}+C \Rightarrow C=-\frac{1}{2}$
Putting $C=-\frac{1}{2}$ in (i), we obtain
$y\left(x^2+1\right)+2 x=0$ .....(ii)
Putting $x=-\frac{1}{2}$ in (ii), we obtain $y=\frac{4}{5}$.
Hence, $f\left(-\frac{1}{2}\right)=\frac{4}{5}$