y×ydx+(3xy-1)dy=0

Question

y×y=y2

Answer 1

y2 dx = (1 - 3xy) dy

dx/dy = 1/(y2) - 3x/y

dx/dy + 3x/y = 1/y2

This is a first order linear differential equation of the form x' + P(y)x = Q(y) where the integrating factor = e(∫P(y) dy) = e(∫3/y dy) = y3

Multiplying both sides by integrating factor:

y3(dx/dy) + 3y2x = y

Note that the left hand side equals d/dy[x*y^3], so: d/dy[xy3] = y

Integrating both sides:

xy3 = (y2)/2 + C

x = 1/(2y) + C/(y3)