APM3701 Assignment 1 (COMPLETE ANSWERS) 2025 (608471) - DUE 29 May 2025

APM3701 Assignment 1
(COMPLETE ANSWERS) 2025
(608471) - DUE 29 May 2025
100% GUARANTEEED

,APM3701 Assignment 1 (COMPLETE
ANSWERS) 2025 (608471) - DUE 29 May
2025
QUESTION 1 Solve the following (initial)-boundary
value problem, a. uxy (x, y) = xy3, x, y 0. u (x, 0)
= f (x) , and uy (0, y) = g (y) . Determine u (x, y) ,
if f (x) = cosx and g (y) = y+sin y. (Check your
answer by substituting, and explain all the steps
clearly) (15 Marks) b. xux + yuy = yu u (2x2, x) =
x2 − 1. (1) (Check your answer by substituting
before applying the boundary condition, and
explain all the steps clearly) (15 Marks) [30 Marks]
Let's solve both parts of the problem step by step.


Part (a): Solving u xy=x y 3
We have the partial differential equation:
3
u xy =x y , x , y ≥ 0

with boundary conditions:
u ( x , 0 )=f ( x )=cos x , u y ( 0 , y )=g ( y )= y +sin y .

Step 1: Integrate w.r.t. x

Integrating u xy=x y 3 with respect to x :
2
3 x 3
u y =∫ x y dx= y +C ( y ) ,
2

where C ( y ) is an arbitrary function of y .

, Step 2: Apply u y ( 0 , y )=g ( y )
We substitute x=0 :
2
0 3
u y ( 0 , y )= y +C ( y ) =C ( y )= y +sin y .
2

Thus, C ( y ) = y+ sin y, and we now have:
x2 3
uy = y + y +sin y .
2

Step 3: Integrate w.r.t. y

Integrating both sides with respect to y :

u=∫ ( x2 3
2 )
y + y +sin y dy .

2 4 2
x y y
u= ⋅ + −cos y + D ( x ) .
2 4 2
x2 4 y2
u= y + −cos y + D ( x ) .
8 2

Step 4: Apply u ( x , 0 )=f ( x )=cosx
Substituting y=0:
2
x2 4 ( 0)
u ( x , 0 )= (0 ) + −cos ( 0 )+ D ( x )=D ( x )−1 .
8 2

Since u ( x , 0 )=cos x , we equate:
D ( x ) −1=cos x ⇒ D ( x ) =cos x+ 1.

Final Solution:
x2 4 y2
u ( x , y )= y + −cos y +cos x +1 .
8 2

Verification:
 Compute u xy and verify it satisfies u xy=x y 3.
 Check boundary conditions.