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 March 19th, 2011, 09:07 PM #1 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 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. March 20th, 2011, 06:05 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 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 March 20th, 2011, 09:38 PM #3 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 Re: Intergrating factor for inexact ODE I followed that right up until the multiplying by 3/2 part. Can you please explain that bit? March 21st, 2011, 02:36 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Intergrating factor for inexact ODE Integrating and multiplying by 3/2 (replaced the constant) Therefore Tags factor, inexact, intergrating, ode ,

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