y''-y=x求通解

特征方程r² - 1 = 0
r = ±1
y1 = c1*e^x
y2 = c2*e^(-x)
设特解yp = ax + b
yp' = a,yp'' = 0,代入方程
0 - (ax + b) = x
-a = 1 => a = -1
b = 0
yp = -x
通解为y = y1 + y2 + yp
即y = c1*e^x + c2*e^(-x) - x
copy的,