Inverse Laplace transform of (s - 3)/(s^2 + 9)

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

Step by step

Question: find the inverse Laplace transform of (s − 3)/(s² + 9).

ℒ⁻¹{s − 3s² + 9} =

The inverse transform goes back: which function of t has this transform?

We read the table backwards.

ℒ⁻¹: back from s to t

Split the top: the s part and the number part, each over s² + 9.

= ss² + 9 − 3s² + 9

Each piece is already a row of the table.

cos(bt) → ss² + b²

sin(bt) → bs² + b²

Read the table backwards, piece by piece: cos(3t) − sin(3t).

f(t) = cos(3t) − sin(3t)

Let's check: the table forwards turns our answer back into s/(s² + 9) − 3/(s² + 9).

Right!

forwards: cos(3t) − sin(3t) → ss² + 9 − 3s² + 9

Let's look back.

First: split it up.

Then: use the table backwards.

Last: check.

Answer

f(t) = cos(3t) − sin(3t)

Try this one yourself, then check:

ℒ⁻¹{s − 2s² + 4}

Show the answer

f(t) = cos(2t) − sin(2t)