Step by step
The question
Question: find the Laplace transform of e^(2t), from the definition.
ℒ{e^(2t)} =
1 Use the definition
The definition: multiply by e to the minus st, and integrate from 0 to infinity.
ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt
Here f(t) is e^(2t).
ℒ{e^(2t)} = ∫ from 0 to ∞ of e^(−st)e^(2t) dt
Join the powers: −st + 2t is −(s − 2)t.
= ∫ from 0 to ∞ of e^(−(s − 2)t) dt
e^(a) × e^(b) = e^(a + b)
2 Integrate
Integrate: divide by the number in front of t, which is −(s − 2).
= [−e^(−(s − 2)t)s − 2] from 0 to ∞
When t goes to infinity, e to a big minus power goes to 0.
That needs s > 2.
t → ∞: e^(−(s − 2)t) → 0 when s > 2
Careful!
If s is 2 or less, the power is not negative, and the integral never stops growing.
Careful: only for s > 2
3 Top minus bottom
At infinity: 0.
At 0: e⁰ = 1, so −1/(s − 2).
Top minus bottom.
= 0 − (−1s − 2)
So ℒ = 1/(s − 2), for s > 2.
= 1s − 2
4 Check
Let's check with the table row: e to the at gives 1/(s − a), with a = 2.
The same.
Right!
table: e^(at) → 1s − a, a = 2
Let's look back.
First: use the definition.
Then: integrate.
Then: top minus bottom.
Last: check.
Answer
= 1s − 2