Mathematics Resit Coursework - Section A, Questions 1-3

This document presents the solutions to the mathematics section (Section A) of a resit coursework. The solutions cover three questions, focusing on partial derivatives and differential equations. Question 1 involves finding the partial derivative of a given function with respect to time and evaluating it at a specific point, along with an interpretation of the result. Question 2 requires solving a differential equation using integration and finding a particular solution given initial conditions. Question 3 involves solving a linear differential equation using an integrating factor, and then determining the particular solution based on provided initial conditions. All steps and calculations are shown in detail, providing a comprehensive understanding of the solution process. This coursework is designed to assess the student's understanding of calculus and differential equations.

Q 1.
Answer
T = 24e−0.5tcos (0.4x) + te−0.2tsin x
1.
Partial derivative with respect to time
∂T
∂t = cos (0.4x)
∂
∂t (24e−0.5t) + sin x
∂
∂t (te−0.2t)
[Applying Product rule where needed]
= cos (0.4x)(−12e−0.5t) + sin x(e(−0.2t)(1) + t(−0.2e−0.2t)
= −12e−0.5tcos (0.4x) + sin xe−0.2t(1 − 0.2t)
At x = 0 and t = 1.9
∂T
∂t x=0,t=1.9= −12e−0.5×1.9cos (0) + sin (0)e−0.2×1.9
(1 − 0.2 × 1.9)
= −12e−0.95 = −4.641
2.
Partial derivative with respect to time represents the temporal rate of change
of temperature at a location.x = 0 is a boundary location.The solution
for T (x, t) indicates that temperature at the boundary x = 0 is function of
1

time and the partialderivative of T (w.r.t.time) indicates that boundary
temperature is decreasing with time.
Answer
1.
ydy
dx = e− 2x
19 + cos 0.1x
19
y
Taking all the y terms to the left side and all the x terms to the right
y2dy = e− 2x
19 + cos 0.1x
19 dx
[Integrating both sides]
Z
y2dy =
Z
e− 2x
19 + cos 0.1x
19 dx
y3
3 = −19
2 e− 2x
19 + 190 sin0.1x
19 + C0
∴ y = 3
s
−28.5e− 2x
19 + 570 sin0.1x
19 + C
2.
2

When x = 0, y = 0.Therefore,
0 = 3
p −28.5e0 + 570 sin (0) + C
=⇒ 0 = −28.5 + C
∴ C = 28.5
The solution is then given by
y = 3
s
−28.5e− 2x
19 + 570 sin0.1x
19 + 28.5
When x = 4.75
y(4.75) =3
s
−28.5e− 2×4.75
19 + 570 sin0.1 × 4.75
19 + 28.5
= 3
s
−28.5e− 1
2 + 570 sin 1
40 + 28.5
= 2.94
Answer
1.
dy
dx − 4xy = e2x2
3

Linear differential equation of the form
dy
dx + p(x)y = q(x)
has an integrating factor μ(x)
μ(x) = e
R p(x) dx
and solution in the form
y(x) =
R μ(x)q(x) dx + C
μ(x) (1)
Comparing the given D.E. with the standard form, we have
p(x) = −4x
Therefore, the integrating factor is
μ(x) = e
R −4x dx = e−2x2
2.
General solution is (refer (1))
y =
R e−2x2
· e2x2
dx + C
e−2x2
=
R dx + C
e−2x2
y = e2x2
(x + C)
3.
When x = 0, y = 9.5.Therefore,
y(0) = 9.5 = e0(0 + C) =⇒ C = 9.5
The particular solution then is
y = e2x2
(x + 9.5)
4