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:(adsbygoogle = window.adsbygoogle || []).push({});

[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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# Homework Help: Need Help with Differential Equation (integrating factor)

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