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

$x = 2y^2 + Cy$

The correct answer is Option (1) → $x = 2y^2 + Cy$ ##

The given differential equation can be written as

$\frac{dx}{dy} - \frac{x}{y} = 2y$

This is a linear differential equation of the type $\frac{dx}{dy} + P_1x = Q_1$, where $P_1 = -\frac{1}{y}$ and $Q_1 = 2y$. Therefore

$\text{I.F.} = e^{\int -\frac{1}{y} \, dy} = e^{-\log y} = e^{\log(y)^{-1}} = \frac{1}{y}$

Hence, the solution of the given differential equation is

$x \cdot \frac{1}{y} = \int (2y) \left( \frac{1}{y} \right) dy + C$

$\text{or } \frac{x}{y} = \int (2 \, dy) + C$

$\text{or } \frac{x}{y} = 2y + C$

$\text{or } x = 2y^2 + Cy$

which is a general solution of the given differential equation.