The general solution of the differential equation $ydx-(x+2y^2)dy = 0$

$x = 2y^2+ Cy$, Where C is constant of integration

The correct answer is Option (4) → $x = 2y^2+ Cy$, Where C is constant of integration

Given differential equation:

$y\,dx-(x+2y^{2})\,dy=0$

Write in the form:

$y\,dx=(x+2y^{2})\,dy$

Consider $x$ as a function of $y$.

$\frac{dx}{dy}=\frac{x+2y^{2}}{y}$

$\frac{dx}{dy}-\frac{1}{y}x=2y$

This is a linear equation in $x$ with respect to $y$.

Integrating factor:

$IF=e^{\int -\frac{1}{y}dy}=e^{-\log y}=\frac{1}{y}$

Multiply throughout by $\frac{1}{y}$:

$\frac{1}{y}\frac{dx}{dy}-\frac{1}{y^{2}}x=2$

$\frac{d}{dy}\left(\frac{x}{y}\right)=2$

Integrate:

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

$x=2y^{2}+Cy$

Final answer: $x=2y^{2}+Cy$