What is the integrating factor for #0 = (3x^2 + 3y^2)dx + x(x^2 + 3y + 6y)dy#?

Question

Paisley Johnson · Accepted Answer

I got: #mu(y) = e^y# The point of an integrating factor is to turn an inexact differential into an exact one. One physical application of this is to turn a path function into a state function in chemistry (such as dividing by #T# to turn #q_"rev"#, a path function, into #S#, a state function, entropy). I assume that the second terms include #3y^2#, not #3y# (it would be odd to not simply write #9y#). Two options to find the special integrating factor, as defined by Nagle, are: #bb(mu(x) = "exp"[int (((delM)/(dely))_x - ((delN)/(delx))_y)/(N(x,y))dx])#, #bb(mu(y) = "exp"[int (((delN)/(delx))_y - ((delM)/(dely))_x)/(M(x,y))dy])#, for the differential #bb(dF(x,y) = ((delF)/(delx))_y dx + ((delF)/(dely))_xdy)#, where #M = ((delF)/(delx))_y# and #N = ((delF)/(dely))_x#.  For now, let's find the partial derivatives. For your differential: #color(green)(M(x,y) = 3x^2 + 3y^2)#
#color(green)(N(x,y) = x^3 + 3xy^2 + 6xy)# Therefore: #color(green)(((delM)/(dely))_x = 6y)# #color(green)(((delN)/(delx))_y = 3x^2 + 3y^2 + 6y)# which are clearly not equal, so the current differential is inexact. So, let us divide by #N(x,y)# and see if the integral with respect to #x# is reasonable to do: #lnmu(x) = int (6y - 3x^2 - 3y^2 - 6y)/(x^3 + 3xy^2 + 6xy)dx# #= int (-3x^2 - 3y^2)/(x^3 + 3xy^2 + 6xy)dx# This doesn't look all that nice (it cannot be readily factored to eliminate #y# terms), so what if we try integrating with respect to #y# instead, and dividing by #M(x,y)# instead? Then: #mu(y) = "exp"[int (((delN)/(delx))_y - ((delM)/(dely))_x)/(M(x,y))dy]# and: #lnmu(y) = int(3x^2 + 3y^2 + 6y - 6y)/(3x^2 + 3y^2)dy# #= intcancel((3x^2 + 3y^2)/(3x^2 + 3y^2))^(1)dy# And this integral is easy. It's just #y#. Therefore, #color(blue)(mu(y) = e^y)# would be your integrating factor. Let's test it out: #3e^y(x^2 + y^2)dx + xe^y(x^2 + 3y^2 + 6y)dy = 0# #(3x^2e^y + 3y^2e^y)dx + (x^3e^y + 3xy^2e^y + 6xye^y)dy = 0# Checking for exactness, we obtain: #((delM)/(dely))_x stackrel(?" ")(=) ((delN)/(delx))_y# #3x^2e^y + 3(y^2e^y + 2ye^y) stackrel(?" ")(=) 3x^2e^y + 3y^2e^y  + 6ye^y# #3x^2e^y + 3y^2e^y + 6ye^y = 3x^2e^y + 3y^2e^y  + 6ye^y# #color(blue)(sqrt"")# so we know our integrating factor is correct!

Emma Aldridge · Answer

undefined The integrating factor for the given differential equation $0 = (3x^2 + 3y^2)dx + x(x^2 + 3y + 6y)dy$ is $e^{\int P(x) \, dx}$, where $P(x)$ is the coefficient of $dy$ after dividing the equation by $dx$. In this case, $P(x) = \frac{x(x^2 + 3y + 6y)}{3x^2 + 3y^2}$.

Ezra Adams · Answer

undefined The integrating factor for the given first-order linear ordinary differential equation $0 = (3x^2 + 3y^2)dx + x(x^2 + 3y + 6y)dy$ is $\mu(x, y) = e^{\int P(x)dx}$, where $P(x)$ is the coefficient of $dy$ after rearranging the equation into the standard form $M(x, y)dx + N(x, y)dy = 0$. In this case, $P(x) = \frac{dN}{dy} - \frac{dM}{dx}$. After calculation, we find $P(x) = 1$, so the integrating factor is $\mu(x, y) = e^{\int dx} = e^x$.