If xdy = y(dx + ydy), y 0 and y(1) = 1, the | vedclass

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(B) Given the differential equation $xdy = y(dx + ydy)$.Dividing both sides by $y^2$,we get $\frac{xdy - ydx}{y^2} = dy$.This can be written as $-d(\f

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$3$

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(B) Given the differential equation $xdy = y(dx + ydy)$.Dividing both sides by $y^2$,we get $\frac{xdy - ydx}{y^2} = dy$.This can be written as $-d(\frac{x}{y}) = dy$.Integrating both sides,we get $-\frac{x}{y} = y + C$,which simplifies to $\frac{x}{y} + y = C$.Given $y(1) = 1$,substituting $x=1$ and $y=1$ gives $\frac{1}{1} + 1 = C$,so $C = 2$.The equation becomes $\frac{x}{y} + y = 2$,or $x + y^2 = 2y$,which is $y^2 - 2y + x = 0$.For $x = -3$,the equation is $y^2 - 2y - 3 = 0$.Factoring the quadratic,we get $(y - 3)(y + 1) = 0$.Thus,$y = 3$ or $y = -1$.Since the condition $y > 0$ is given,we have $y = 3$.

If $xdy = y(dx + ydy), y > 0$ and $y(1) = 1,$ then $y(-3)$ is equal to

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