Homogeneous Equation: 5/2 Degree?

[coffeebean51]

Hi, can a homogeneous equation be homogeneous to the 5/2 degree? Must it be a integer degree?

 

[coomast]

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:

F(x,y,y',...,y^{(n)})=0

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:

y''+p(x)y'+q(x)y=g(x)

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?

 

[HallsofIvy]

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 \lambda x and \lambda y results in exactly the same equation (i.e. the \lambda'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 \lambda x and \lambda y in g and h results in \lambda^\alpha g(x,y) and \lambda^\alpha h(x,y), then clearly the \lambda cancels and the equation is homogenous (here of degree \alpha). Yes, \alpha can be any real number and there can be equations that are "homogeneous of degree 5/2.

 

[coffeebean51]

thanks halls. you were right. i just started this class and wasnt talking about the first homogeneity mentioned. thanks!