Question

the inverse Laplace transformation of (3s-12)/(s^2-4s+8)

Answer 1

Agh.

Answer 2

Calc 2?

Answer 3

ok lets do it u know the basics right?

Answer 4

you have
\[L^{-1}\frac{ 3s-12 }{ s^2-4s+8 }=L^{-1}\frac{ 3(s-2)-6 }{ (s-2)^2+4}\\=3L^{-1}\frac{ s-2 }{ (s-2)^2+4 }-6L^{-1}\frac{ 1 }{ (s-2)^2+4 }=3*e^{2t}\cos(2t)-3e^{2t}\sin(2t)\]

Answer 5

hope this helps @exanroner318