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

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

\alpha = \alpha (r)

\alpha ' + \frac{n-1}{2r} \alpha =0

\alpha '' + \frac{n-1}{r} \alpha ' = 0


The attempt at a solution

The solution to the first one is

\alpha = r^{\frac{-(n-1)}{2}

The solution to the second one is

\alpha '= r^{-(n-1)}

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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