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Find the inverse Laplace transform?


Answer: u5(t)[-e^(-2(t-5))+e^(-(t-5))]

How do I get the answer?

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Meagan W. | Biology, Mathematics and Statistics TutoringBiology, Mathematics and Statistics Tuto...
5.0 5.0 (24 lesson ratings) (24)

Typically, it is extremely complicated to directly solve for the inverse laplace transform of a function. This is why we use tables. With that being said, we want to manipulate the function to match an equation in the table of laplace transforms.

e^(-5s)/(s^2+3s+2) = e^(-5s)*(1/(s^2+3s+2) = e^(-5s)*(1/s(s+3+2/s)) = (e^(-5s)/s)*(1/(s+3+2/s)) = (e^(-5s)/s)*(s/(s+1)(s+2))

Now, doing partial fractions:

H(s) = A/(s+1) + B/(s+2)

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

s = A(s+1)+B(s+2)

B = -1

A = 1

H(s) = -1/(s+2) + 1/(s+1) 

inverse laplace transform of H(s) = -e^(-2t) + e^(-t)

applying inverse laplace trasform of e^(-5s)/s gives the heavy-side function us(t) = u(t-c) where us(t) = t-5

plugging in we get:


I think this is correct.