“…where y 0 (x) = 1, y 1 (x) = a 1,0 x + a 1,1 . In terms of the hypergeometric functions, the polynomial solutions in the case of a 1,0 a 2,1 − a 2,0 a 1,1 = 0 and a 2,1 > 0 are y n a 2,0 a 2,1 0 a 1,0 a 1,1 x = (−1) n a n 2,1 a 1,0 a 2,1 − a 2,0 a 1,1 a 2,0 a 2,1 n 2 F 1 (−n, n − 1 + a 1,0 a 2,0 ; a 1,0 a 2,1 − a 2,0 a 1,1 a 2,0 a 2,1 ; a 2,0 a 2,1 x + 1) (33) which can be easily obtained from equation (16) as a limit case of a 2,2 → 0, while for a 1,0 a 2,1 − a 2,0 a 1,1 = 0, the polynomial solution is simplified to…”