Step by step
The question
Question: find the inverse Laplace transform of (2s + 1)/(s² + 4).
ℒ⁻¹{2s + 1s2 + 4} =
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 Split it up
Split the top: the s part and the number part, each over s² + 4.
= 2ss2 + 4 + 1s2 + 4
2 Make it fit the table
Make each piece look exactly like a row of the table.
sin needs b on top.
The extra number waits in front.
ℒ⁻¹ = 2ss2 + 4 + 12 × 2s2 + 4
cos(bt) → ss² + b²
sin(bt) → bs² + b²
3 Use the table backwards
Careful!
sin(2t) has 2 on top.
With 1 on top we get 1/2 sin(2t).
Careful: not 1/(s² + 4) → 1sin(2t) ✗
Read the table backwards, piece by piece: 2cos(2t) + ½sin(2t).
f(t) = 2cos(2t) + 12sin(2t)
4 Check
Let's check: the table forwards turns our answer back into 2s/(s² + 4) + 1/(s² + 4).
Right!
forwards: 2cos(2t) + 12sin(2t) → 2ss2 + 4 + 1s2 + 4
Let's look back.
First: split it up.
Then: make it fit the table.
Then: use the table backwards.
Last: check.
Answer
f(t) = 2cos(2t) + 12sin(2t)