y''+2y'-3y=3xe^x求通解.

Question

y''+2y'-3y=3xe^x 求通解.

Answer 1

特征方程：
λ² + 2λ - 3 = 0
λ = -3 or 1
y₁= C₁e^(-3x),y₂= C₂e^x
右边是(3x)e^x,所以设
yp = x(e^x)(Ax + B) = Ax²e^x + Bxe^x
yp' = A(2xe^x + x²e^x) + B(e^x + xe^x) = Ax²e^x + (2A + B)xe^x + Be^x
yp'' = A(2xe^x + x²e^x) + (2A + B)(e^x + xe^x) + Be^x
代入y'' + 2y' - 3y = 3xe^x中：
[Ax²e^x + (2A + B)xe^x + Be^x] + 2[A(2xe^x + x²e^x) + (2A + B)(e^x + xe^x) + Be^x]
- 3(Ax²e^x + Bxe^x) = 3xe^x
(4A + 2B)e^x + 8Bxe^x = 3xe^x
4A + 2B = 0,8B = 3
解得B = 3/8,A = -3/16
C₁e^(-3x) + C₂e^x - (3/16)xe^x + (3/8)x²e^x