Question

Chapter 8 Practice:

Compute the inverse Laplace transform of

`F(s) = (s² +1)/((s-1)(s+1)(s - 2))`

(​Cover up method)

Answer 1

Continue the given solution and solve for B and C :

`(s+1)(s^2+1)/((s-1)(s+1)(s-2)) = (s+1)\left(-\frac1{s-1} + \frac B{s+1} + \frac C{s-2}\right)`

`(s^2+1)/((s-1)(s-2)) = -(s+1)/(s-1) + B + (C(s+1))/(s-2)`

`s=-1 \implies B = \frac13`

Similar steps lead to C = 5/3, and so

`(s^2+1)/((s-1)(s+1)(s-2)) = -\frac1{s-1} + \frac13*\frac1{s+1} + \frac53*\frac1{s-2}`

Now take the inverse transform of each term.

`Y(s) = -\frac1{s-1} \implies y(t) = -e^t L^(-1)\left\{\frac1s\right\} = -e^t`

`Y(s) = \frac13*\frac1{s+1} \implies y(t) = \frac{e^(-t)}3 L^(-1)\left\{\frac1s\right\} = \frac{e^(-t)}3`

`Y(s) = \frac53*\frac1{s-2} \implies y(t) = \frac53e^(2t) L^(-1)\left\{\frac1s\right\} = \frac{5e^(2t)}3`

Then the inverse transform of F(s) is

`F(s) = (s^2+1)/((s-1)(s+1)(s-2)) \implies f(t) = -e^t + \frac{e^(-t)+5e^(2t)}3`

`\implies f(t) = \boxed{(5e^(3t) - 3e^(2t) + 1)/(3e^t)}`