15
$\begingroup$

I'm not finding any resource or description or systemic methodology to find integrating factors when the integrating factor will be a function of both x and y.

I'm on this problem,

$$ ( y - xy^2 ) dx + (x + x^2y^2) dy = 0 $$

In its present form its not exact, but multiplying through by an integrating factor $ \frac{1}{x^2y^2} $ does make it exact.

Now I want to know how you find that, when the integrating factor is both a function of x and y.

The technique on this page for finding an integrating factor $u(x)$ fails.

Which means there's no function of purely x that will make this equation exact.

So I'm trying something like this:

1. Multiply the original equation by a function u(x,y):

$$ u(x,y)( y - xy^2 )dx + u(x,y)(x + x^2y^2)dy = 0 $$

2. Attempt to set $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $

$$ \frac{\partial}{\partial y} u(x,y)( y - xy^2 ) = \frac{\partial}{\partial x} u(x,y)(x + x^2y^2) $$

(The two sides must be equal for u(x,y) to make the original equation exact, since $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $ for an exact equation)

3. Now you have:

$$ u_y(x,y)(y-xy^2) + u(x,y)(1-2xy) = u_x(x,y)(x+x^2y^2) + u(x,y)(1+2xy^2) $$

I'm kind of stuck now. What is the next step, or is this an incorrect start?

  • 0
    Usually when trying to solve first order PDEs like the one you have in step 3, one would use the method of characteristics. However, it seems that this doesn't give a tractable problem in your case; your equation has a particular solution $u(x,y)=(xy)^{-2}$ which appears to be much simpler than the general solution. If someone knows a systematic procedure for finding this particular solution, I would be interested in learning about it too.2011-05-07
  • 0
    @HansLundmark I have added a recent answer to this question. :)2012-05-01
  • 0
    @Gerry Myerson any reason why the differential geometry tag was not included? I think it's definitely an appropriate tag considering my answer. Cheers2012-05-01
  • 0
    I removed it because there was no differential geometry in the statement of the question, and because students meeting integrating factors for the first time are unlikely to know anything about 1-forms, integral manifolds, etc., etc. But I take your point, and I'll put the tag back.2012-05-01
  • 0
    @GerryMyerson Ah, I understand your point as well. Thanks!2012-05-01

2 Answers 2