# Diff Eq thinger

1. May 10, 2006

### daveed

Hey...

So the question is as stated:

Show that
$$\frac{1} {M_x + N_y}$$, where $$M_x+N_y$$ is not identically zero, is an integrating factor of the homogeneous equation $$M(x, y)dx+N(x, y)dy=0$$ of degree n.

So I am not too sure where to go with this. I suppose what it's saying is, that I'm supposed to show that with the integrating factor, it's an exact equation, so differentiating the $$\frac{M(x, y)} {M_x + N_y}$$ term with respect to y should equal the value from differentiating $$\frac{N(x, y)} {M_x + N_y}$$ with respect to x...
but that doesn't work,
and I'm not sure what else will.

I'm looking at my book, which says that only sometimes will an integrating factor make an equation like this exact; however, it does say that $$M(x, y)dx+N(x, y)dy=0$$ has degree n. Does that imply that they are polynomial equations, and if so, how would this help me? Does anyone have any suggestions?

Last edited: May 10, 2006
2. May 11, 2006

### Tom Mattson

Staff Emeritus
Are you sure about the problem statement? Differential equations don't have a degree, they have an order.

3. May 12, 2006

### J77

I think the key is that the equation is homogeneous...

ie. $$\frac{dy}{dx}=\frac{-M(x,y)}{N(x,y)}$$

but because of the homogenity,

* rethinking *

The right-hand side must be a function of $$\frac{x}{y}$$ or $$\frac{y}{x}$$...

* probably best to ignore me on this one *

Last edited: May 12, 2006
4. May 12, 2006

### daveed

I asked my teacher about that part, and she says that it implies the functions $$M$$ and $$N$$ are algebraic functions of degree $$n$$
However, I don't know how to incorporate that into a proof(because such a strategy for finding an integrating factor wouldn't be true in general, so the being a polynomial part must be important)

J77,
I am not quite sure what you mean. The problem states that that thing IS the integrating factor, so I assumed that by multiplying it out and "showing the result is exact" would be sufficient.

Do you think that I could get it in the form of $$F(y/x)$$ in every form? Because if so... I guess I could attempt a proof. The problem is, that the very next problem says to use the process from this problem, and , when we try it, the integrating factor still doesn't make the expression exact(in the second problem).

5. May 12, 2006

### J77

yeah - I was a bit confused, sorry.

Sticking that solution back in, I end up with...

$$M_y(M_x+N_y)-M(M_{xy}+N_{yy})=N_x(M_x+N_y)-N(M_{xx}+N_{yx})$$

Last edited: May 12, 2006
6. May 12, 2006

### daveed

That's about where I stopped too... heh

7. May 12, 2006

### J77

If you multiply those brackets out, is there not a clear reversing of the product rule that comes out - my brain's in Friday meltdown... :tongue:

8. May 12, 2006

### HallsofIvy

Staff Emeritus
Saying that the d.e. is "homogeneous of degree n" means that if you replace both x and y by $\lambda x$ and $\lambda y$ the only effect is to multiply the equation by $\lambda^n$. That, itself, means that you can write the equation completely in terms of $\left(\frac{y}{x}\right)^n$ or that M(x,y) and N(x,y) involve only terms of the form $x^iy^j$ where i+ j= n.

9. May 15, 2006

### daveed

It's been a few days, and I still haven't made much headway on this. Our teacher gave us a hint, which was that we need to get it into the form
$$(NM_x - MN_x)x + (NM_y - MN_y)y$$

Which I can see equalling 0, by the definition given by HallsofIvy above, because
$$N(xM_x + yM_y) - M(xM_x + yN_y) = nNM - nNM = 0$$

Though, I'm at a loss as to how to get to that form...

Last edited: May 15, 2006