The question
Question: find the Laplace transform of 1.
ℒ{1} =
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.
1 → 1s
2 Use the table
Each part by its row: 1 gives 1 over s.
ℒ{1} = 1s
3 Check
Let's check: read the table backwards.
1/s comes from 1: the question.
Right!
backwards: 1s → 1
Let's look back.
First: the idea.
Then: use the table.
Last: check.
▶ Watch it solved
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.
▶ Watch it solved
The question
Question: find the Laplace transform of e^(2t), from the definition.
ℒ{e^(2t)} =
1 Use the definition
The definition: multiply by e to the minus st, and integrate from 0 to infinity.
ℒ{f(t)} = ∫ from 0 to ∞ of e^(−st) f(t) dt
Here f(t) is e^(2t).
ℒ{e^(2t)} = ∫ from 0 to ∞ of e^(−st)e^(2t) dt
Join the powers: −st + 2t is −(s − 2)t.
= ∫ from 0 to ∞ of e^(−(s − 2)t) dt
e^(a) × e^(b) = e^(a + b)
2 Integrate
Integrate: divide by the number in front of t, which is −(s − 2).
= [−e^(−(s − 2)t)s − 2] from 0 to ∞
When t goes to infinity, e to a big minus power goes to 0.
That needs s > 2.
t → ∞: e^(−(s − 2)t) → 0 when s > 2
Careful!
If s is 2 or less, the power is not negative, and the integral never stops growing.
Careful: only for s > 2
3 Top minus bottom
At infinity: 0.
At 0: e⁰ = 1, so −1/(s − 2).
Top minus bottom.
= 0 − (−1s − 2)
So ℒ = 1/(s − 2), for s > 2.
= 1s − 2
4 Check
Let's check with the table row: e to the at gives 1/(s − a), with a = 2.
The same.
Right!
table: e^(at) → 1s − a, a = 2
Let's look back.
First: use the definition.
Then: integrate.
Then: top minus bottom.
Last: check.
▶ Watch it solved
