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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:

[tex] \frac{dy}{dx} + P(x)y = Q(x) [/tex]

and if I understood my prof's remark correctly thathomogeneouslinear first-order d.e.'s are those for which [itex] Q(x) = 0 [/itex], then all homogeneous first-order linear differential equations are actuallyseparablebecause:

[tex] \frac{dy}{dx} + P(x)y = 0 [/tex]

[tex] \frac{dy}{dx} = -P(x)y [/tex]

Which can be solved as follows:

[tex] \int{\frac{dy}{y}} = -\int{P(x)dx} [/tex]

^^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 theconverseis not always true right? A separable first-order d.e. is not necessarily a homogeneous linear d.e. right? After all, [itex] \frac{dy}{dx} = \frac{6x^2}{2y + cosy} [/itex] is not linear in [itex] y [/itex] 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) ...

[tex] \ln|y| = -\int{P(x)dx} [/tex]

[tex] |y| = e^{-\int {P(x)dx}} [/tex]

Now, the most general solution for [itex] y [/itex] must include the most general antiderivative, so we'll have a [itex] C [/itex] stuck in there if and when we solve the integral:

[tex] y = \pm e^{-\int {P(x)dx} + C} = \pm e^{-\int {P(x)dx}}e^C = \pm Ae^{-\int {P(x)dx}} [/tex]

Whoah, cool! So in a homogeneous first order d.e. the solution takes the form of some constant [itex] \pm A [/itex] times thereciprocalof the integrating factor [itex] I(x) [/itex]?! Is this always true?

EDIT: Yeah actually I can see why that is. Take the usual statement about the integrating factor:

[tex] (I(x)y)^{\prime} = I(x)Q(x) [/tex]

If [itex] Q(x) = 0 [/itex], then

[tex] (I(x)y)^{\prime} = 0 [/tex]

[tex] I(x)y = \pm A [/tex]

[tex] y = \frac{\pm A}{I(x)} [/tex]

Awesome! I discovered a lot more writing this post than I expected. I hope someone will correct me if I've made any errors.

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# Does Homogeneous = Separable?

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