Proof for a problem (ordinary differential equations)

[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!

 

[lippyka]

It looks like you are trying to use the second hint to prove that 1/(dM/dx + dN/dy) is the integrating factor for the equation. To do this, you need to show that multiplying the equation by the integrating factor and then differentiating it yields an exact equation. To start with, you can multiply the equation M(x,y)dx + N(x,y)dy = 0 by the integrating factor 1/(dM/dx + dN/dy) to getM(x,y)/(dM/dx + dN/dy) dx + N(x,y)/(dM/dx + dN/dy) dy = 0. Now, you can differentiate this equation using the product rule and the chain rule to get (dM/dx)(1/(dM/dx + dN/dy)) dx + (dN/dx)(1/(dM/dx + dN/dy)) dx + (M(x,y)/(dM/dx + dN/dy))(-(d^2M/dx^2 + d^2N/dxdy)) dx - (dM/dy)(1/(dM/dx + dN/dy)) dy - (dN/dy)(1/(dM/dx + dN/dy)) dy - (N(x,y)/(dM/dx + dN/dy))(-(d^2M/dy^2 + d^2N/dydx)) dy = 0. Now, you can rearrange this equation to the form P(x,y)dx + Q(x,y)dy = 0 where P(x,y) and Q(x,y) are polynomials of degree n. Then, you can use the hint about homogeneous equations of degree n to show that P(lambda*x, lambda*y) = lambda^n*Q(x,y) and Q(lambda*x, lambda*y) = lambda^n*P(x,y). This implies that the equation is exact, and therefore 1/(dM/dx + dN/dy) is the integrating factor for the equation