Steply

Inverse Laplace transform of 5/((s - 2)(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 5 over the product (s − 2)(s + 3).

ℒ⁻¹{5(s − 2)(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.

5(s − 2)(s + 3) = As − 2 + Bs + 3

Times both sides by the whole bottom (s − 2)(s + 3).

Each letter keeps the factors it does not have.

5 = A(s + 3) + B(s − 2)

Put s = 2: every other letter is multiplied by 0.

5 = 5 × A.

s = 2: 5 = 5 × A

So A = 1.

A = 1

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

5 = −5 × B.

s = −3: 5 = −5 × B

So B = −1.

B = −1

So it splits into 1/(s − 2) − 1/(s + 3).

5(s − 2)(s + 3) = 1s − 2 − 1s + 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.

e^(at) → 1s − a

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

f(t) = e^(2t) − e^(−3t)

3 Check

Let's check: the table forwards turns our answer back into 1/(s − 2) − 1/(s + 3).

Right!

forwards: e^(2t) − e^(−3t) → 1s − 2 − 1s + 3

And the split is right: at s = 1, the question and the pieces both give −1.25.

Right!

s = 1: −1.25 and −1.25

Let's look back.

First: partial fractions.

Then: use the table backwards.

Last: check.

Answer
f(t) = e^(2t) − e^(−3t)

Your turn

Try this one yourself, then check:

ℒ⁻¹{7(s − 2)(s + 5)}
Show the answer
f(t) = e^(2t) − e^(−5t)

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