# Thread: Is this an homogeneous differential equation?

1. ## Is this an homogeneous differential equation?

I read that a differential equation is homogenous if every term of the equation contains the indepedent variable (unknown). I was wondering if the following equation:

${dy \over dx } - \sin{y} = 0$

is homogenous, since it's not the function $y$, but it sine what appears in a term of the equation.

Just curiosity :P

Thanks

2. ## Re: Is this an homogeneous differential equation?

I wonder where you saw that definition of "homogeneous". There are, in fact, two types of "homogenous" defined for differential equations. When we are talking specifically about first order equations, we say that f(x,y)dy+ g(x,y)dx= 0 is "homogenous" if and only if [tex]\frac{f(\lambda x, \lambda y)}{g(\lambda x, \lambda y)}= \frac{f(x, y)}{g(x, y)}[/itex] that is the same as saying that if we were to write this as dy/dx= F(x,y), F would actually depend only on y/x, it cannot be a general function of x, y and we can simplify the equation by substituting u= y/x. This equation, while first order, is clearly not "homogeneous" in that sense.

There is also a definition of "homogeneous" for linear differential equations of any order. This equation is not linear so that does not apply. This equation is not "homogeneous' in either sense.

(And the "unknown" is the dependent variable, not independent variable.)

3. ## Re: Is this an homogeneous differential equation?

Many thanks HallsofIvy!

I'm trying to become familar with all the different differential equation types.

I understood that the term homogeneous is only applied if the differential equation is linear...

Yes, the unknown is the dependent variable, fast fingers here ...:P

4. ## Re: Is this an homogeneous differential equation?

Originally Posted by jguzman
I read that a differential equation is homogenous if every term of the equation contains the indepedent variable (unknown). I was wondering if the following equation:

${dy \over dx } - \sin{y} = 0$

is homogenous, since it's not the function $y$, but it sine what appears in a term of the equation.

Just curiosity :P

Thanks
If your plan is to solve this DE it will be easiest to note that it is separable.