If $x d y = y (d x + y d y), y (1) = 1$ and $y (x) > 0$ . Then, $y (- 3)$ is equal to

3

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Solution

The correct option is A $3$
Given, $x d y = y (d x + y d y), y > 0$
$\Rightarrow x d y - y d x = y^{2} d y$
$\Rightarrow \frac{x d y - y d x}{y^{2}} = d y$
$\Rightarrow - d (\frac{x}{y}) = d y$
On integrating both sides, we get
$\frac{x}{y} = - y + C$ .... (i)
As $y (1) = 1 \Rightarrow x = 1, y = 1$
$∴ C = 2$
Equation (i) becomes $\frac{x}{y} + y = 2$
Again, for $x = - 3$
$\Rightarrow - 3 + y^{2} = 2 y$
$\Rightarrow y^{2} - 2 y - 3 = 0$
$\Rightarrow (y + 1) (y - 3) = 0$
As $y > 0$ , we take $y = 3$ , (neglecting $y = - 1)$ .

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