APM3701 Assignment 1 (COMPLETE ANSWERS) 2025 (608471) - DUE 29 May 2025;100% CORRECT AND TRUSTED SOLUTIONS

,APM3701 Assignment 1 (COMPLETE ANSWERS) 2025 (608471) - DUE 29
May 2025;100% CORRECT AND TRUSTED SOLUTIONS

QUESTION 1
Solve the following (initial)-boundary value problem,

a.
uxy(x,y)=xy3,x,y≥0.u_{xy}(x, y) = xy^3, \quad x, y \geq 0.uxy
(x,y)=xy3,x,y≥0.
u(x,0)=f(x),u(x, 0) = f(x),u(x,0)=f(x), and uy(0,y)=g(y).u_y(0, y) =
g(y).uy(0,y)=g(y).
Determine u(x,y),u(x, y),u(x,y), if f(x)=cos⁡xf(x) = \cos xf(x)=cosx
and g(y)=y+sin⁡y.g(y) = y + \sin y.g(y)=y+siny.
(Check your answer by substituting, and explain all the steps
clearly)
(15 Marks)
𝑮𝒊𝒗𝒆𝒏:

 𝑢𝑥𝑦(𝑥, 𝑦) = 𝑥𝑦3𝑢{𝑥𝑦}(𝑥,𝑦) = 𝑥𝑦 3𝑢𝑥𝑦(𝑥,𝑦) = 𝑥𝑦3, 𝑓𝑜𝑟 𝑥, 𝑦 ≥ 0𝑥, 𝑦 ≥
0𝑥, 𝑦 ≥ 0
 𝑢(𝑥, 0) = 𝑓(𝑥) = cos 𝑥𝑢(𝑥, 0) = 𝑓(𝑥) = \𝑐𝑜𝑠 𝑥𝑢(𝑥, 0) = 𝑓(𝑥) = 𝑐𝑜𝑠𝑥
 𝑢𝑦(0, 𝑦) = 𝑔(𝑦) = 𝑦 + sin 𝑦𝑢𝑦(0,𝑦) = 𝑔(𝑦) = 𝑦 + \𝑠𝑖𝑛 𝑦𝑢𝑦(0, 𝑦) =
𝑔(𝑦) = 𝑦 + 𝑠𝑖𝑛𝑦

𝑊𝑒 𝑎𝑟𝑒 𝑡𝑜 𝒇𝒊𝒏𝒅 𝑢(𝑥, 𝑦)𝑢(𝑥, 𝑦)𝑢(𝑥, 𝑦), 𝑎𝑛𝑑 𝒗𝒆𝒓𝒊𝒇𝒚 𝑡ℎ𝑒 𝑟𝑒𝑠𝑢𝑙𝑡.



✅ 𝑺𝒕𝒆𝒑 − 𝒃𝒚 − 𝑺𝒕𝒆𝒑 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏

𝑺𝒕𝒆𝒑 𝟏: 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒆 𝒕𝒉𝒆 𝑷𝑫𝑬

𝑊𝑒 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑚𝑖𝑥𝑒𝑑 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒:

𝑢𝑥𝑦(𝑥, 𝑦) = 𝑥𝑦3𝑢{𝑥𝑦}(𝑥,𝑦) = 𝑥𝑦 3𝑢𝑥𝑦(𝑥,𝑦) = 𝑥𝑦3

𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒚𝒚𝒚 𝑓𝑖𝑟𝑠𝑡:

, 𝑢𝑥(𝑥, 𝑦) = ∫ 𝑢𝑥𝑦(𝑥, 𝑦) 𝑑𝑦 = ∫ 𝑥𝑦3 𝑑𝑦 = 𝑥∫ 𝑦3 𝑑𝑦 = 𝑥 ⋅ 𝑦44 + 𝐴(𝑥)𝑢𝑥(𝑥,𝑦)
= ∫ 𝑢{𝑥𝑦}(𝑥,𝑦), 𝑑𝑦 = ∫ 𝑥𝑦 3 , 𝑑𝑦 = 𝑥 ∫ 𝑦 3 , 𝑑𝑦
= 𝑥 ⋅\𝑓𝑟𝑎𝑐{𝑦 4 }{4} + 𝐴(𝑥)𝑢𝑥(𝑥, 𝑦) = ∫ 𝑢𝑥𝑦(𝑥, 𝑦)𝑑𝑦 = ∫ 𝑥𝑦3𝑑𝑦
= 𝑥∫ 𝑦3𝑑𝑦 = 𝑥 ⋅ 4𝑦4 + 𝐴(𝑥)

𝑆𝑜,

𝑢𝑥(𝑥, 𝑦) = 14𝑥𝑦4 + 𝐴(𝑥)𝑢𝑥(𝑥,𝑦) = \𝑓𝑟𝑎𝑐{1}{4}𝑥𝑦 4 + 𝐴(𝑥)𝑢𝑥(𝑥, 𝑦)
= 41𝑥𝑦4 + 𝐴(𝑥)


𝑺𝒕𝒆𝒑 𝟐: 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒆 𝒘𝒊𝒕𝒉 𝒓𝒆𝒔𝒑𝒆𝒄𝒕 𝒕𝒐 𝒙𝒙𝒙

𝑁𝑜𝑤 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑢𝑥(𝑥, 𝑦)𝑢𝑥(𝑥,𝑦)𝑢𝑥(𝑥,𝑦)𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥𝑥𝑥 𝑡𝑜 𝑔𝑒𝑡 𝑢(𝑥, 𝑦)𝑢(𝑥, 𝑦)𝑢(𝑥, 𝑦):

𝑢(𝑥, 𝑦) = ∫ 𝑢𝑥(𝑥, 𝑦) 𝑑𝑥 = ∫ (14𝑥𝑦4 + 𝐴(𝑥))𝑑𝑥𝑢(𝑥, 𝑦) = ∫ 𝑢𝑥(𝑥,𝑦), 𝑑𝑥
= ∫ ( \𝑓𝑟𝑎𝑐{1}{4}𝑥𝑦 4 + 𝐴(𝑥))𝑑𝑥𝑢(𝑥, 𝑦) = ∫ 𝑢𝑥(𝑥, 𝑦)𝑑𝑥
= ∫ (41𝑥𝑦4 + 𝐴(𝑥))𝑑𝑥

𝐵𝑟𝑒𝑎𝑘 𝑖𝑡 𝑖𝑛𝑡𝑜 𝑡𝑤𝑜 𝑝𝑎𝑟𝑡𝑠:

𝑢(𝑥, 𝑦) = 14𝑦4∫ 𝑥 𝑑𝑥 + ∫ 𝐴(𝑥) 𝑑𝑥 = 14𝑦4 ⋅ 𝑥22 + ∫ 𝐴(𝑥) 𝑑𝑥 + 𝐵(𝑦)𝑢(𝑥, 𝑦)
= \𝑓𝑟𝑎𝑐{1}{4}𝑦 4 ∫ 𝑥 , 𝑑𝑥 + ∫ 𝐴(𝑥), 𝑑𝑥
= \𝑓𝑟𝑎𝑐{1}{4}𝑦 4 ⋅\𝑓𝑟𝑎𝑐{𝑥 2 }{2} + ∫ 𝐴(𝑥), 𝑑𝑥 + 𝐵(𝑦)𝑢(𝑥, 𝑦)
= 41𝑦4∫ 𝑥𝑑𝑥 + ∫ 𝐴(𝑥)𝑑𝑥 = 41𝑦4 ⋅ 2𝑥2 + ∫ 𝐴(𝑥)𝑑𝑥 + 𝐵(𝑦)

𝑆𝑜:

𝑢(𝑥, 𝑦) = 18𝑥2𝑦4 + 𝐹(𝑥) + 𝐵(𝑦)𝑢(𝑥, 𝑦)
4
= \𝑓𝑟𝑎𝑐{1}{8}𝑥 2𝑦 + 𝐹(𝑥) + 𝐵(𝑦)𝑢(𝑥, 𝑦)
= 81𝑥2𝑦4 + 𝐹(𝑥) + 𝐵(𝑦)

𝑊ℎ𝑒𝑟𝑒:

 𝐹(𝑥) = ∫ 𝐴(𝑥) 𝑑𝑥𝐹(𝑥) = ∫ 𝐴(𝑥), 𝑑𝑥𝐹(𝑥) = ∫ 𝐴(𝑥)𝑑𝑥
 𝐵(𝑦)𝐵(𝑦)𝐵(𝑦)𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦𝑦𝑦 𝑜𝑛𝑙𝑦 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤. 𝑟. 𝑡. 𝑥𝑥𝑥)