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

##### 1 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

I assume that the second terms include

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

#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

#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

#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!