Question about homogeneous dif. equations

  1. I've been confused for a little while as my teacher in dif. eq.'s taught us about homogeneous equations, where you can call an eq. homogeneous if all terms are of the same degree. But then, when seeing linear dif. equations we were taught that an homogeneous eq. is one with certain form and equal to zero. Ok to get to the point, I'm confused because I am sometimes asked whether an eq. is homogeneous or not...but I mean, it could not be homogeneous according to the first definition that I was taught and it could be homogeneous according to the second one...so what is a correct answer here??


    Thanks for any help and I hope I got myself clear
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,367
    Staff Emeritus
    Science Advisor

    Yes, it is unfortunate that the same term is used for two completely different things!

    However, the first use of "homogeneous" applies only to first order equations. The second use applies only to linear equations.

    A first order differential equation that was "homogeneous" in the second sense would have to be of the form dy/dx= f(x)y and would be a relatively simple separable equation.
     
  4. Thanks a lot!
    ...but wait, you said that an equation that would satisfy both definitions would be of the form dy/dx= f(x)y , but wouldn't it depend on f(x) to whether it satisfies the first definition(in which all terms are of the same degree)?
    dy/dx has 0 degree right? y has a degree of 1, so f(x) would require a degree of -1 for the eq. to be homogeneous according to our first definition...or am I missing something?
     
  5. HallsofIvy

    HallsofIvy 40,367
    Staff Emeritus
    Science Advisor

    No, you are right- I mispoke. A first order d.e. is "homogenous" in the first sense if dy/dx= f(x,y) and f(x,y) can be written as a function of the single expression y/x. That means that a linear, first order equation that is "homogenous" in both senses must be dy/dx= y/x!
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?