Linear System of ODEs: Solving for n=1 or n=3

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SUMMARY

The discussion focuses on solving a system of ordinary differential equations (ODEs) defined by the equations α' + (n-1)/(2r) α = 0 and α'' + (n-1)/r α' = 0. The solutions derived are α = r^{-(n-1)/2} for the first equation and α' = r^{-(n-1)} for the second. The key conclusion is that the equations yield consistent solutions specifically for n=3, while n=1 presents complications due to singular behavior. Careful algebraic manipulation is essential to reconcile the solutions across different values of n.

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McCoy13
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Homework Statement
I'm trying to solve the following system of ODEs.

[tex]\alpha = \alpha (r)[/tex]

[tex]\alpha ' + \frac{n-1}{2r} \alpha =0[/tex]

[tex]\alpha '' + \frac{n-1}{r} \alpha ' = 0[/tex]


The attempt at a solution

The solution to the first one is

[tex]\alpha = r^{\frac{-(n-1)}{2}[/tex]

The solution to the second one is

[tex]\alpha '= r^{-(n-1)}[/tex]

Ultimately the goal is to show that n=1 or n=3 (it's a problem dealing with wave attenuation and distortion, but I'm just having problems with this step). I really can't reconcile these answers, even using arbitrary scalar factors against my solutions. When I tried substituting one equation into the other all that happened was I ended up with a factor of sqrt(2) that wasn't consistent with either equation individually.
 
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The solutions to these equations do match when n=3, just be a bit more careful with the algebra. The equations themselves are not well-behaved when n=1, so you should be careful when trying to match the solutions for arbitrary n to the limit n->1. It happens to work for the first equation, but there is an additional solution of the second that is not a solution to the first.
 
Gah, I see. I think I was forgetting that r was to a negative power when I was thinking about it in the case n = 3 and in the case n = 1 I wasn't looking at the equations, just the solutions, so I missed the fact that they sort of become singular. Thanks for elucidating.
 

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