APM3701 Assignment 1 (COMPLETE ANSWERS) 2025 (608471) - DUE 29 May 2025; 100% TRUSTED Complete, trusted solutions and explanations.

,APM3701 Assignment 1 (COMPLETE ANSWERS) 2025
(608471) - DUE 29 May 2025; 100% TRUSTED Complete,
trusted solutions and explanations.. Ensure your success
with us...

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]




Question 1a: Solving the Boundary Value Problem

Problem Statement:
Solve the boundary value problem:
uxy(x,y)=xy3,x,y≥0uxy(x,y)=xy3,x,y≥0

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

, Determine u(x,y)u(x,y) and verify the solution by
substitution.



Solution:

Step 1: Integrate uxy(x,y)=xy3uxy(x,y)=xy3 with respect
to xx

We start by integrating the given PDE with respect to xx:
uy(x,y)=∫xy3 dx=x2y32+C(y),uy(x,y)=∫xy3dx=2x2y3+C(y),

where C(y)C(y) is an arbitrary function of yy (the constant
of integration with respect to xx).

Step 2: Apply the boundary condition uy(0,y)=g(y)uy
(0,y)=g(y)

Substitute x=0x=0 into uy(x,y)uy(x,y):
uy(0,y)=C(y)=g(y)=y+sin⁡y.uy(0,y)=C(y)=g(y)=y+siny.

Thus, C(y)=y+sin⁡yC(y)=y+siny, and:
uy(x,y)=x2y32+y+sin⁡y.uy(x,y)=2x2y3+y+siny.

Step 3: Integrate uy(x,y)uy(x,y) with respect to yy

Now, integrate uy(x,y)uy(x,y) with respect to yy: