# Proof for a problem (ordinary differential equations)

1. May 22, 2006

### jahz

Gah, I've tried to solve this so many times, but I can't get on the right track, so I need help, naturally.

I want to show that 1/(dM/dx + dN/dy), where dM/dx + dN/dy is not identically zero, is an integrating factor for the homogeneous equation M(x,y)dx + N(x,y)dy = 0 of degree n.

I've done the obvious, which is multiplying the equation by the integrating factor and then differentiating parts of it to check for an exact equation.

So far, what I have is

((dM/dy)*(dM/dx + dN/dy) - d(dM/dx + dN/dy)/dy*M)/((dM/dx + dN/dy)^2) - ((dN/dx)*(dM/dx + dN/dy) - d(dM/dx + dN/dy)/dx*N)/((dM/dx + dN/dy)^2), and I think I should try to show that this is equal to zero and therefore is an exact solution.

The problem is, I don't know where to go from here. I've gotten two hints from my teacher, that is:

1) "Homogeneous equations of the degree n" means that

M(lambda*x, lambda*y) = lambda^n * N(x, y) and
N(lambda*x, lambda*y) = lambda^n * M(x, y)

2)

x(n*dM/dx - M*dN/dx) + y(N*dM/dy - M*dN/dy) turns into
N(x*dM/dx + y*dM/dy) - M(x*dN/dx + y*dN/dy), which in turn becomes
N*mu*M- M*mu*N, which = 0 and therefore shows that the equation is exact. mu, I assume, is the integrating factor, though I can't figure out how the second-to-last equation turns into the last one.

Anyway, I've got no idea how to use these two hints. For 2), I 've tried and failed to get the equation into the form specified in the first equation. As for 1), I've tried using that to get all M or N on one side of the equation, but when differentiating, I have to use the chain rule, so I ended up with incompatible differentials such as dM/du + dM/dy, where u = lambda * x.

So, any help would be greatly appreciated!

Edit: And sorry for not using the fancy latex thingys that give nice-looking math equations... I can't figure out how to do it, though that's not my main problem right now!

Last edited: May 22, 2006