Step by step
The question
Question: find the Laplace transform of 2cos(4t) − 5t.
ℒ{2cos(4t) − 5t} =
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.
cos(bt) → ss² + b²
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.
= 2ℒ{cos(4t)} − 5ℒ{t}
ℒ{af + bg} = aℒ{f} + bℒ{g}
3 Use the table
Each part by its row: cos(4t) gives s/(s² + 16), t gives 1/s².
= 2 × ss2 + 16 − 5 × 1s2
4 Tidy up
Multiply the numbers: 2s/(s² + 16) − 5/s².
= 2ss2 + 16 − 5s2
5 Check
Let's check: read the table backwards.
2s/(s² + 16) − 5/s² comes from 2cos(4t) − 5t: the question.
Right!
backwards: 2ss2 + 16 − 5s2 → 2cos(4t) − 5t
Let's look back.
First: the idea.
Then: split it up.
Then: use the table.
Then: tidy up.
Last: check.
Answer
= 2ss2 + 16 − 5s2