Steply

Inverse Laplace transform of 1/((s - 1)(s - 2))

Step by step, with the reason for every step · The inverse Laplace transform · Differential Equations

Step by step

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)

Your turn

Try this one yourself, then check:

ℒ⁻¹{3(s − 3)(s − 6)}
Show the answer
f(t) = −e^(3t) + e^(6t)

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