微分方程 y''-2y'+y=0， 特征根 是 1， 1通解是 y = (C1+C2x)e^x, 则 y' = (C1+C2+C2x)e^xy(2) = 1, y'(2) = -2, 代入上两式得C1+2C2 = e^(-2)C1+3C2 = -2e^(-2)解得 C2 = -3e^(-2), C1 = 7e^(-2)则特解为 y = (7-3x)e^(x-2)
