Step by step
The question
Question: find the Laplace transform of 3e^(2t) + 4.
ℒ{3e^(2t) + 4} =
1 The idea
The Laplace transform turns a function of t into a function of s.
It is an integral, but we almost never work it out: we use a table made from it.
ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt
The table rows we need.
e^(at) → 1s − a
1 → 1s
2 Split it up
The transform of a sum is the sum of the transforms.
A number in front waits outside.
= 3ℒ{e^(2t)} + 4ℒ{1}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: e^(2t) gives 1/(s − 2), 1 gives 1 over s.
= 3 × 1s − 2 + 4 × 1s
Careful!
e to the at gives s minus a.
Here a is 2, so it is s − 2.
Careful: not ℒ{e^(2t)} = 1s + 2 ✗
4 Tidy up
Multiply the numbers: 3/(s − 2) + 4 over s.
= 3s − 2 + 4s
5 Check
Let's check: read the table backwards.
3/(s − 2) + 4/s comes from 3e^(2t) + 4: the question.
Right!
backwards: 3s − 2 + 4s → 3e^(2t) + 4
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 3s − 2 + 4s