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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.
thanks your your reply. so, whats the integrating factor for xdx+ (x+ 2y)dy= 0?

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