# Coveting ODE to polar coordinates

1. Sep 23, 2007

### gtabmx

Hi, I was wondering how to go about converting a homogeneous ODE of the form M(x,y)dx+N(x,y)dy=0 (where, by definition of a homogeneous ODE, M(tx,ty)=(t^a)M(x,y) and N(tx,ty)=(t^a)N(x,y) ) to polar coordinates. I wan to do this because using substitution of y/x=u and dy/dx=u+xdu/dx to make the ODE separable does not always result in the easiest integration towards the final steps. I figure by making x=rcosθ and y=rsinθ, I can completely isolate and remove r and make the ODE separable in terms of r and θ. I am completely able to convert half of the equation, but have very little idea how to transform dy/dx into something in terms of r and θ. Can anyone explain a little on how to to this?

Thanks,
Mike

2. Sep 23, 2007

### HallsofIvy

You are "coveting" a differential equation? Oh, you shouldn't do that!

You convert from one variable to another (one coordinate system to another) by using the chain rule.

Since $y= r sin(\theta)$, $dy= sin(\theta)dr+ r cos(\theta)d\theta$ and since $x= r cos(\theta)$, $dx= cos(\theta)dr- r sin(\theta)d\theta$. Also, of course, convert x and y in M(x,y) and N(x,y) into r and $\theta$.

I am wondering why you think converting to polar coordinates will allow you to "completely isolate and remove r" without giving specific M and N. Obviously whether you can, in fact, remove r, depends on what M and N are.

3. Sep 23, 2007

### gtabmx

In the case I'm working with, r will be canceled out, obviously except for the r which has come from the product rule when obtaining the conversion of dx and dy. If you would like to see a particular case then I would be more than glad to offer one.

(2*y^4 - 9*x^2*y^2-20*x^4)dx - (3*x*y^3)dy = 0

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