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Inverse Laplace transform of 6/(s(s + 3))

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 6/(s(s + 3)).

ℒ⁻¹{6s(s + 3)} =

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.

6s(s + 3) = As + Bs + 3

Times both sides by the whole bottom s(s + 3).

Each letter keeps the factors it does not have.

6 = A(s + 3) + Bs

Put s = 0: every other letter is multiplied by 0.

6 = 3 × A.

s = 0: 6 = 3 × A

So A = 2.

A = 2

Put s = −3: every other letter is multiplied by 0.

6 = −3 × B.

s = −3: 6 = −3 × B

So B = −2.

B = −2

So it splits into 2/s − 2/(s + 3).

6s(s + 3) = 2s − 2s + 3

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.

1 → 1s
e^(at) → 1s − a

Read the table backwards, piece by piece: 2 − 2e^(−3t).

f(t) = 2 − 2e^(−3t)

3 Check

Let's check: the table forwards turns our answer back into 2/s − 2/(s + 3).

Right!

forwards: 2 − 2e^(−3t) → 2s − 2s + 3

And the split is right: at s = 1, the question and the pieces both give 1.5.

Right!

s = 1: 1.5 and 1.5

Let's look back.

First: partial fractions.

Then: use the table backwards.

Last: check.

Answer
f(t) = 2 − 2e^(−3t)

Your turn

Try this one yourself, then check:

ℒ⁻¹{10s(s + 5)}
Show the answer
f(t) = 2 − 2e^(−5t)

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