Solve the differential equation: $ydx + (x - y^2) dy = 0$

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

The correct answer is Option (2) → $xy = \frac{y^3}{3} + C$ ##

$ydx + (x - y^2) dy = 0$

Reducing the given differential equation to the form $\frac{dx}{dy} + Px = Q$

we get, $\frac{dx}{dy} + \frac{x}{y} = y$

$I.F. = e^{\int P dy} = e^{\int \frac{1}{y} dy} = e^{\log y} = y$

The general solution is given by

$x \cdot I.F. = \int Q \cdot I.F. \, dy \Rightarrow xy = \int y^2 dy$

$\Rightarrow xy = \frac{y^3}{3} + C, \text{ which is the required general solution}$