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  4. If y dx + y2 dy = xdy, x ∈ R, y > 0 and y(1) = 1, then y(- 3) =

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Q. If $y\, dx + y^2\, dy = xdy$, $x \in R$, $y > 0$ and $y(1) = 1$, then $y(- 3) =$

Differential Equations

Solution:

We have, $\frac{ydx-xdy}{y^{2}}+dy=0$
$\Rightarrow d\left(\frac{x}{y}+y\right)=0$
On integration, we get $\frac{x}{y}+y=c$
$x=1$,
$y=1$
$\Rightarrow c=2$
$\therefore \frac{x}{y}+y=2$
when $x = -3$
$\Rightarrow -\frac{3}{y}+y=2$
$\Rightarrow y^{2}-2y-3=0$
$\Rightarrow y=3$, $-1$