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 March 26th, 2011, 01:24 AM #1 Member   Joined: Jul 2010 Posts: 34 Thanks: 0 Another Inexact ODE Okay so I have another working check (2x+y^2)dx+ 4xy dy = 0 its not exact and we get an integrating factor of 1/sqrt(x) u = integrate N(x,y) dy + g(x) =2*sqrt(x)*y^2 + g(x) also partial derivative du/dx = M(x,y) = (2x+y^2)/sqrt(x) and also = y^2/sqrt(x)+g'(x) so g'(x) = 2x and g(x) = 2x^2/2 = x^2 so implicit solution is 2*sqrt(x)*y^2+x^2 Can someone please confirm or deny.
 March 26th, 2011, 05:40 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Another Inexact ODE You have: $$$2x+y^2$$dx+$$4xy$$dy=0$ $\frac{\delta M}{\delta y}=2y$ $\frac{\delta N}{\delta x}=4y$ You are right, this is not exact, so we consider: $\frac{\frac{\delta M}{\delta y}-\frac{\delta N}{\delta x}}{N}=\frac{2y-4y}{4xy}=-\frac{1}{2x}$ Since this is a function of just x, we obtain an integrating factor by: $\mu(x)=e^{-\frac{1}{2}\int\frac{1}{x}\,dx}=\frac{1}{\sqrt{x}}$ This gives the exact equation: $$$\frac{2x+y^2}{\sqrt{x}}$$dx+$$4\sqrt{x}y$$dy=0$ thus: $\frac{\delta F}{\delta x}=\frac{2x+y^2}{\sqrt{x}}$ Integrating with respect to x, we get: $F(x,y)=\int\frac{2x+y^2}{\sqrt{x}}\,dx+g(y)=\frac{ 4}{3}x^{\frac{3}{2}}+2y^2x^{\frac{1}{2}}+g(y)$ To determine g(y), we take the partial derivative with respect to y and substitute N for $\frac{\delta F}{\delta y}$: $4\sqrt{x}y=4y\sqrt{x}+g#39;(y)\:\therefore\:g(y) = C$ thus the solution is given implicitly by: $\frac{4}{3}x^{\frac{3}{2}}+2y^2x^{\frac{1}{2}}=C$

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