If $y$ is a function of $x$ given by $y d x + y^{2} d y = x d y$ where $y > 0$ and $y (1) = 1$ , then $y (- 3)$ is:

3

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2

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1

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5

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Solution

The correct option is A $3$

If $y$ is a function of $x$ given by $y d x + y^{2} d y = x d y$ where $y > 0$ and $y (1) = 1$
$y d x - x d y = - y^{2} d y$
$d (\frac{x}{y}) = \frac{y d x - x d y}{y^{2}}$

$d (\frac{x}{y}) = - d y$

Integrating with respect to $x$ we get,

$\frac{x}{y} = - y + c$

We have when $x = 1$ , we get $y = 1$

So, $1 = - 1 + c$

$c = 2$

$x = - y^{2} + 2 y$

When $x = - 3$ ,

$y^{2} - 2 y - 3 = 0$

$y = - 1$ and $y = 3$ as $y > 0$ .

Hence, $y = 3$

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