19/09/28 03:29:59.55 flE+CrWr.net
>>848
xy '= Σ[n=1,∞] 2/(n･C[2n,n])・x^(2n),
y " = Σ[n=0,∞] 2(2n+1)/((n+1)･C[2n+2,n+1])・x^(2n)
(4-xx) y " = Σ[n=1,∞] {8(2n+1)/((n+1)･C[2n+2,n+1]) - 2(2n-1)/(n･C[2n,n])}・x^(2n)
= Σ[n=1,∞] {4/C[2n,n] - 2(2n-1)/(n･C[2n,n])}・x^(2n)
= Σ[n=1,∞] 2/(n･C[2n,n])・x^(2n),
よって
(4-xx) y " - xy ' = 0.
x = 2sinθ とおくと
(d/dθ) = √(4-xx)･(d/dx)
題意より
(d/dθ)^2 y = 4,
y = 2θ^2,