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(A) The given differential equation is $y dx - (x + 2y^2) dy = 0$. Dividing by $dy$,we get $y \frac{dx}{dy} - x - 2y^2 = 0$. Rearranging the terms,we

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$x = 2y^2 + Cy$

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(A) The given differential equation is $y dx - (x + 2y^2) dy = 0$. Dividing by $dy$,we get $y \frac{dx}{dy} - x - 2y^2 = 0$. Rearranging the terms,we have $y \frac{dx}{dy} - x = 2y^2$. Dividing by $y$,we get $\frac{dx}{dy} - \frac{1}{y} x = 2y$. This is a linear differential equation of the form $\frac{dx}{dy} + P_1 x = Q_1$,where $P_1 = -\frac{1}{y}$ and $Q_1 = 2y$. The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P_1 dy} = e^{\int -\frac{1}{y} dy} = e^{-\log y} = e^{\log(y^{-1})} = \frac{1}{y}$. The general solution is given by $x(I.F.) = \int (Q_1 	imes I.F.) dy + C$. Substituting the values,we get $x \cdot \frac{1}{y} = \int (2y \cdot \frac{1}{y}) dy + C$. $\frac{x}{y} = \int 2 dy + C$. $\frac{x}{y} = 2y + C$. $x = 2y^2 + Cy$.

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

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