Suppose $y = y\left(x\right)$ satisfies the differential
equation $ydx+y^{2}dy=xdy\text{.}$ If $y\left(x\right)>0\mathrm{\forall }x\in \mathbf{R}$ and $y\left(1\right)=1$

then $y(–3)$ equals .

$\frac{ydx-xdy}{y^2}=-dy$

$\Rightarrow d\left(\frac{x}{y}\right)=-dy\Rightarrow x/y=-y+c$

since $y\left(1\right)=1$so $c=2$ , Thus $x/y=-y+2$

when $x=-3,-\frac{3}{v}=-y+2$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-3=-y^2+2y\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\Rightarrow (y-1{)}^2=4\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}y\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=1\pm 2=3 or -1\end{array}$

As $y\left(x\right) > 0$ for all $x\in \mathbf{R}$ so $y(-3)=3$