Q.430. Homogeneous differential equation.

Question 430.
Solve and get the general solution of the differential equation:
xy dx - (x^2 + 3y^2) dy = 0.

Answer 430.
xy dx - (x^2 + 3y^2) dy = 0
=> dy/dx = xy/(x^2 + 3y^2)
This is a homogeneous differential eqn. of the first order.
Let y = vx
=> dy/dx = v + x dv/dx
=> v + x dv/dx = vx^2 / (x^2 + 3v^2x^2)
=> x dv/dx = v/(1 + 3v^2) - v
=> x dv/dx = (v - v - 3v^3) / (1 + 3v^2)
=> x dv/dx = - 3v^3 / (1 + 3v^2)
=> (1 + 3v^2) / 3v^3 dv + dx/x = 0
=> (1/3) ∫ dv/v^3 + ∫ dv/v + ∫ dx/x = 0
=> - (1/6v^2) + lnv + lnx = lnc
=> - (x^2/6y^2) + lny = lnc
=> ln(y/c) = x^2/6y^2
=> y = c e^(x^2/6y^2).

Link to YA!