Solve Legendre Polynomial using Method of Frobenius

feoxx
Messages
2
Reaction score
0
Not sure how this can be done. can anyone help?
 

Attachments

  • Frobenious method.jpg
    Frobenious method.jpg
    15.6 KB · Views: 921
Physics news on Phys.org
Have you made an attempt?

If you don't know what the method of Frobenius is, then you should look it up. It usually comes under power series solutions of ODEs.
 
I tried, i know that y''P(x)y'+Q(x)y = 0.
P(x) = (1-x)^2
Q(x) = (n^2+n)

but I am not sure what to do in terms of legendre polynomial.
 
The Legendre polynomials are the solutions to this type of ODE. For the moment, forget Legendre Polynomials.

What you must do is use the method of Frobenius to solve the given ODE. If you do it correctly, your solution should come out equal to the series expression given. Note: these solutions are series with a finite number of terms.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Back
Top