,APM3701 Assignment 1 (COMPLETE ANSWERS) 2025
(608471) - DUE 29 May 2025; 100% TRUSTED Complete,
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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]
Part (a) - Solving the PDE uxy(x,y)=xy3u_{xy}(x, y) =
xy^3uxy(x,y)=xy3, with boundary conditions:
Given:
1. uxy(x,y)=xy3u_{xy}(x, y) = xy^3uxy(x,y)=xy3, where
uxyu_{xy}uxy represents the mixed second derivative of
u(x,y)u(x, y)u(x,y) with respect to xxx and yyy.
2. Boundary conditions:
o u(x,0)=f(x)=cos(x)u(x, 0) = f(x) =
\cos(x)u(x,0)=f(x)=cos(x)
o uy(0,y)=g(y)=y+sin(y)u_y(0, y) = g(y) = y +
\sin(y)uy(0,y)=g(y)=y+sin(y)
We need to find u(x,y)u(x, y)u(x,y).
Step 1: Integrate with respect to yyy
, We know that uxy(x,y)=xy3u_{xy}(x, y) = xy^3uxy(x,y)=xy3.
To find ux(x,y)u_x(x, y)ux(x,y), we integrate both sides of the
equation with respect to yyy:
∫uxy(x,y) dy=∫xy3 dy\int u_{xy}(x, y) \, dy = \int xy^3 \, dy∫uxy
(x,y)dy=∫xy3dy ux(x,y)=xy44+h(x)u_x(x, y) = \frac{xy^4}{4} +
h(x)ux(x,y)=4xy4+h(x)
where h(x)h(x)h(x) is an arbitrary function of xxx that arises
from the integration with respect to yyy.
Step 2: Integrate with respect to xxx
Next, we integrate ux(x,y)u_x(x, y)ux(x,y) with respect to xxx:
u(x,y)=∫(xy44+h(x)) dxu(x, y) = \int \left( \frac{xy^4}{4} + h(x)
\right) \, dxu(x,y)=∫(4xy4+h(x))dx
This gives:
u(x,y)=x2y48+H(x)+C(y)u(x, y) = \frac{x^2 y^4}{8} + H(x) +
C(y)u(x,y)=8x2y4+H(x)+C(y)
where H(x)H(x)H(x) is an arbitrary function of xxx, and
C(y)C(y)C(y) is an arbitrary function of yyy.
Step 3: Apply the boundary condition
u(x,0)=f(x)=cos(x)u(x, 0) = f(x) = \cos(x)u(x,0)=f(x)=cos(x)
We know that u(x,0)=cos(x)u(x, 0) = \cos(x)u(x,0)=cos(x).
Using the general form of u(x,y)u(x, y)u(x,y):
(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]
Part (a) - Solving the PDE uxy(x,y)=xy3u_{xy}(x, y) =
xy^3uxy(x,y)=xy3, with boundary conditions:
Given:
1. uxy(x,y)=xy3u_{xy}(x, y) = xy^3uxy(x,y)=xy3, where
uxyu_{xy}uxy represents the mixed second derivative of
u(x,y)u(x, y)u(x,y) with respect to xxx and yyy.
2. Boundary conditions:
o u(x,0)=f(x)=cos(x)u(x, 0) = f(x) =
\cos(x)u(x,0)=f(x)=cos(x)
o uy(0,y)=g(y)=y+sin(y)u_y(0, y) = g(y) = y +
\sin(y)uy(0,y)=g(y)=y+sin(y)
We need to find u(x,y)u(x, y)u(x,y).
Step 1: Integrate with respect to yyy
, We know that uxy(x,y)=xy3u_{xy}(x, y) = xy^3uxy(x,y)=xy3.
To find ux(x,y)u_x(x, y)ux(x,y), we integrate both sides of the
equation with respect to yyy:
∫uxy(x,y) dy=∫xy3 dy\int u_{xy}(x, y) \, dy = \int xy^3 \, dy∫uxy
(x,y)dy=∫xy3dy ux(x,y)=xy44+h(x)u_x(x, y) = \frac{xy^4}{4} +
h(x)ux(x,y)=4xy4+h(x)
where h(x)h(x)h(x) is an arbitrary function of xxx that arises
from the integration with respect to yyy.
Step 2: Integrate with respect to xxx
Next, we integrate ux(x,y)u_x(x, y)ux(x,y) with respect to xxx:
u(x,y)=∫(xy44+h(x)) dxu(x, y) = \int \left( \frac{xy^4}{4} + h(x)
\right) \, dxu(x,y)=∫(4xy4+h(x))dx
This gives:
u(x,y)=x2y48+H(x)+C(y)u(x, y) = \frac{x^2 y^4}{8} + H(x) +
C(y)u(x,y)=8x2y4+H(x)+C(y)
where H(x)H(x)H(x) is an arbitrary function of xxx, and
C(y)C(y)C(y) is an arbitrary function of yyy.
Step 3: Apply the boundary condition
u(x,0)=f(x)=cos(x)u(x, 0) = f(x) = \cos(x)u(x,0)=f(x)=cos(x)
We know that u(x,0)=cos(x)u(x, 0) = \cos(x)u(x,0)=cos(x).
Using the general form of u(x,y)u(x, y)u(x,y):
