Solve the differential equation $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$
The given differential equation is $\;$ $y \; dx - \left(x + 2 y^2\right) \; dy = 0$
i.e. $y \; \dfrac{dx}{dy} - \left(x + 2y^2\right) = 0$
i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} - 2y = 0$
i.e. $\dfrac{dx}{dy} - \dfrac{x}{y} = 2 y$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is a linear differential equation in the form $\;$ $\dfrac{dx}{dy} + P \left(y\right)x = Q \left(y\right)$
Here $P \left(y\right) = -\dfrac{1}{y}$ $\;\;\; \cdots \; (2a)$; $\;$ $Q \left(y\right) = 2y$ $\;\;\; \cdots \; (2b)$
Integrating Factor (I.F) $= e^{\int P \left(y\right) \; dy}$
i.e. $I.F = e^{\int -\frac{1}{y} \; dy} = e^{- \log \left|y\right|} = e^{\log \left|\frac{1}{y}\right|} = \dfrac{1}{y}$ $\;\;\; \cdots \; (3)$
According to the general solution,
$x \; e^{\int P \left(y\right) \; dy} = \displaystyle \int Q \left(y\right) \; e^{\int P \left(y\right) \; dy} \; dy$
$\therefore$ $\;$ In view of equations $(2a)$, $(2b)$ and $(3)$, the general solution of equation $(1)$ is
$\dfrac{x}{y} = \displaystyle \int 2 y \times \dfrac{1}{y} \; dy$
i.e. $\dfrac{x}{y} = 2 \displaystyle \int dy$
i.e. $\dfrac{x}{y} = 2y + c$ $\;\;\; \cdots \; (4)$
Equation $(4)$ is the general solution of the given differential equation.