# Need Help with Differential Equation (integrating factor)

by minger
Tags: differential, equation, factor, integrating
 Sci Advisor P: 1,497 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: $y^2+y-xy'=0$ I then divided by x (i prefer it this way), so $\frac{y^2+y}{x} + y' = 0$ then, letting u be a integration factor: $\frac{duM}{dy} = \frac{duN}{dx} \\\\\Rightarrow \frac{du(y^2+y)x^{-1}}{dy} = -\frac{du}{dx}$ $(\frac{du}{dy})(\frac{y^2+y}{x}) + u\frac{2y+1}{x} = -\frac{du}{dx}$ Assuming that u is a function of only x, we get rid of the u partial with y term, so: $\frac{du}{dx} = -u\frac{2y+1}{x} \\\mbox{now doing some dividing on both sides and integrating} \\ \int\frac{\delta\mu}{\mu} = -\int\frac{2y+1}{x}dx$ 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.
 P: 97 Why not just make the case separable? Seems to me a lot easier this way $$y^2+y - x y^{\prime} = 0 \Rightarrow \frac{y^{\prime}}{y^2+y}=\frac{1}{x} \Rightarrow \frac{dy}{y^2+y} = \frac{dx}{x}$$ and this is an easy integral to do.
 PF Patron Sci Advisor Thanks Emeritus P: 38,429 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 $$\frac{1}{x(y^2+ y)}$$ is an integrating factor.)