APM3701 ASSIGNMENT 01 SOLUTIONS 2026

APM3701 ASSIGNMENT 01 SOLUTIONS 2026
UNISA
Module: Partial Differential Equations
ASSIGNMENT 01 CHAPTER 1 – CHAPTER 5 OF STUDY GUIDE Assignment Unique
number : 192480 Due date: 25 May 2026



TAKE NOTE OF THE FOLLOWING: CHECK THIS IN YOUR SG • All numbers and sections
in bracket refer to the Study Guide (SG) and to the Prescribe Book (PB), unless specified
otherwise. • Please avoid repeating proofs of formulae and theorems already done in
the Study Guide and Prescribed Book, use or apply them directly instead. • No mark will
be awarded if you copy solution from past assignments and exam solutions or repeat
proof of formulae already done in the Study Guide and Prescribed Book

DUE: 25 MAY 2026

,QUESTION 1(a)

Solve

∂3 𝑢
(𝑥, 𝑦, 𝑡) = 2𝑥𝑡
∂𝑥 ∂𝑦 ∂𝑡


subject to

𝑦𝑡 2 𝑡 𝑦
𝑢(1, 𝑦, 𝑡) = + + +1
2 2 4
∂𝑢 𝑥𝑦
(𝑥, 𝑦, 0) =
∂𝑥 2
2
∂ 𝑢
(𝑥, 0, 𝑡) = 2𝑥𝑡 + 𝑥 − 𝑡
∂𝑥 ∂𝑡




Integrate with respect to 𝒙

Given

𝑢𝑥𝑦𝑡 = 2𝑥𝑡


Integrate w.r.t. 𝑥:

𝑢𝑦𝑡 = ∫ 2𝑥𝑡 𝑑𝑥 = 𝑥 2 𝑡 + 𝑓1 (𝑦, 𝑡)


where 𝑓1 (𝑦, 𝑡)is an arbitrary function.



Integrate with respect to 𝒚

𝑢𝑡 = ∫ (𝑥 2 𝑡 + 𝑓1 (𝑦, 𝑡)) 𝑑𝑦
𝑢𝑡 = 𝑥 2 𝑡𝑦 + 𝐹(𝑦, 𝑡) + 𝑓2 (𝑥, 𝑡)


where

𝐹𝑦 (𝑦, 𝑡) = 𝑓1 (𝑦, 𝑡)




Integrate with respect to 𝒕

, 𝑢 = ∫ (𝑥 2 𝑡𝑦 + 𝐹(𝑦, 𝑡) + 𝑓2 (𝑥, 𝑡)) 𝑑𝑡
𝑥 2 𝑦𝑡 2
𝑢= + 𝐺(𝑦, 𝑡) + 𝐻(𝑥, 𝑡) + 𝑓3 (𝑥, 𝑦)
2


We now use the conditions.



Use condition:

𝑦𝑡 2 𝑡 𝑦
𝑢(1, 𝑦, 𝑡) = + + +1
2 2 4


Substitute 𝑥 = 1:

𝑦𝑡 2 𝑦𝑡 2 𝑡 𝑦
+ 𝐺(𝑦, 𝑡) + 𝐻(1, 𝑡) + 𝑓3 (1, 𝑦) = + + +1
2 2 2 4


Thus,
𝑡 𝑦
𝐺(𝑦, 𝑡) + 𝐻(1, 𝑡) + 𝑓3 (1, 𝑦) = + +1
2 4


Take
𝑡 𝑦
𝐺(𝑦, 𝑡) = , 𝑓3 (1, 𝑦) = , 𝐻(1, 𝑡) = 1
2 4




Use condition:
𝑥𝑦
𝑢𝑥 (𝑥, 𝑦, 0) =
2


Differentiate:

𝑢𝑥 = 𝑥𝑦𝑡 2 + 𝐻𝑥 (𝑥, 𝑡) + 𝑓3𝑥 (𝑥, 𝑦)


Set 𝑡 = 0:

𝑢𝑥 (𝑥, 𝑦, 0) = 𝐻𝑥 (𝑥, 0) + 𝑓3𝑥 (𝑥, 𝑦)


Hence,