Question

Chapter 8 Practice:

Compute the inverse Laplace transform of
$F(s) = \frac{s² +1}{(s-1)(s+1)(s - 2)}$

(​Cover up method)

Answer 1

Continue the given solution and solve for B and C :

$(s+1)\dfrac{s^2+1}{(s-1)(s+1)(s-2)} = (s+1)\left(-\dfrac1{s-1} + \dfrac B{s+1} + \dfrac C{s-2}\right)$

$\dfrac{s^2+1}{(s-1)(s-2)} = -\dfrac{s+1}{s-1} + B + \dfrac{C(s+1)}{s-2}$

$s=-1 \implies B = \dfrac13$

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

$\dfrac{s^2+1}{(s-1)(s+1)(s-2)} = -\dfrac1{s-1} + \dfrac13\times\dfrac1{s+1} + \dfrac53\times\dfrac1{s-2}$

Now take the inverse transform of each term.

$Y(s) = -\dfrac1{s-1} \implies y(t) = -e^t L^{-1}\left\{\dfrac1s\right\} = -e^t$

$Y(s) = \dfrac13\times\dfrac1{s+1} \implies y(t) = \dfrac{e^{-t}}3 L^{-1}\left\{\dfrac1s\right\} = \dfrac{e^{-t}}3$

$Y(s) = \dfrac53\times\dfrac1{s-2} \implies y(t) = \dfrac53e^{2t} L^{-1}\left\{\dfrac1s\right\} = \dfrac{5e^{2t}}3$

Then the inverse transform of F(s) is

$F(s) = \dfrac{s^2+1}{(s-1)(s+1)(s-2)} \implies f(t) = -e^t + \dfrac{e^{-t}+5e^{2t}}3$

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