工程數學求解 y"+2y'+y=(12x^2+6x)e^-x?

y"+2y'+y=(12x^2+6x)e^(-x)

對應的齊次方程式:
y" + 2y' + y = 0
即 (D^2+2D+I)y = 0 或 (D+I)^2 y = 0
其解
y_h = (C1 x + C2)e^(-x)

因此, 原方程式一特解為
y = y_p = (A x^4 + B x^3 + C x^2)e^(-x)
則
y' = [-(Ax^4+B x^3+Cx^2)+(4Ax^3+3Bx^2+2Cx)]e^(-x)
    = [-Ax^4+(4A-B)x^3+(3B-C)x^2+2Cx]e^(-x)
y" = {-[-Ax^4+(4A-B)x^3+(3B-C)x^2+2Cx]
       +[-4Ax^3+3(4A-B)x^2+2(3B-C)x+2C]}e^(-x)
   = [Ax^4+(-8A+B)x^3+(12A-6B+C)x^2+(6B-4C)x+2C]e^(-x)
所以, 在 x^4, x^3 項抵消後
  y"+2y'+y
    = {[(12A-6B+C)x^2+(6B-4C)x+2C]
              +2[(3B-C)x^2+2Cx] + Cx^2}e^(-x)
    = (12Ax^2+6Bx+2C)e^(-x)
    = (12x^2+6x)e^(-x)
∴ A = 1, B = 1, C = 0
即 y_p = (x^4+x^3)e^(-x)

∴ y = y_h + y_p
     = (x^4 + x^3 + C1 x + C2)e^(-x)


[另法]
原方程式以算子表示: 
    (D+I)^2 y  = (12x^2+6x)e^(-x)
令 z = (D+I)y, 則得
    (D+I)z = (12x^2+6x)e^(-x)
故
    z = (D+I)^(-1) (12x^2+6x)e^(-x)
       = e^(-x) ∫ e^x (12x^2+6x)e^(-x) dx
       = e^(-x) ∫ (12x^2+6x) dx
       = (4x^3+3x^2+C1)e^(-x)
       = (D+I)y
∴ y = (D+I)^(-1) (4x^3+3x^2+C1)e^(-x)
      = e^(-x) ∫ e^x (4x^3+3x^2+C1)e^(-x) dx
      = (x^4 +x^3 +C1 x +C2)e^(-x)