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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

  • math -

    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 ∞
    (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
    ∑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.

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