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 January 16th, 2014, 04:56 PM #1 Newbie   Joined: Jan 2014 Posts: 4 Thanks: 0 Inexact differential how do you solve this inexact differential. dz=xdx+(x+2y)dy. I get e^(y/x) as the integrating factor. Is it correct? help would be appreciated!
 January 17th, 2014, 04:51 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Inexact differential Well, if $e^{y/x}$ is an integrating factor then $e^{y/x}xdx+ e^{y/x}(x+ 2y)dy$ would be an "exact" differential, meaning that there is some function, F, of x and y, such that $dF= e^{y/x}xdx+ e^{y/x}(x+ 2y)dy$ which would mean that $\frac{\partial F}{\partial x}= e^{y/x}x$ and $\frac{\partial F}{\partial y}= e^{y/x}(x+ 2y)$. That would mean that the two mixed second derivatives $\frac{\partial^2 F}{\partial x\partial y}= \frac{1}{x}e^{y/x}x= e^{y/x}$ and $\frac{\partial^2 F}{\partial y\partial x}= -\frac{y/x^2}e^{y/x}(x+ 2y)+ e^{y/x}= e^{y/x}\left(1- \frac{y}{x}- 2y^2}{x^2}\right)$ must be equal. But they are not so that is NOT an exact differential so $e^{y/x}$ is NOT an integrating factor. However, I don't know what you mean by "solve $dz= xdx+ (x+ 2y)dy$" or even by $dz= xdx+ (x+ 2y)dy$" itself! Because $xdx+ (x+ 2y)dy$ is not an exact differential, there is NO function, z, such that dz is equal to it. So what is the problem you are trying to solve, really? If it were $xdx+ (x+ 2y)dy= 0$ that would make sense because you could multiply by the integrating factor on both sides and the right side would still be 0.
January 17th, 2014, 06:39 AM   #3
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Re: Inexact differential

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 Originally Posted by HallsofIvy Well, if $e^{y/x}$ is an integrating factor then $e^{y/x}xdx+ e^{y/x}(x+ 2y)dy$ would be an "exact" differential, meaning that there is some function, F, of x and y, such that $dF= e^{y/x}xdx+ e^{y/x}(x+ 2y)dy$ which would mean that $\frac{\partial F}{\partial x}= e^{y/x}x$ and $\frac{\partial F}{\partial y}= e^{y/x}(x+ 2y)$. That would mean that the two mixed second derivatives $\frac{\partial^2 F}{\partial x\partial y}= \frac{1}{x}e^{y/x}x= e^{y/x}$ and $\frac{\partial^2 F}{\partial y\partial x}= -\frac{y/x^2}e^{y/x}(x+ 2y)+ e^{y/x}= e^{y/x}\left(1- \frac{y}{x}- 2y^2}{x^2}\right)$ must be equal. But they are not so that is NOT an exact differential so $e^{y/x}$ is NOT an integrating factor. However, I don't know what you mean by "solve $dz= xdx+ (x+ 2y)dy$" or even by $dz= xdx+ (x+ 2y)dy$" itself! Because $xdx+ (x+ 2y)dy$ is not an exact differential, there is NO function, z, such that dz is equal to it. So what is the problem you are trying to solve, really? If it were $xdx+ (x+ 2y)dy= 0$ that would make sense because you could multiply by the integrating factor on both sides and the right side would still be 0.

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