y' = dy/dx = 3y/(1+x)dy/y = 3dx/(1+x)两边同时积分，得到：∫dy/y = ∫3dx/(1+x)lny = 3ln(1+x) + c把 y(0) = 1 代入上式：ln1 = 3ln(1+0) + c所以， c = 0那么，lny = 3ln(1+x)或 y = (1+x)³
