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 March 19th, 2011, 08:07 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, 05:05 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Intergrating factor for inexact ODE $\frac{2y-4y}{2xy}=\frac{-2y}{4xy}=-\frac{1}{2x}$ and $-\frac12\int\frac1x\mathrm{d}x= -\frac12\log x$ So $e^h= e^{-\frac12\log x} = x^{-\frac12}=\frac{1}{\sqrt{x}}$ is your integrating factor. $(2\sqrt{x}+y^2x^{-1/2})\mathrm{d}x+(4\sqrt{x}y)\mathrm{d}y= 0$ Integrating the dx term with respect to x: $\frac43x\sqrt{x}+2y^2\sqrt{x}+f(y)$ Integrating the dy term with respect to y: $2y^2\sqrt{x}+g(x)$ Hence $\int\mathrm{d}(\frac43x\sqrt{x}+2y^2\sqrt{x})= \int 0$ Integrating and multiplying by 3/2 you get $2x\sqrt{x}+3y^2\sqrt{x}= c$
 March 20th, 2011, 08:38 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, 01:36 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Intergrating factor for inexact ODE $\int\mathrm{d}(\frac43x\sqrt{x}+2y^2\sqrt{x})= \int 0$ Integrating $\frac43x\sqrt{x}+2y^2\sqrt{x}= k$ and multiplying by 3/2 $2x\sqrt{x}+3y^2\sqrt{x}= \frac32k = c$ (replaced the constant) Therefore $y^2= \frac{c}{3\sqrt{x}}-\frac23x$

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