y0′′=(x2−1)y0, so α=1.
Substitution and cancellation give f′′−2xf′+(α−1)f=0.
Equating
powers gives
(n+2)(n+1)an+2+(α−1−2n)an=0.
y′(0)=0 forces the odd
series to vanish. The even recurrence terminates at degree
2N precisely when
α=4N+1.
After
division by
e−x2/2,
f′′−2xf′+8f=2. The initial
data and evenness give
f=1−3x2+x4, hence
y=e−x2/2(1−3x2+x4).
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