I'll give it a try.
(1-x)y"+y=0 at x0=0
A. Check for singular points.
Write as y"+(1/1-x)y=0
and find Lim x->0 (1/1-x) = 1
Therefore x0=0 is not a singular point.
B. Assume y=∑an x^n for n=0 to ∞
then
(1-x)y"=∑n=0 to ∞ (n)(n-1)an x^(n-2)
=∑n=2 to ∞ (n)(n-1)an x^(n-2) -∑n=1 to ∞ (n)(n-1)an x^(n-1)
Now switch indices
=∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) -∑n=0 to ∞ (n+1)(n)an+1 x^(n)
Add to y gives
y(x)
∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) -∑n=0 to ∞ (n+1)(n)an+1 x^(n)+∑ n=0 to ∞ an x^n =0
This gives
an+1 = [(n)(n+1)an+1 - an]/[(n+2)(n+1)]
The coefficients an are dependent on values of a0 and a1, which are arbitrary constants depending on the given initial conditions.
Check my arithmetic.
Seek power series solution of the given differential equation about the given point x0; find the recurrence relation.
(1-x)y'' + y = 0; x0 = 0
1 answer