If $x d y=y(d x+y d y), y(1)=1$ and $y(x)>0$. Then, $y(-3)=$

3

We have,

$x d y=y(d x+y d y)$

$\Rightarrow x d y-y d x=y^2 d y \Rightarrow-\left(\frac{y d x-x d y}{y^2}\right)=d y \Rightarrow-d\left(\frac{x}{y}\right)=d y$

On integrating, we get

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

It is given that $y(1)=1$ i.e., $y=1$ when $x=1$

∴  $-1=1+C \Rightarrow C=-2$

Putting $C=-2$ in (i), we get $\frac{-x}{y}=y-2$

Putting $x=-3$, we get

$\frac{3}{y}=y-2$

$\Rightarrow y^2-2 y-3=0 \Rightarrow(y-3)(y+1)=0 \Rightarrow y=3$        [∵ y > 0]