The general solution of the differential equation ydx+(x-y^3)dy=0 is :

The correct answer is Option (3) → \( xy=\displaystyle\frac{y^4}{4}+c \): c is an arbitrary constant.

The equation is linear in x when x is taken as the dependent variable. Dividing by y dy,

    \( \displaystyle \frac{dx}{dy}+\frac{x}{y}=y^2 \)

Here \( P=\frac{1}{y} \), so the integrating factor is \( e^{\int\frac{1}{y}dy}=y \).

    \( \displaystyle x\cdot y=\int y^2\cdot y\,dy \)

    \( \displaystyle xy=\frac{y^4}{4}+c \)

Shortcut: The equation can be written as \( y\,dx+x\,dy=y^3\,dy \), i.e. \( d(xy)=y^3\,dy \). Integrating gives \( xy=\frac{y^4}{4}+c \) at once.