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 engininja March 26th, 2011 01:24 AM

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.

 MarkFL March 26th, 2011 05:40 AM

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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