Is this inverse laplace transforms correct?

Question

Is the following correct?

6/(3s + 2) = (6/3)*e^(-2t)

Answer 1

(6 / 3) [1 / [ s - (- 2/3) ]
2 [ 1 / s - ( - 2/3) ]
Transform is:-
2 e^[- (- 2/3) t ]
2 e^ [ (- 2/3) t ]

Answer 2

6/(3s + 2) = 6/3 * 1/(s + 2/3)

=2* 1/(s + 2/3)

The inverse of this is given from the transform...

e^{-a*t} * u(t) has the transform 1/ (s+a) s > - a

So you get 2e^{-2t/3} * u(t) as the inverse transform where u(t) is the heavy side step function

Answer 3

=2*1/(s+2/3)
The inverse would be 2*e^-2/3 x

Answer 4

those are the striking formula, be cautious with the a interior the numerator of sinh(at). The h stands for "hyperbolic" and it particularly is there to tell apart it from sin(x). they are distinctive applications that have no longer something to do with one yet another, different than the actuality that they ensue to have a number of the comparable residences. sinh(x) := (a million/2)(exp(x) - exp(-x)) ...........^ ability "defined as" further cosh(x) := (a million/2)(exp(x) + exp(-x))