Step by step
The question
Question: find the Laplace transform of 6e^(−t) − 2.
ℒ{6e^(−t) − 2} =
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.
= 6ℒ{e^(−t)} − 2ℒ{1}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: e^(−t) gives 1/(s + 1), 1 gives 1 over s.
= 6 × 1s + 1 − 2 × 1s
Careful!
e to the at gives s minus a.
Here a is −1, so it is s + 1.
Careful: not ℒ{e^(−t)} = 1s − 1 ✗
4 Tidy up
Multiply the numbers: 6/(s + 1) − 2 over s.
= 6s + 1 − 2s
5 Check
Let's check: read the table backwards.
6/(s + 1) − 2/s comes from 6e^(−t) − 2: the question.
Right!
backwards: 6s + 1 − 2s → 6e^(−t) − 2
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 6s + 1 − 2s