求解2題Laplace 反 轉換(20點)

Q1:
s^4- 6s^3+32s= s(s^3-6s^2+32)= s(s+2)(s^2-8s+16)= s(s+2)(s-4)^2
1/(s^4-6s^3+32s) (改為部分分式)
= a / s + b/ (s+2) + c/(s-4) + d/(s-4)^2
a= 1/32, b= -1/72, c= -5/288 , d= 1/24
=>Laplace inverse = a+b e^(-2t) + c e^(4t)+ d t e^(4t)
= 1/32 - e^(-2t) / 72 - 5 e^(4t)/ 288 + t e^(4t) / 24
Q2:
L{ t^a } = Γ(a+1)/ s^(a+1)
取 a= 1.5 = 3/2, Γ(5/2)=(3/2)Γ(3/2)=(3/2)(1/2)Γ(1/2)=(3/4)√π
=>Laplace inverse of 1/s^2.5 = (4/3) t^1.5 /√π

1. 分母=s(s-4)(s^2-2s-8)=s(s-4)[(s-1)^2-3^2]

2. Use公式 L^-1{s^(-a)}=t^(a-1)/Gamma(a), for a大於0. Here you have a=2.5.