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If $x d y=y(d x+y d y) ; y(1)=1$ and $y(x)>0$, then what is $y(-$ 3) equal to? $\quad$
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Verified Answer
The correct answer is:
3
$x d y=y d x+y^{2} d y$
$\Rightarrow x \cdot \frac{d y}{d x}=y+y^{2} \cdot \frac{d y}{d x} \Rightarrow \frac{y-x \cdot \frac{d y}{d x}}{y^{2}}=\frac{-d y}{d x}$
Integrating both sides
$\frac{x}{y}=-y+c$
Given, $x=1, y=1$
$\Rightarrow \frac{1}{1}=-1+c \quad \Rightarrow c=2$
$\therefore \frac{\mathrm{x}}{\mathrm{y}}+\mathrm{y}=2$
$\Rightarrow x+y^{2}=2 y$
$\Rightarrow-3+y^{2}-2 y=0$
$\Rightarrow y^{2}-2 y-3=0$
$y=3,-1$
$\therefore y=3$
$\Rightarrow x \cdot \frac{d y}{d x}=y+y^{2} \cdot \frac{d y}{d x} \Rightarrow \frac{y-x \cdot \frac{d y}{d x}}{y^{2}}=\frac{-d y}{d x}$
Integrating both sides
$\frac{x}{y}=-y+c$
Given, $x=1, y=1$
$\Rightarrow \frac{1}{1}=-1+c \quad \Rightarrow c=2$
$\therefore \frac{\mathrm{x}}{\mathrm{y}}+\mathrm{y}=2$
$\Rightarrow x+y^{2}=2 y$
$\Rightarrow-3+y^{2}-2 y=0$
$\Rightarrow y^{2}-2 y-3=0$
$y=3,-1$
$\therefore y=3$
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