2K 8 !2 3 15 2k+1 (2k+1)!! 原方程的通解 k=1,2,3, 2n 2n+1 ∑ a ∑ n=02m!=0(2n+1) (an,a是任意常数), 3 1 3 a a = , 15 1 5 a a =  , (2 1)!! 1 2 1 + + = k a a k k = 1,2,3,  , 2 0 2 a a = , 8 0 4 a a =  , ! 2 0 2k k k a a = 原方程的通解    =  + = + = + 0 2 1 1 0 2 0 2 ! n (2 1)!! n n n n n x a n x y a ( , ) a0 a1是任意常数