Find the inverse Laplace transform?

Question

Find the inverse Laplace transform of F(s)=(8s^2-4s+12)/(s(s^2+4)).

Answer: f(t)=3-2 sin 2t+5 cos 2t

A/s+(Bs+C)/(s^2+4)

8s^2-4s+12=A(s^2+4)+(Bs+C)(s)=As^2+4A+Bs^2+Cs=s^2(A+B)+Cs+4A

8=A+B

C=-4

A=3

B=5

L^-1 (3/s)+L^-1 ((5s-4)/(s^2+4))

=3+ (now I'm stucked)

Answer 1

Write [(8s² - 4s +12)/(s(s² + 4)) = A/s + (Bs + C)/(s² + 4)].

Multiplying both sides of equation in square brackets above by s(s² + 4) gives

(8s² - 4s +12) = A(s² + 4) + (Bs + C)s equal to (A + B)s² +Cs + 4A.

Then A + B = 8

C = -4

4A = 12

A = 3

B = 5

Next, obtain (8s² - 4s + 12)/(s(s² + 4)) = 3/s + (5s - 4)/(s² + 4).

Finally, write L^-1{ 3/s + 5s/(s² + 4) - 4/(s² + 4) } which gives 3L^-1{1/s} + 5L^-1{s/(s² + 4)} - 2L^-1{2/(s² + 4)}.

A table of Laplace Transforms will give the Inverse Laplace Transform as 3 + 5cos 2t - 2sin2t.