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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, here's 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.
[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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