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(B) Given the differential equation: $y dx + (x - y^2) dy = 0$Rearranging the terms,we get: $y dx = (y^2 - x) dy$Dividing by $dy$ and $y$,we get: $\fr

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$x = \frac{y^3}{3} + \frac{C}{y}$

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(B) Given the differential equation: $y dx + (x - y^2) dy = 0$Rearranging the terms,we get: $y dx = (y^2 - x) dy$Dividing by $dy$ and $y$,we get: $\frac{dx}{dy} = \frac{y^2 - x}{y} = y - \frac{x}{y}$This can be rewritten as: $\frac{dx}{dy} + \frac{1}{y} x = y$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{y}$ and $Q = y$.Now,calculate the Integrating Factor $(I.F.)$:$I.F. = e^{\int P dy} = e^{\int \frac{1}{y} dy} = e^{\ln |y|} = y$The general solution is given by: $x(I.F.) = \int (Q 	imes I.F.) dy + C$Substituting the values: $x(y) = \int (y \cdot y) dy + C$$xy = \int y^2 dy + C$$xy = \frac{y^3}{3} + C$Dividing by $y$,we get: $x = \frac{y^2}{3} + \frac{C}{y}$

Find the general solution of the differential equation: $y dx + (x - y^2) dy = 0$

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