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Cauchy Euler Equation

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

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What is the characteristic equation you've found? [And ask yourself whether you really cant see any way of finding at least one root of it...]
Reply 2
a(a-1) + 2ra -n(n+1)

where a = alpha.
That's not quite right. Check your working.
Reply 4
Original post by SwimGood
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
(edited 9 years ago)
Reply 5
Original post by DFranklin
That's not quite right. Check your working.


I can't see where i'm quite going wrong. I found dV/dr and then d2V/dr^2, then substituted in the values, also subbing in V = r^a

(this is for the trial solution)

I get a(a-1)r^a + 2r*ar^a - n(n+1)*r^a = 0
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.
(edited 9 years ago)
Reply 7
ar^r-1 ?

Ermmm i'm not sure. I'm really confused. So, take a step back,

I get, r^2(a-1)ar^a-2 + 2r*ar^a-1 -n(n+1)r^a

that's just subbing V, dV and d2V into the equation, with V = r^a, dv = ar^a-1 and d2V as (a-1)ar^a-2 in
Reply 8
Sorry i've read your hint and it's confused me even more. my brains not with it today.
SwimGood
..
What you've just written (post 8) is correct (but you should now take out the common factor of r^a).

If you look at your previous post (post 6), you'll find you gave a different equation there (as I said, the error is in the middle term).
(edited 9 years ago)
Reply 10
Original post by SwimGood
ar^r-1 ?

Ermmm i'm not sure. I'm really confused. So, take a step back,

I get, r^2(a-1)ar^a-2 + 2r*ar^a-1 -n(n+1)r^a

that's just subbing V, dV and d2V into the equation, with V = r^a, dv = ar^a-1 and d2V as (a-1)ar^a-2 in


Fellow sufferer of "Latex Ignorance"

if you are using y=r^a as your trial, you should get a quadratic in r which will also have constants n at the end.

the term r^a will appear in every term, but will "cancel out" or factorize out if you want.

I just looked at the answer you had a few posts back and you are almost correct.

a(a-1)+2a-n(n-1) = 0 or thereabouts .
Original post by TeeEm
Fellow sufferer of "Latex Ignorance"
There is a LaTeX guide: http://www.thestudentroom.co.uk/wiki/LaTeX

I do strongly recommend learning at least the basics if you intend to post here a lot - it is much easier for the reader!

a(a-1)+2a-n(n-1) = 0 or thereabouts .
This still isn't quite right (the constant term should be n(n+1)).
Reply 12
I will leave you in the capable hands of DFranklin, because it is confusing with both of us.

if you find your way round I got a=n or a= -(n+1)

you auxillary probably is better by completing the square
Reply 13
Original post by DFranklin
There is a LaTeX guide: http://www.thestudentroom.co.uk/wiki/LaTeX

I do strongly recommend learning at least the basics if you intend to post here a lot - it is much easier for the reader!

This still isn't quite right (the constant term should be n(n+1)).


another typo sorry ... and probably I used this to solve auxillary... past my bedtime so I am off:smile:
Reply 14
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?
Original post by SwimGood
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 2rdVdr2r \dfrac{dV}{dr}. V = r^a, so dVdr=ara1\dfrac{dV}{dr} = a r^{a-1}. Multiply by 2r and the final expression for the middle term is 2a r^a (i.e.2ara2ar^a). [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+12a r^{a+1}). Which is wrong (and means you get an unsolvable equation).
Reply 16
Oh apologies, I see! forgot the power.

a(a-1) + 2a - n(n+1) = characteristic equation.

do I solve this in terms of alpha??
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\alpha, then the 2nd root is given by B/αB / \alpha.

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 B±B24AC2A\dfrac{-B \pm \sqrt{B^2-4AC}}{2A}, 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.
(edited 9 years ago)
Reply 18
a = n, -n-1 ?
Reply 19
Original post by SwimGood
a = n, -n-1 ?


correct

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