Question

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

Answer 1

I got:

`mu(y) = e^y`

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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!