Question

write the ordinary differential equation y dx + ( xy + x - 3y ) dy = 0 in the linear form, and hence find its solution.

Answer 1

ODE Ordinary differential equation of 1st order 

y dx + (xy + x - 3y) dy = 0

(y dx + x dy) + xy  dy  - 3 y dy = 0

Let  xy = u ,    x dy + y dx = du

So    du + u dy - 3 y dy = 0
        du / dy + u = 3 y              ---- Linear form of ODE

This ODE is in the form of   u' + p(y) u = q(y)
    Function   p(y) = 1 ,   and  q(y) = 3 y
    Integrating factor:   μ(y) = exp{Integral p(y) dy } = e^y 
    So the solution:   u(y) = 1/μ(y) * [ integral μ(y) q(y) dy  + C]

=>  u(y) = e^(-y) * [ integral 3 y e^y  + C ]
              = e^(-y) * (3 y e^y - 3 e^y + C ]
              = 3 y - 3 + C e^(-y)             

=>   x y  = 3 y - 3  + C e^(-y)
        x - 3 = -3 / y + C /[ y e^y] 
        x - 3 = [ C - 3 e^y ] / [ y e^y ]