Solve the Euler equation (r^2)d2V/dr +(2r)dV/dr - n(n+1)V =0
Solve using the Euler transformation r=e^t
y = V in this question, and your constants a = r^2 and b = 2r and c = -n(n+1)
I also have to solve using trial solution V = r^(a) where a = alpha
For the trial solution, I managed to get the characteristic equation, but I don't know how to solve it (it's a quadratic) because of the -n(n+1) term
Any help is appreciated
Using the standard transformation for this ODE or a trial solution of the form r^m will involve having to solve a quadratic in n, but a "constant term in terms of constants".
either see if you can find an obvious solution by inspection and then work back your factorization or use the quadratic formula
In the "middle" term, where has a r^a come from? And what is the derivative of r^a?
Hint/piece of advice: the whole point of the trial solution / auxiliary equation method is to end up with find a solution that works for all r. You can basically only do this if you can rewrite your auxiliary equation in the form f(a)g(r) = 0, where f(a) does not depend on r.
i dont see why the middle term (from post 6) is wrong?
The constants of the equation are, lets say, A, B, C
the question states that A = r^2 B= 2r C = -n(n+1)
dv = ar^a-1
So all I've done is multiplied that by 2r which is B ? middle term is BdV/dr is it not?
The middle term is 2rdrdV. V = r^a, so drdV=ara−1. Multiply by 2r and the final expression for the middle term is 2a r^a (i.e.2ara). [To be clear, I don't think this is any different from what you've just said, I've just made sure all the expressions are provided clearly and unambiguously].
The middle term you provided in post 6 is 2r*ar^a (exact quote), which equals 2a r^{a+1} (i.e. 2ara+1). Which is wrong (and means you get an unsolvable equation).
Yes, solve for a. Combine a(a-1) with 2a and it should be obvious what one root is. Once you know one root, use the fact that if you have a quadratic x^2+Ax+B = 0, with a known root α, then the 2nd root is given by B/α.
Edit: note that if you really can't spot a root, you can still use the quadratic formula even if some of the coefficients are in terms of other variables. e.g. with the formula 2A−B±B2−4AC, you'd need to set C = -n(n+1) here. It will still work, but the algebra may be tedious, and at some point you're going to end up with \sqrt{stuff} where stuff will be a polynomial and you'll need to recognize it's actually the square of a simpler polynomial, which might be tricky for you. It's useful to know you *can* do this though, because sometimes it's hard to see a better plan.