Does Homogeneous = Separable?

**cepheid** · Sep12-04, 01:47 PM

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:

$LaTeX Code: \\frac{dy}{dx} + P(x)y = Q(x)$

and if I understood my prof's remark correctly that homogeneous linear first-order d.e.'s are those for which

, then all homogeneous first-order linear differential equations are actually separable because:

$LaTeX Code: \\frac{dy}{dx} + P(x)y = 0$

$LaTeX Code: \\frac{dy}{dx} = -P(x)y$

Which can be solved as follows:

$LaTeX Code: \\int{\\frac{dy}{y}} = -\\int{P(x)dx}$

^^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, $LaTeX Code: \\frac{dy}{dx} = \\frac{6x^2}{2y + cosy}$ is not linear in

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) ...

$LaTeX Code: \\ln|y| = -\\int{P(x)dx}$

$LaTeX Code: |y| = e^{-\\int {P(x)dx}}$

Now, the most general solution for

must include the most general antiderivative, so we'll have a

stuck in there if and when we solve the integral:

$LaTeX Code: y = \\pm e^{-\\int {P(x)dx} + C} = \\pm e^{-\\int {P(x)dx}}e^C = \\pm Ae^{-\\int {P(x)dx}}$

Whoah, cool! So in a homogeneous first order d.e. the solution takes the form of some constant $LaTeX Code: \\pm A$ times the reciprocal of the integrating factor

?! Is this always true?

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

$LaTeX Code: (I(x)y)^{\\prime} = I(x)Q(x)$

If

, then

$LaTeX Code: (I(x)y)^{\\prime} = 0$

$LaTeX Code: I(x)y = \\pm A$

$LaTeX Code: y = \\frac{\\pm A}{I(x)}$

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

**Muzza** · Sep12-04, 01:53 PM

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** · Sep12-04, 02:51 PM

Good point. In fact, I guess I could have gotten rid of the +/- way up here:

$LaTeX Code: \\pm (e^{-\\int {P(x)dx}})(e^C) = Ae^{-\\int {P(x)dx}}$

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

**HallsofIvy** · Sep12-04, 06:48 PM

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).