If x , dy = y(dx + y , dy),y(1) = 1,y(x) 0, | vedclass

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(C) Given equation: $x \, dy = y(dx + y \, dy)$ Rearranging the terms: $x \, dy = y \, dx + y^2 \, dy$ $y \, dx = (x - y^2) \, dy$ Dividing by $y \, d

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$3$

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(C) Given equation: $x \, dy = y(dx + y \, dy)$ Rearranging the terms: $x \, dy = y \, dx + y^2 \, dy$ $y \, dx = (x - y^2) \, dy$ Dividing by $y \, dy$: $\frac{dx}{dy} = \frac{x}{y} - y$ $\frac{dx}{dy} - \frac{1}{y} x = -y$ This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = -y$. Integrating Factor ($I$.$F$.) $= e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln y} = \frac{1}{y}$. The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) \, dy + C$. $x \cdot \frac{1}{y} = \int (-y) \cdot \frac{1}{y} \, dy + C$ $\frac{x}{y} = \int -1 \, dy + C$ $\frac{x}{y} = -y + C$ Given $y(1) = 1$,substitute $x = 1$ and $y = 1$: $\frac{1}{1} = -1 + C \Rightarrow C = 2$. So,$\frac{x}{y} = -y + 2$. To find $y(-3)$,substitute $x = -3$: $\frac{-3}{y} = -y + 2$ $-3 = -y^2 + 2y$ $y^2 - 2y - 3 = 0$ $(y - 3)(y + 1) = 0$ Since $y(x) > 0$,we have $y = 3$.

If $x \, dy = y(dx + y \, dy)$,$y(1) = 1$,$y(x) > 0$,then $y(-3)$ is

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