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March 19th, 2011, 09:07 PM   #1
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Intergrating factor for inexact ODE

I wrote this out in Wolfram so excuse the poor annotation.
I'm trying to find the integrating factor for the in exact equation below.

(2 x + y^2) dx + 4xy dy =0

[Delta]M/[Delta]y = 2 x + y^2 dx
= 2 y
[Delta]N/[Delta]x = 4 x y dy
= 4 y
[Delta]M/[Delta]y != [Delta]N/[Delta]x Therefore equation is not exact
So to find intergrating factor
1/N[ [Delta]M/[Delta]y - [Delta]N/[Delta]x] = h (x)
= 1/4 xy [2 y - 4 y]
= -2/4 x is a function of h (x)
so intergrating factor = exp^[Integral]h (x)
= 1/[Sqrt]x
(I think I'm going wrong somewhere around here)
[Delta]M/[Delta]y = (2 x + y^2)/(1/[Sqrt]x) dx
= 2 y/([Sqrt]x)
[Delta]N/[Delta]x = 4 x y dy
!= 2 y/(\[Sqrt]x)
So therefore intergrating factor is not correct
Help.
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March 20th, 2011, 06:05 AM   #2
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Re: Intergrating factor for inexact ODE

and

So is your integrating factor.



Integrating the dx term with respect to x:
Integrating the dy term with respect to y:

Hence

Integrating and multiplying by 3/2 you get
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March 20th, 2011, 09:38 PM   #3
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Re: Intergrating factor for inexact ODE

I followed that right up until the multiplying by 3/2 part. Can you please explain that bit?
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March 21st, 2011, 02:36 AM   #4
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Re: Intergrating factor for inexact ODE



Integrating


and multiplying by 3/2

(replaced the constant)

Therefore
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