Question

find a three - term recurrence relation for solutions of the form ( y=sum_{n = 0}^{infty}c_{n}x^{n} ). then find the first three nonzero terms in each of two linearly independent solutions
( (x^{2}-3)y+2xy+2xy = 0 )
the three - term recurrence relation is ( c_{2}=0,c_{n + 2}=square ) for ( ngeq1 ).

Explanation:

Step1: Find \(y'\) and \(y''\)

If \(y=\sum_{n = 0}^{\infty}c_{n}x^{n}\), then \(y'=\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}\) and \(y''=\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\).

Step2: Substitute into the differential equation

\((x^{2}-3)\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}+2x\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}+2x\sum_{n = 0}^{\infty}c_{n}x^{n}=0\)
First, expand \((x^{2}-3)\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\):
\(\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n}-3\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\)
For \(2x\sum_{n = 1}^{\infty}nc_{n}x^{n - 1}\), we have \(2\sum_{n = 1}^{\infty}nc_{n}x^{n}\)
For \(2x\sum_{n = 0}^{\infty}c_{n}x^{n}\), we have \(2\sum_{n = 0}^{\infty}c_{n}x^{n + 1}\)
Change the index of \(-3\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n - 2}\) to \(m=n - 2\), then \(n=m + 2\) and it becomes \(-3\sum_{m = 0}^{\infty}(m + 2)(m + 1)c_{m+2}x^{m}\)
Change the index of \(2\sum_{n = 0}^{\infty}c_{n}x^{n + 1}\) to \(k=n + 1\), then \(n=k - 1\) and it becomes \(2\sum_{k = 1}^{\infty}c_{k - 1}x^{k}\)
The equation becomes:
\(\sum_{n = 2}^{\infty}n(n - 1)c_{n}x^{n}-3\sum_{n = 0}^{\infty}(n + 2)(n + 1)c_{n+2}x^{n}+2\sum_{n = 1}^{\infty}nc_{n}x^{n}+2\sum_{n = 1}^{\infty}c_{n - 1}x^{n}=0\)

Step3: Equate the coefficients of \(x^{n}\)

For \(n = 0\):
\(-3\times2\times1c_{2}=0\Rightarrow c_{2}=0\)
For \(n\geq1\):
\(n(n - 1)c_{n}-3(n + 2)(n + 1)c_{n+2}+2nc_{n}+2c_{n - 1}=0\)
\(c_{n+2}=\frac{(n^{2}+n)c_{n}+2c_{n - 1}}{3(n + 2)(n + 1)}=\frac{n(n + 1)c_{n}+2c_{n - 1}}{3(n + 1)(n + 2)}=\frac{nc_{n}+2c_{n - 1}}{3(n + 2)}\)

Answer:

\(c_{n+2}=\frac{nc_{n}+2c_{n - 1}}{3(n + 2)}\)