Find the Laplace transforms of the specified functions.
(a) f(t) = 6e5t + et3 - 9
(b) g(t) = 4cos(4t) - 9sin(4t) + 2cos(10t)
(c) h(t) = 3sinh(2t) + 3sin(2t)
(d) g(t) = et3 + cos(6t) - et3 cos(6t)
Okay, there's not actually a complete lot to do now other than go to the table, transform the particular functions up, place any constants back in and after that add or subtract the results.
We'll do these illustrations in a little more detail than is classically used as it is the first time we're using the tables.
(a) f(t) = 6e5t + et3 - 9
F(s) = (6 ×1/(s- (-5))) + (1/(s-3)) + (5 × 3!/(s3 + 1)0 - (9 × (1/5))
= (6 /(s + 5)) + (1/(s - 3)) + (30/s4) - (9/5)
(b) g(t) = 4cos(4t) - 9sin(4t) + 2cos(10t)
G(s) = (4 ×1/(s2 + 5)) - (9× (4/(s2- 42)) + (2 × (s/(s2 + 102)))
= (4s/(s2 + 16)) - (36/(s2 + 16)) + (2s/(s2 + 100))
(c) h(t) = 3sinh(2t) + 3 sin(2t)
H(s) = (3× (2/(s2- 22)) +(3× (2/(s2- 22))
= 6/(s2 - 4) + 6/(s2 - 4)
(d) g(t) = e3t + cos(6t) - e3t cos(6t)
G(t) = (1/(s- 3) + (s/(s2 + 62) - ((s-3)/((s-3) + 62))
= (1/(s- 3) + (s/(s2 + 36) - ((s-3)/((s-3) + 36)
Ensure that you pay attention to the difference among a "normal" trig function and hyperbolic functions. The simple difference among them is the "+ a2" for the "normal" trig functions turns into a "- a2" in the hyperbolic function! This is very simple to find in a hurry and not notice and grab the wrong formula. Whether you don't recall the definition of the hyperbolic functions notice the notes for the table.