L^-1((4s^2-3s+5)/((s-1)(s^2-3s+2)))

Question

use partial fraction method to solve this & then use laplace inverse transform.

search in some other reference book & tell me the answer.pls

Answer 1

s²-3s+2=(s-2)(s-1).

(4s²-3s+5)/((s-1)²(s-2))=A/(s-1)+B/(s-1)²+C/(s-2),

4s²-3s+5=As²-3As+2A+Bs-2B+Cs²-2Cs+C,

(1) A+C=4, so C=4-A, (2) -3A+B-2C=-3, (3) 2A-2B+C=5.

2(2)+(3)=-4A-3C=-1, 4A+3C=1

4A+3(4-A)=1,

4A+12-3A=1,

A=-11, C=15, 2A-2B+C=-22-2B+15=5,

2B=-12, B=-6.

(4s²-3s+5)/((s-1)(s²-3s+2))=-11/(s-1)-6/(s-1)²+15/(s-2).

ℒ{e^attⁿ}=n!/(s-a)ⁿ⁺¹; when a=n=1, ℒ{e^tt}=1/(s-1)².

ℒ^-1{(4s²-3s+5)/((s-1)(s²-3s+2))}=-11e^t-6te^t+15e^2t.