第1个回答 2022-12-15
y''-y'-2y=x^2
The aux. equation
r^2-r-2=0
(r-2)(r+1)=0
r=-1 or 2
let
yg=Ae^(-x) +Be^(2x)
yp=Cx^2+Dx+E
yp'=2Cx+D
p''=2C
yp''-yp'-2yp=x^2
2C -(2Cx+D) -2(Cx^2+Dx+E) = x^2
-2Cx^2 +(-2C-2D)x +(2C-D-2E) =x^2
=> C=-1/2
-2C-2D =0
1-2D=0
D=1/2
2C-D-2E=0
-1-1/2-2E=0
E=-3/4
yg=Cx^2+Dx+E = -(1/2)x^2 +(1/2)x -3/4
通解
y=yg+yp=Ae^(-x) +Be^(2x) -(1/2)x^2 +(1/2)x -3/4
第2个回答 2022-12-15
y"-y'-2y = x^2, 特征方程 r^2-r-2 = 0, r = -1, 2
设特解 y = ax^2+bx+c, 代入微分方程, 得
2a - 2ax - b - 2ax^2 - 2bx - 2c = x^2
a = -1/2, b = 1/2, c = -3/4
通解 y = C1e^(-x) + C2e^(2x) - (1/4)(2x^2-2x+3)