(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?
(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?
You simply add A/(s+2) and B/(s+1) back together. You will get A(s+1)+B(s+2) in numerator, and (s^{2}+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.
Comments
Notice that the coefficients of the polynomial, i.e. 0s + 1, in the numerator on the left hand side correspond to the linear coefficients on the right hand side, i.e. (A+B)s + (A+2B), when forming the linear system...A+B=0 and A+2B=1.
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