Step by step
The question
Question: find the Laplace transform of t³ + 2t.
ℒ{t3 + 2t} =
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.
tn → n!sn + 1
2 Split it up
The transform of a sum is the sum of the transforms.
A number in front waits outside.
= ℒ{t3} + 2ℒ{t}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: t³ gives 6/s⁴, t gives 1/s².
= 3!s4 + 2 × 1s2
Careful!
The top is n!: 3!
= 6, not 1.
Careful: not ℒ{t³} = 1s4 ✗
4 Tidy up
Multiply the numbers: 6/s⁴ + 2/s².
= 6s4 + 2s2
5 Check
Let's check: read the table backwards.
6/s⁴ + 2/s² comes from t³ + 2t: the question.
Right!
backwards: 6s4 + 2s2 → t3 + 2t
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 6s4 + 2s2