# Does Homogeneous = Separable?

by cepheid
Tags: homogeneous, separable
 Emeritus Sci Advisor PF Gold P: 5,196 Hi, Please bear with me, I've only had the first sort of "pseudo-lecture" in ordinary d.e.'s this past week, and I was doing some reading ahead. It occurred to me that if linear first-order differential equations are those that can be written in the general form: $$\frac{dy}{dx} + P(x)y = Q(x)$$ and if I understood my prof's remark correctly that homogeneous linear first-order d.e.'s are those for which $Q(x) = 0$, then all homogeneous first-order linear differential equations are actually separable because: $$\frac{dy}{dx} + P(x)y = 0$$ $$\frac{dy}{dx} = -P(x)y$$ Which can be solved as follows: $$\int{\frac{dy}{y}} = -\int{P(x)dx}$$ ^^Looks separable to me. I just wanted to make sure I was getting the basics before I moved on. So is the thread title statement true then? Oh yeah, and I'm guessing that the converse is not always true right? A separable first-order d.e. is not necessarily a homogeneous linear d.e. right? After all, $\frac{dy}{dx} = \frac{6x^2}{2y + cosy}$ is not linear in $y$ right? But it looks like it could be homogeneous non-linear (if such a thing exists, I don't know???). In that case, would the statement homogeneous = separable be true in the most general sense? Moving on with the original example (I'm just solving while "LaTeXing" to see what comes out) ... $$\ln|y| = -\int{P(x)dx}$$ $$|y| = e^{-\int {P(x)dx}}$$ Now, the most general solution for $y$ must include the most general antiderivative, so we'll have a $C$ stuck in there if and when we solve the integral: $$y = \pm e^{-\int {P(x)dx} + C} = \pm e^{-\int {P(x)dx}}e^C = \pm Ae^{-\int {P(x)dx}}$$ Whoah, cool! So in a homogeneous first order d.e. the solution takes the form of some constant $\pm A$ times the reciprocal of the integrating factor $I(x)$?! Is this always true? EDIT: Yeah actually I can see why that is. Take the usual statement about the integrating factor: $$(I(x)y)^{\prime} = I(x)Q(x)$$ If $Q(x) = 0$, then $$(I(x)y)^{\prime} = 0$$ $$I(x)y = \pm A$$ $$y = \frac{\pm A}{I(x)}$$ Awesome! I discovered a lot more writing this post than I expected. I hope someone will correct me if I've made any errors.
 Emeritus Sci Advisor PF Gold P: 5,196 Good point. In fact, I guess I could have gotten rid of the +/- way up here: $$\pm (e^{-\int {P(x)dx}})(e^C) = Ae^{-\int {P(x)dx}}$$ just combining everything into "A", which would remain as A for the ensuing discussion.