
First order ODE
xy' + y  2x = 0
We need to learn the method where you rearrange too
N(x,y)dy + M(x,y)dx = 0
Then solve somehow getting d(y2x)/dy = 1 etc...
Sorry I do not know the name of this method, but I really could do with a step by step sort of guide on how to do it for any example, as I can solve this in many other ways, but don't understand this particular method of solving 1st Order ODE's
Thanks.

I think you're talking about exact equations. See Chris's tutorial, post # 2 (scroll down a little bit).

I think you're right, thanks for the help.

An alternative:
$\displaystyle \displaystyle x\,\frac{dy}{dx} + y  2x = 0$
$\displaystyle \displaystyle x\,\frac{dy}{dx} + y = 2x$
$\displaystyle \displaystyle \frac{d}{dx}(x\,y) = 2x$
$\displaystyle \displaystyle x\,y = \int{2x\,dx}$
$\displaystyle \displaystyle x\,y = x^2 + C$
$\displaystyle \displaystyle y = x + \frac{C}{x}$.

Thanks for trying with the help prove it, I already have the Integrating Factor, Seperation of Variables and also the (y=uv) Method, just needed to learn this one for the course incase it appears on the exam :)
Thanks anyway tho, brilliant sig btw ! lol