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If $\mathrm{xdy}=\mathrm{y} \mathrm{dx}+\mathrm{y}^{2} \mathrm{dy}, \mathrm{y}>0$ and $\mathrm{y}(1)=1$, then what is
$\mathrm{y}(-3)$ equal to?
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$\mathrm{y}(-3)$ equal to?
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Verified Answer
The correct answer is:
3 only
Given, $x d y=y d x+y^{2} d y$
$\Rightarrow 1=\frac{y}{x} \frac{d x}{d y}+\frac{y^{2}}{x}$
$\Rightarrow \frac{d x}{d y}+y=\frac{x}{y}$
$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=-y$
$P=-\frac{1}{y}, Q=-y$
$\mathrm{IF}=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}$
Now, Solution of d.E. $\mathrm{x}(\mathrm{I.F})=\int(\mathrm{Q} . \mathrm{I.F}) \mathrm{dy}$
$\frac{x}{y}=\int \frac{1}{y}(-y) d y+C$
$\Rightarrow \frac{x}{y}=\int-1 d y+C$
$\Rightarrow \frac{x}{y}=-y+C$
$y(1)=1$
$\frac{1}{1}=-1+C \Rightarrow C=2$
$\Rightarrow \frac{x}{y}=-y+2 \Rightarrow x=-y^{2}+2 y$
$\Rightarrow y(-3) \Rightarrow-3=-y^{2}+2 y$
$\Rightarrow y^{2}-2 y-3=0$
$\Rightarrow y=\frac{+2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2}$
$\Rightarrow y=3,-1$
Since $y>0$ so $y=3$.
$\Rightarrow 1=\frac{y}{x} \frac{d x}{d y}+\frac{y^{2}}{x}$
$\Rightarrow \frac{d x}{d y}+y=\frac{x}{y}$
$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=-y$
$P=-\frac{1}{y}, Q=-y$
$\mathrm{IF}=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}$
Now, Solution of d.E. $\mathrm{x}(\mathrm{I.F})=\int(\mathrm{Q} . \mathrm{I.F}) \mathrm{dy}$
$\frac{x}{y}=\int \frac{1}{y}(-y) d y+C$
$\Rightarrow \frac{x}{y}=\int-1 d y+C$
$\Rightarrow \frac{x}{y}=-y+C$
$y(1)=1$
$\frac{1}{1}=-1+C \Rightarrow C=2$
$\Rightarrow \frac{x}{y}=-y+2 \Rightarrow x=-y^{2}+2 y$
$\Rightarrow y(-3) \Rightarrow-3=-y^{2}+2 y$
$\Rightarrow y^{2}-2 y-3=0$
$\Rightarrow y=\frac{+2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2}$
$\Rightarrow y=3,-1$
Since $y>0$ so $y=3$.
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