Does Homogeneous = Separable?

by cepheid
Tags: homogeneous, separable
cepheid is offline
Sep12-04, 12:47 PM
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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 that homogeneous linear first-order d.e.'s are those for which [itex] Q(x) = 0 [/itex], then all homogeneous first-order linear differential equations are actually separable because:

[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 the converse is 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 the reciprocal of 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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Muzza is offline
Sep12-04, 12:53 PM
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Just wondering, but do you really need to state that the constant of integration is +/- A, instead of just A (in the part where you talk about integrating factors)?
cepheid is offline
Sep12-04, 01:51 PM
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Good point. In fact, I guess I could have gotten rid of the +/- way up here:

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

just combining everything into "A", which would remain as A for the ensuing discussion.

HallsofIvy is offline
Sep12-04, 05:48 PM
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Does Homogeneous = Separable?

In answer to the original question, yes, a first order d.e. that is "homogeneous" in this sense: P(x) dy/dx+ Q(x)y= 0 is trivially seperable: dy/y= -(Q(x)/P(x))dx.

But be careful: in the limited area of FIRST ORDER d. e.s, the term "homogeneous" is often used in quite a different way (the d.e. A(x,y)dx+ B(x,y)dy= 0 is "homogeneous" if and only if B(&lamda;x,&lamda;y)/A(λx, λy)= B(x,y)/A(x,y) (essentially that means that the total exponent of x and y in each term is the same).

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