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Q.
If $x d y=y(d x+y d y), y(1)=1$ and $y\left(x_0\right)=-3$, then $x_0=$
Differential Equations
Solution:
$x d y=y d x+y^2 d y \Rightarrow \frac{x d y-y d x}{y^2}=d y$
$ \Rightarrow -d\left(\frac{x}{y}\right)=d y $
$ -\frac{x}{y}=y+c $ put $ x=1 \,\, y=1 \Rightarrow c=-2$
$ -\frac{x}{y}=y-2 $ put $ y=-3 $
$ \therefore \frac{x}{3}=-5 \Rightarrow x=-15$