题不是特别清楚,反正二阶导数为0就是.设y=xe^（x/2）y‘ = x'e^(x/2) + x[e^(x/2)]' = e^(x/2) + xe^(x/2) *(1/2) = e^(x/2) (1 + x/2)y'' = [e^(x/2)]'(1 + x/2) + e^(x/2) (1+ x/2)'= (1/2)e^(x/2)(1 + x/2 + 1)= ...