给出一个常系数二阶微分方程:y''+3y'+2y=e^t,当y(0)=y'(0)=1时:

对等式两边取拉氏转换:[s^2Y(s)-sy(0)-y'(0)]+3[sY(s)-3y(0)]+2Y(s)=1/s+1

带入y(0)=y'(0)=1,整理得到:Y(s)=(s^2+3s-3)/(s-1)(s^2+3s+2)

拆分得到:(A/S-1)+(B/S+1)+(C/S-2)=Y(S)

反拉氏转换:

A=(s^2+3s-3)/(S+1)(S-2)|s=1 =1/6
B=(s^2+3s-3)/(S-1)(S-2)|s=-1 =5/2
C=(s^2+3s-3)/(S-1)(S+1)|s=2 =-5/3

由此可得
y(t)=(1/6)e^t+(5/2)e^-t-(5/3)e^-2t