First order ODE

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December 8th 2010, 12:14 PM
Monkens

First order ODE

xy' + y - 2x = 0

We need to learn the method where you re-arrange too

N(x,y)dy + M(x,y)dx = 0

Then solve somehow getting d(y-2x)/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.
December 8th 2010, 01:07 PM
Ackbeet

I think you're talking about exact equations. See Chris's tutorial, post # 2 (scroll down a little bit).
December 10th 2010, 02:59 AM
Monkens

I think you're right, thanks for the help.
December 10th 2010, 03:04 AM
Prove It

An alternative:

$\displaystyle x\,\frac{dy}{dx} + y - 2x = 0$

$\displaystyle x\,\frac{dy}{dx} + y = 2x$

$\displaystyle \frac{d}{dx}(x\,y) = 2x$

$\displaystyle x\,y = \int{2x\,dx}$

$\displaystyle x\,y = x^2 + C$

$\displaystyle y = x + \frac{C}{x}$ .
December 10th 2010, 03:07 AM
Monkens

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

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