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!