Step by step
The question
Question: find the Laplace transform of 3t² − 4e^(5t) + 2.
ℒ{3t2 − 4e^(5t) + 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.
tn → n!sn + 1
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{t2} − 4ℒ{e^(5t)} + 2ℒ{1}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: t² gives 2/s³, e^(5t) gives 1/(s − 5), 1 gives 1 over s.
= 3 × 2!s3 − 4 × 1s − 5 + 2 × 1s
Careful!
The top is n!: 2!
= 2, not 1.
Careful: not ℒ{t²} = 1s3 ✗
4 Tidy up
Multiply the numbers: 6/s³ − 4/(s − 5) + 2 over s.
= 6s3 − 4s − 5 + 2s
5 Check
Let's check: read the table backwards.
6/s³ − 4/(s − 5) + 2/s comes from 3t² − 4e^(5t) + 2: the question.
Right!
backwards: 6s3 − 4s − 5 + 2s → 3t2 − 4e^(5t) + 2
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 6s3 − 4s − 5 + 2s