Which of the following is a homogeneous differential equation ?

The correct answer is Option (3) → \( x^2\displaystyle\frac{dy}{dx}=xy-2x^2+y^2 \)

A differential equation \( \frac{dy}{dx}=F(x,y) \) is homogeneous when F(x, y) is a homogeneous function of degree zero, that is, \( F(\lambda x,\lambda y)=F(x,y) \); equivalently F can be written as a function of \( \frac{y}{x} \).

For Option (3):

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

    \( \displaystyle =\frac{y}{x}-2+\left(\frac{y}{x}\right)^2 \)

This is a function of \( \frac{y}{x} \) only, so the equation is homogeneous. (Every term of \( xy-2x^2+y^2 \) has degree 2, the same as \( x^2 \).)

Why the other options are incorrect

• Option (1) \( y\,dx+(x-y^2)dy=0 \) — Incorrect. \( \frac{dy}{dx}=\frac{-y}{x-y^2} \); the terms x and \( y^2 \) have different degrees, so F is not homogeneous.

• Option (2) \( \displaystyle\frac{dy}{dx}+2y\tan x=2x \) — Incorrect. F(x, y) = 2x − 2y tan x contains tan x, which cannot be written as a function of \( \frac{y}{x} \) (this is a linear equation).

• Option (4) \( (x^3-y^3)\,dx+3xy\,dy=0 \) — Incorrect. \( x^3-y^3 \) has degree 3 but \( 3xy \) has degree 2, so \( \frac{dy}{dx}=\frac{y^3-x^3}{3xy} \) is homogeneous of degree 1, not 0.