Find the simple inverse Laplace transform?

Question

(1/2)/(s^2+3s+2)

Answer: (-1/2)e^(-2t)+(1/2)e^(-t)

Here's the work:

(1/2)(1/(s^2+3s+2))=(1/2)(A/(s+2)+B/(s+1))

How do I find A and B?

Answer 1

You simply add A/(s+2) and B/(s+1) back together. You will get A(s+1)+B(s+2) in numerator, and (s²+3s+2) in denominator. Collect similar terms, you will get s(A+B)+(A+2B) in numerator. This has to equal 1 (see left side of equation). Thus you have to have A+B=0 and A+2B=1. Two equations, two unknowns. Solve them for A and B--problem solved.