Question

Find the inverse Laplace transforms of the following functions:

(a)

s + 3 /(s^2 + 6s + 13)

(b)

2s + 3/(s^2+6s+13)

Answer 1

(a)

$$L^{-1}(\dfrac{s+3}{s^2+6s+13})=L^{-1}(\dfrac{s-(-3)}{(s-(-3))^2+2^2})$$

$$=e^{-3t}\cos(2t)$$

(b)

$$L^{-1}(\dfrac{2s+3}{s^2+6s+13})=L^{-1}(\dfrac{2(s-(-3))-\dfrac{3}{2}(2)}{(s-(-3))^2+2^2})$$

$$=2L^{-1}(\dfrac{s-(-3)}{(s-(-3))^2+2^2})$$

$$-\dfrac{3}{2}L^{-1}(\dfrac{2}{(s-(-3))^2+2^2})$$

$$=2e^{-3t}\cos(2t)-\dfrac{3}{2}e^{-3t}\sin(2t)$$