APM2611 Assignment 3 Solutions 2026

APM2611 Assignment 3 Solutions 2026

Full Solutions By TA tutor iQ level

UNISA

DUE DATE: 12-08-2026

,APM2611 Assignment 03 – Full Solutions



QUESTION 1 — Laplace Transforms

1.1 f(t) = te⁻³ᵗ
𝑛!
Use the First Shifting Theorem: ℒ{𝑡 𝑛 𝑒 𝑎𝑡 } = (𝑠−𝑎)𝑛+1

Here n = 1, a = −3:
1! 1
ℒ{𝑡𝑒 −3𝑡 } = =
( )1+1 (𝑠 + 3)2
1
ℒ{𝑡𝑒 −3𝑡 } =
(𝑠 + 3)2




𝒕
1.2 𝒇(𝒕) = ∫𝟎 𝝉 𝒆−𝟐𝝉 𝐬𝐢𝐧⁡(𝟑𝝉) 𝒅𝝉

Step 1: Find ℒ{𝜏𝑒 −2𝜏 sin⁡(3𝜏)}first.
3
Start with ℒ{sin⁡(3𝑡)} = 𝑠2 +9

Apply First Shifting Theorem (replace s with s+2):
3
ℒ{𝑒 −2𝑡 sin⁡(3𝑡)} =
(𝑠+2)2+9


𝑑
Apply the multiplication-by-t formula ℒ{𝑡 𝑔(𝑡)} = − 𝑑𝑠 𝐺(𝑠):

𝑑 3
ℒ{𝑡𝑒 −2𝑡 sin⁡(3𝑡)} = − [ ]
𝑑𝑠 (𝑠+2)2 +9
−2(𝑠 + 2) 6(𝑠 + 2)
= −3 ⋅ 2 2
=
[(𝑠 + 2) + 9] [(𝑠 + 2)2 + 9]2


𝑡 𝐺(𝑠)
Step 2: Apply the Convolution/Integration theorem: ℒ {∫0 𝑔 (𝜏) 𝑑𝜏} =
𝑠

𝑡
6(𝑠 + 2)
ℒ {∫ 𝜏 𝑒 −2𝜏 sin⁡(3𝜏) 𝑑𝜏} =
0 𝑠[(𝑠 + 2)2 + 9]2

, 1.3 f(t) = te³ᵗcos(4t)
𝑠
Step 1: ℒ{cos⁡(4𝑡)} = 𝑠2 +16

Step 2: First shifting theorem (replace s → s−3):
𝑠−3
ℒ{𝑒 3𝑡 cos⁡(4𝑡)} =
(𝑠 − 3)2 + 16


𝑑
Step 3: Multiply by t: ℒ{𝑡𝑔(𝑡)} = − 𝑑𝑠 𝐺(𝑠)

𝑑 𝑠−3
ℒ{𝑡𝑒 3𝑡 cos⁡(4𝑡)} = − [ ]
𝑑𝑠 (𝑠 − 3)2 + 16

𝑢
Let u = s−3. Differentiate 𝑢2 +16using quotient rule:

𝑑 𝑢 (𝑢2 + 16) − 𝑢 ⋅ 2𝑢 16 − 𝑢2
[ 2 ]= =
𝑑𝑢 𝑢 + 16 (𝑢2 + 16)2 (𝑢2 + 16)2
16 − (𝑠 − 3)2 (𝑠−3)2−16
ℒ{𝑡𝑒 3𝑡 cos⁡(4𝑡)} = − =
[(𝑠 − 3)2 + 16]2 [(𝑠 − 3)2 + 16]2
(𝑠−3)2−16
ℒ{𝑡𝑒 3𝑡 cos⁡(4𝑡)} =
[(𝑠 − 3)2 + 16]2




1.4 f(t) = (t − 3)U(t − 3)

Use the Second Shifting Theorem: ℒ{𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)} = 𝑒 −𝑎𝑠 𝐺(𝑠)
1
Here g(t−3) = (t−3), so g(t) = t, and 𝐺(𝑠) = ℒ{𝑡} = 𝑠2 , with a = 3:

𝑒 −3𝑠
ℒ{(𝑡 − 3)𝑈(𝑡 − 3)} = 2
𝑠


𝒕
1.5 𝒉(𝒕) = 𝒆𝟐𝒕 ∫𝟎 𝐬𝐢𝐧⁡(𝝉) 𝒅𝝉
𝑡 1 1 1
Step 1: Find ℒ {∫0 sin ⁡(𝜏) 𝑑𝜏} = 𝑠 ⋅ 𝑠2 +1 = 𝑠(𝑠2 +1)

Step 2: Apply First Shifting Theorem (multiply by e²ᵗ → replace s with s−2):

1
ℒ{ℎ(𝑡)} =
(𝑠 − 2)[(𝑠 − 2)2 + 1]