The question
Question: find the inverse Laplace transform of 1 over s.
ℒ⁻¹{1s} =
The inverse transform goes back: which function of t has this transform?
We read the table backwards.
ℒ⁻¹: back from s to t
1 Use the table backwards
Each piece is already a row of the table.
1 → 1s
Read the table backwards, piece by piece: 1.
f(t) = 1
2 Check
Let's check: the table forwards turns our answer back into 1 over s.
Right!
forwards: 1 → 1s
Let's look back.
First: use the table backwards.
Last: check.
▶ Watch it solved
The question
Question: find the inverse Laplace transform of 5/s³.
ℒ⁻¹{5s3} =
The inverse transform goes back: which function of t has this transform?
We read the table backwards.
ℒ⁻¹: back from s to t
1 Make it fit the table
Make each piece look exactly like a row of the table.
tⁿ needs n!
on top.
The extra number waits in front.
ℒ⁻¹ = 52 × 2!s3
tn → n!sn + 1
2 Use the table backwards
Careful!
2!
= 2 belongs on top.
So the number in front is 5/2.
Careful: not 5/s³ → 5t² ✗
Read the table backwards, piece by piece: 5/2 × t².
f(t) = 52t2
3 Check
Let's check: the table forwards turns our answer back into 5/s³.
Right!
forwards: 52t2 → 5s3
Let's look back.
First: make it fit the table.
Then: use the table backwards.
Last: check.
▶ Watch it solved
The question
Question: find the inverse Laplace transform of 1 over the product (s − 1)(s − 2).
ℒ⁻¹{1(s − 1)(s − 2)} =
The inverse transform goes back: which function of t has this transform?
We read the table backwards.
ℒ⁻¹: back from s to t
1 Partial fractions
Split it into simple fractions, one for each factor at the bottom.
1(s − 1)(s − 2) = As − 1 + Bs − 2
Times both sides by the whole bottom (s − 1)(s − 2).
Each letter keeps the factors it does not have.
1 = A(s − 2) + B(s − 1)
Put s = 1: every other letter is multiplied by 0.
1 = −1 × A.
s = 1: 1 = −1 × A
So A = −1.
A = −1
Put s = 2: every other letter is multiplied by 0.
1 = 1 × B.
s = 2: 1 = 1 × B
So B = 1.
B = 1
So it splits into −1/(s − 1) + 1/(s − 2).
1(s − 1)(s − 2) = −1s − 1 + 1s − 2
Careful!
A product of transforms is not the transform of a product.
Split it first.
Careful: not ℒ⁻¹{1(s − a)(s − b)} = e^(at) × e^(bt) ✗
2 Use the table backwards
Each piece is already a row of the table.
e^(at) → 1s − a
Read the table backwards, piece by piece: −e^(t) + e^(2t).
f(t) = −e^(t) + e^(2t)
3 Check
Let's check: the table forwards turns our answer back into −1/(s − 1) + 1/(s − 2).
Right!
forwards: −e^(t) + e^(2t) → −1s − 1 + 1s − 2
And the split is right: at s = 3, the question and the pieces both give 0.5.
Right!
s = 3: 0.5 and 0.5
Let's look back.
First: partial fractions.
Then: use the table backwards.
Last: check.
Answer f(t) = −e^(t) + e^(2t)
▶ Watch it solved
