>>26
(x+1) y' + y = x(x+1)
(d/dx) { (x+1) y} = x^2 + x
xで積分して
(x+1) y = (1/3)x^3 + (1/2)x^2 +c
cは積分定数

x^3 + 1 = (x+1)(x^2 -x+1)
x^2 - 1 = (x+1)(x-1)
だから

(x+1) y = (1/3)x^3 + (1/2)x^2 +c = (x+1) { (1/3)(x^2 -x+1) + (1/2)(x-1)} -(1/3) + (1/2) +c
= {(x+1)/6} { 2x^2 +x -1} + (1/6) +c
c_0 = (1/6) +c とおいて

y = (1/6) (2x^2 +x-1) + c_0 { 1/(x+1)}
ここで
1/(1+x) = Σ (-1)^k x^k