Power series solution of (1−x²)y''−2xy'+2y=0

Equation: (1 − x²)y″ − 2xy′ + 2y = 0. This is Legendre's equation with n(n + 1) = 2, i.e. n = 1. x = 0 is an ordinary point; take y = Σ aₖxᵏ.

Recurrence

Coefficient of xᵏ: (k+2)(k+1)aₖ₊₂ − k(k−1)aₖ − 2k aₖ + 2aₖ = 0.

Since k(k−1) + 2k − 2 = k² + k − 2 = (k − 1)(k + 2),

aₖ₊₂ = (k − 1)(k + 2) aₖ / ((k + 1)(k + 2)) = (k − 1) aₖ / (k + 1)

Coefficients (a₀, a₁ arbitrary)

k=0:  a₂ = −a₀
k=1:  a₃ = 0       (so all further odd terms vanish)
k=2:  a₄ = a₂/3 = −a₀/3
k=4:  a₆ = 3a₄/5 = −a₀/5
k=6:  a₈ = 5a₆/7 = −a₀/7

Even part: 1 − x² − x⁴/3 − x⁶/5 − x⁸/7 − …

Solution

y = a₀(1 − x² − x⁴/3 − x⁶/5 − …) + a₁x

The odd solution y = x (= P₁) is a polynomial; the even solution is an infinite series (convergent for |x| < 1) = 1 − x·tanh⁻¹x.