Question

Inveres laplace transform (3s-14)/s^2-4s+8​

Answer 1

Complete the square in the denominator.

`s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4`

Rewrite the given transform as

`(3s-14)/(s^2-4s+8) = (3(s-2) - 8)/((s-2)^2+4) = 3*(s-2)/((s-2)^2+2^2) - 4*(2)/((s-2)^2+2^2)`

Now take the inverse transform:

`L^(-1)_t\left\{3*(s-2)/((s-2)^2+2^2) - 4*(2)/((s-2)^2+2^2)\right\} \\\\ 3L^(-1)_t\left\{(s-2)/((s-2)^2+2^2)\right\} - 4L^(-1)_t\left\{(2)/((s-2)^2+2^2)\right\} \\\\ 3e^(2t) L^(-1)_t\left\{\frac s{s^2+2^2}\right\} - 4e^(2t) L^(-1)_t\left\{(2)/(s^2+2^2)\right\} \\\\ \boxed{3e^(2t) \cos(2t) - 4e^(2t) \sin(2t)}`