特解 y = (ax^2+bx)e^(2x)
y' = (2ax+b)e^(2x) + 2(ax^2+bx)e^(2x)
= [2ax^2+2(a+b)x+b]e^2x
y'' = [4ax+2(a+b)]e^2x + 2[2ax^2+2(a+b)x+b]e^2x
= [4ax^2+4(2a+b)x+2a+4b]e^2x
代入微分方程得
[4ax^2+4(2a+b)x+2a+4b] - 5 [2ax^2+2(a+b)x+b] + 6 (ax^2+bx) = x
则 -2a = 1,2a - b = 0,解得 a = -1/2, b = -1
追问谢谢你。太感谢了!