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

Question

find laplace transform of that sum these quesion from 2 semester of bscit

Answer 1

Split into partial fractions:

(4s+5)/[(s-1)²(s+2)]=A/(s-1)+B/(s-1)²+C/(s+2).

So, A(s-1)(s+2)+B(s+2)+C(s-1)²=4s+5,

As²+As-2A+Bs+2B+Cs²-2Cs+C=4s+5. Match coefficients:

A+C=0, C=-A;

A+B-2C=4, A+B+2A=4, 3A+B=4, B=4-3A;

-2A+2B+C=5, -2A+8-6A-A=5, 9A=3, A=⅓⇒B=3, C=-⅓.

(4s+5)/[(s-1)²(s+2)]=⅓(1/(s-1))+3/(s-1)²-⅓(1/(s+2)).

Apply inverse Laplace to each term:

⅓e^t+3te^t-⅓e^-2t