# Does Homogeneous = Separable?

1. Sep 12, 2004

### cepheid

Staff Emeritus
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.

2. Sep 12, 2004

### Muzza

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)?

3. Sep 12, 2004

### cepheid

Staff Emeritus
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.

Last edited: Sep 12, 2004
4. Sep 12, 2004

### HallsofIvy

Staff Emeritus
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(&lambda;x, &lambda;y)= B(x,y)/A(x,y) (essentially that means that the total exponent of x and y in each term is the same).