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Need Help with Differential Equation (integrating factor)

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minger
#1
Sep5-06, 12:48 PM
Sci Advisor
P: 1,498
Hi there, I am having a bit of difficulty finding the integration factor for the following problem. The problem lies in taking the integral of a function of two variables. Anyways, heres what I have:

[itex] y^2+y-xy'=0[/itex]
I then divided by x (i prefer it this way), so

[itex] \frac{y^2+y}{x} + y' = 0 [/itex]

then, letting u be a integration factor:

[itex] \frac{duM}{dy} = \frac{duN}{dx}
\\\\\Rightarrow
\frac{du(y^2+y)x^{-1}}{dy} = -\frac{du}{dx}
[/itex]

[itex]
(\frac{du}{dy})(\frac{y^2+y}{x}) + u\frac{2y+1}{x} = -\frac{du}{dx}
[/itex]
Assuming that u is a function of only x, we get rid of the u partial with y term, so:
[itex]
\frac{du}{dx} = -u\frac{2y+1}{x}
\\\mbox{now doing some dividing on both sides and integrating}
\\
\int\frac{\delta\mu}{\mu} = -\int\frac{2y+1}{x}dx
[/itex]

Now this is where I'm stuck. I guess I can't remember how to integrate that right side. Can I just pull the (2y+1) term out of the integral assuming that its contant? Thanks for the help

edit: wow, this latex is really messed up. I'm working on it. The last equation, the integral is correct, and that's the most important part though.
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xman
#2
Sep5-06, 09:59 PM
P: 97
Why not just make the case separable? Seems to me a lot easier this way
[tex] y^2+y - x y^{\prime} = 0 \Rightarrow \frac{y^{\prime}}{y^2+y}=\frac{1}{x} \Rightarrow \frac{dy}{y^2+y} = \frac{dx}{x} [/tex]
and this is an easy integral to do.
HallsofIvy
#3
Sep6-06, 04:55 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,353
You can't integrate that last integral precisely because in has y in it. Your assumption that the integrating factor u could be a function of x only is incorrect.

As xman pointed out, this is obviously a separable equation. (Which is the same as saying that
[tex]\frac{1}{x(y^2+ y)}[/tex]
is an integrating factor.)

minger
#4
Sep6-06, 09:50 AM
Sci Advisor
P: 1,498
Need Help with Differential Equation (integrating factor)

wow, I cannot believe that I missed that. It was in the integrating factor section, so I just assumed it to be of that form.

I feel like an idiot, thanks a lot.
arildno
#5
Sep6-06, 09:55 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Note that usually, non-linear diff.eqs rarely can be solved by methods of finding integrating factors.
That is, the method of finding an integrating factor is mainly of use for solving linear diff. eqs.

On special occasions, though, you may succeed..


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