What is the general solution of the differential equation ? # (6xy - 3y^2+2y) dx + 2(x-y)dy = 0 #

# (6xy - 3y^2+2y) dx + 2(x-y)dy = 0 ..... [A]#

Suppose we have:

# M(x,y) dx = N(x,y) dy #

Then the DE is exact if #M_y-N_x=0#

# M = 6xy - 3y^2+2y => M_y = 6x-6y+2 #
# N= 2(x-y) => N_x = 2 #

# M_y - N_x != 0 => # Not an exact DE

So, we seek an Integrating Factor #mu(u)# such that

# (muM)_y = (muN)_x#

So, we compute::

# (M_y-N_x)/N = (6x-6y+2 - 2)/(2(x-y)) = 3 #

So the Integrating Factor is given by:

# mu(x) = e^(int \ 3 \ dx) #
# \ \ \ \ \ \ \ = e^(3x) #

So when we multiply the DE [A] by the IF we now get an exact equation:

# (6xy - 3y^2+2y)e^(3x) dx + 2(x-y)e^(3x)dy = 0 #

And so if we redefine the function #M# and #N#:

# M = (6xy - 3y^2+2y)e^(3x) #
# N = 2(x-y)e^(3x)#

Then, our solution is given by:

# f_x = M# and #f_y = N# and

If we consider #f_y = N#, then:

# f = int \ 2(x-y)e^(3x) \ dy + g(x) #, where we treat #x# as constant
# \ \ = 2e^(3x) \ int x-y \ dy + g(x) #
# \ \ = 2e^(3x) (xy-1/2y^2) + g(x) #
# \ \ = e^(3x) (2xy-y^2) + g(x) #

And now we consider #f_x = M# and we differentiate the last result to get:

# f_x = (e^(3x))(2y) + (3e^(3x))(2xy-y^2) + g'(x) #
# \ \ \ = (2y + 6xy-3y^2 + g'(x))e^(3x) #

As #f_x=M# then we have:

# 6xy - 3y^2+2y = 2y + 6xy-3y^2 + g'(x) #
# :. g'(x) = 0 => g(x) = K#

Leading to the GS:

# e^(3x) (2xy-y^2) = C #