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

Question

Luke Alvarez · Accepted Answer

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

# (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 #

Ezekiel Alexander · Answer

undefined To find the general solution of the given differential equation, we first check if it is exact. If not, we use an integrating factor to make it exact. Given equation is not exact. So, multiplying the entire equation by the integrating factor exp$$ \left(\int \frac{{M_y - N_x}}{N} dx \right)$$ we can obtain an exact differential equation. After that, we integrate with respect to x and y separately, and we will have the general solution.

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