- #1
rmiller70015
- 110
- 1
Homework Statement
This is for differential equations with nonconstant coefficients and I wasn't so great at series and sequences in calculus so when I came across this example problem I wasn't sure how they got to their final form. If someone could explain it to me that would be really helpful.
Homework Equations
The ODE is:
y"-2xy+2y=0
The Attempt at a Solution
We assume there is some solution of the form:
y(x)=\sum_{n=0}^{\infty}a_nx^n
The derivatives are:
y'(x)=\sum_{n=1}^{\infty}na_{n}x^{n-1}\:\: \text{and}\:\: y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}
A change of index gives
y'(x)=\sum_{n=0}^{\infty}na_nx^{n-1}\:\:\text{and}\:\:y''(x)=\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n
Putting these into the ODE gives:
\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-2x\sum_{n=0}^{\infty}na_nx^{n-1}+2\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}\Big((n+2)(n+1)a_{n+2}-2na_n+2a_n\Big)x^n=0
The Identity Theorem gives:
(n+2)(n+1)a_{n+2}-2na_n+2a_n=0
Solving this for ##a_{n+2}## gives:
a_{n+2}=\frac{2(n-1)a_n}{(n+2)(n+1)}
Odd coefficients are zeros except for ##a_1## because:
a_3=\frac{2(1-1)a_1}{(n+2)(n+1)}=0\:\: \text{and all other coefficients will depend on a multiple of}\:a_3
The even coefficients are given by:
a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}
This is a worked example from the book and the answer is:
\frac{2(2k!)}{k!}
I'm not sure how they got this final form, but I think it has something to do with the even number recursion formula, which can be written as:
a_{2k+2}=\frac{2(2k-1)a_{2k}}{(2k+2)(2k+1)}=\frac{2(2k-1)}{(2k+2)(2k+1)}\frac{2(2k-3)a_{2k-2}}{2k(2k-1)}
This is where I am having the problem. The book says the ##a_{2k+2}## can be written as:
\frac{2^{k+1}(2k-1)(2k-3)...3\cdot2\cdot1}{(2k+2)!}a_0
I understand where the ##2^{k+1}## and ##a_0## are coming from, but the other terms I am unsure of.