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

,APM3701 Assignment 1 (COMPLETE ANSWERS)
2025 (608471) - DUE 29 May 2025; 100% TRUSTED
Complete, trusted solutions and explanations.
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 break the problem down step by step and solve each part
clearly.


Part (a): Solve uxy=xy3u_{xy} = xy^3uxy=xy3 with given
boundary conditions
We are given:
uxy=xy3,x,y≥0u_{xy} = xy^3, \quad x, y \geq 0uxy=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) = \cos x, \quad
u_y(0,y) = g(y) = y + \sin y.u(x,0)=f(x)=cosx,uy(0,y)=g(y)=y+siny.
Step 1: Integrate with respect to yyy
First, integrate uxy=xy3u_{xy} = xy^3uxy=xy3 with respect to
yyy:

,ux=∫xy3 dy=xy44+h(x),u_x = \int xy^3 \, dy = x \frac{y^4}{4} +
h(x),ux=∫xy3dy=x4y4+h(x),
where h(x)h(x)h(x) is an arbitrary function of xxx (since
differentiation with respect to yyy eliminates constants that are
functions of xxx).
Step 2: Integrate with respect to xxx
Now, integrate both sides with respect to xxx:
u=∫(xy44+h(x))dx.u = \int \left( x \frac{y^4}{4} + h(x) \right)
dx.u=∫(x4y4+h(x))dx.
Breaking it down:
u=x22⋅y44+∫h(x)dx+C(y).u = \frac{x^2}{2} \cdot \frac{y^4}{4} + \
int h(x) dx + C(y).u=2x2⋅4y4+∫h(x)dx+C(y).
Simplifying:
u=x2y48+H(x)+C(y),u = \frac{x^2 y^4}{8} + H(x) + C(y),u=8x2y4
+H(x)+C(y),
where H(x)=∫h(x)dxH(x) = \int h(x) dxH(x)=∫h(x)dx and
C(y)C(y)C(y) is an arbitrary function of yyy.
Step 3: Apply Boundary Condition u(x,0)=cos⁡xu(x,0) = \cos
xu(x,0)=cosx
Substituting y=0y = 0y=0:
u(x,0)=x2(0)48+H(x)+C(0)=cos⁡x.u(x, 0) = \frac{x^2 (0)^4}{8} +
H(x) + C(0) = \cos x.u(x,0)=8x2(0)4+H(x)+C(0)=cosx.

, This simplifies to:
H(x)+C(0)=cos⁡x.H(x) + C(0) = \cos x.H(x)+C(0)=cosx.
Let C(0)=C0C(0) = C_0C(0)=C0, then:
H(x)=cos⁡x−C0.H(x) = \cos x - C_0.H(x)=cosx−C0.
Step 4: Apply Boundary Condition uy(0,y)=y+sin⁡yu_y(0,y) = y
+ \sin yuy(0,y)=y+siny
Computing uyu_yuy:
uy=x28⋅4y3+C′(y)=x2y32+C′(y).u_y = \frac{x^2}{8} \cdot 4y^3 +
C'(y) = \frac{x^2 y^3}{2} + C'(y).uy=8x2⋅4y3+C′(y)=2x2y3+C′(y).
Setting x=0x = 0x=0:
C′(y)=y+sin⁡y.C'(y) = y + \sin y.C′(y)=y+siny.
Integrating both sides:
C(y)=∫(y+sin⁡y)dy=y22−cos⁡y+C1.C(y) = \int (y + \sin y) dy = \
frac{y^2}{2} - \cos y + C_1.C(y)=∫(y+siny)dy=2y2−cosy+C1.
Step 5: Solve for u(x,y)u(x, y)u(x,y)
Substituting back:
u(x,y)=x2y48+(cos⁡x−C0)+(y22−cos⁡y+C1).u(x, y) = \frac{x^2 y^4}
{8} + (\cos x - C_0) + \left(\frac{y^2}{2} - \cos y + C_1\
right).u(x,y)=8x2y4+(cosx−C0)+(2y2−cosy+C1).
To satisfy C(0)=C0C(0) = C_0C(0)=C0, we set C0=C1C_0 = C_1C0
=C1, so: