My Math Forum Integrating factor question

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 March 14th, 2012, 05:28 AM #1 Newbie   Joined: Mar 2012 Posts: 12 Thanks: 0 Integrating factor question Hey, I just wanted to double check if my logic is correct for solving this question i've been working on, The question states for the exact equation, $p(x,y)dx+q(x,y)dy=0$ $if\,\,S=\frac{\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}}{xp-yq}\,\,\,\,\,is\,\,\,a\,\,function\,\,of\,\,n,\,\, where\,\,n=xy\,\, \\ \text{then the equation has an integrating factor u as a function of xy such that} \\ \frac{d\mu }{dn}=S(n)\mu \\$ So first I set u(x,y)=z some arbitrary function then $\frac{\partial \mu }{\partial x}=\frac{d\mu }{dz}\frac{\partial z}{\partial x}\,\,\,\,\,and\,\,\,\,\frac{\partial \mu }{\partial y}=\frac{d\mu }{dz}\frac{\partial z}{\partial y}\$ after going through the steps to satisfy an exact equation, I end up with $\frac{d\mu }{dz}=\frac{\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}}{p\frac{\partial z}{\partial y}-q\frac{\partial z}{\partial x}}\mu$ so is it enough to say that by comparison with S, \begin{align} & \frac{\partial z}{\partial y}=x\,\,\,\,\,and\,\,\,\,\frac{\partial z}{\partial x}=y \\ & so\,\,z=xy=n \\ \end{align} so u is a function of xy. Does that seem like the right approach to solve this question? Thanks in advance.
 March 16th, 2012, 08:10 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 That's not quite right. Both S and ? should be functions of n only, where n = xy, rather than arbitrary functions of x and y. If p and q happen to be polynomials, see also the approach I sketched here.

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