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If $x d y=y(d x+y d y), y \gt 0$ and $y(1)=1$, then $y(-3)$ is equal to
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3
$x d y=y(d x+y d y) \Rightarrow \frac{x d y-y d x}{y^2}=d y \Rightarrow-d\left(\frac{x}{y}\right)=d y$
Integrating both sides, we get $\frac{x}{y}+y=c$
$\because y(1)=1 \Rightarrow c=2 ; \therefore \frac{x}{y}+y=2$
For $x=-3$,
$y^2-2 y-3=0 \Rightarrow y=-1$ or $3 \Rightarrow y=3 \quad(\because y\gt0)$
Integrating both sides, we get $\frac{x}{y}+y=c$
$\because y(1)=1 \Rightarrow c=2 ; \therefore \frac{x}{y}+y=2$
For $x=-3$,
$y^2-2 y-3=0 \Rightarrow y=-1$ or $3 \Rightarrow y=3 \quad(\because y\gt0)$
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