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Homogeneity Question

  1. Jan 26, 2008 #1
    Hi, can a homogeneous equation be homogeneous to the 5/2 degree? Must it be a integer degree?
  2. jcsd
  3. Jan 27, 2008 #2
    Hello coffeebean51, welcome to this forum.

    Your question involves non-linear differential equations. The power 5/2 makes the equation non-linear. The definition of a linear differential equation is the following. The equation:


    is linear if F a linear function is of the variables y, y', y'', .... This is not the case with the 5/2 power. The general second order linear differential equation p.e. is:


    In case g(x) equal is to 0, you have a homogeneous equation otherwise it is nonhomogeneous. For non-linear differential equations this is more complicated to define, I should look it up. Is this already helping?
  4. Jan 28, 2008 #3


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    Unfortunately, there are two uses of the word "homogeneous" in differential equations. The one Coomast is giving applies to linear equations and I do not believe that is what is intended here.

    The definition of homogeneous I believe is intended here applies to first order equations: If dy/dx= f(x,y) and replacing both x and y by [itex]\lambda x[/itex] and [itex]\lambda y[/itex] results in exactly the same equation (i.e. the [itex]\lambda[/itex]'s cancel out), then the f can be written in terms of x and y/x and the problem can be simplified by the substitution u= y/x. The "degree" appears when you write the equation as g(x,y)dx+ h(x,y) dy= 0. If replacing x and y by [itex]\lambda x[/itex] and [itex]\lambda y[/itex] in g and h results in [itex]\lambda^\alpha g(x,y)[/itex] and [itex]\lambda^\alpha h(x,y)[/itex], then clearly the [itex]\lambda[/itex] cancels and the equation is homogenous (here of degree [itex]\alpha[/itex]). Yes, [itex]\alpha[/itex] can be any real number and there can be equations that are "homogeneous of degree 5/2.
    Last edited: Jan 28, 2008
  5. Jan 28, 2008 #4
    thanks halls. you were right. i just started this class and wasnt talking about the first homogeneity mentioned. thanks!
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