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Section A: All students
Solutions ii MATH2048W1
A1. [Total for this question: 20 marks]
LO1 Demonstrate knowledge and understanding of ODE eigenvalue problems,
Fourier series and transformations, Laplace transforms and partial differential
equations
LO2 Demonstrate organisational and time-management skills
LO3 Critically analyse and solve some mathematical problems relevant to
engineering
LO4 Perform calculations in simple situations and work through some longer
examples
Unseen but similar to examples seen in class
(a) [4 marks] Explain why the following are incorrect and then correct the error.
In each case, [1 mark] for explaining error, [1 mark for writing down the correct
result].
(i) [2 marks] Let f(t) = δ(t + 3π).
Therefore, the Laplace transform of f(t) is
L[f(t)] = w
∞
0
δ(t + 3π)e
s t dt = e
3πs
for s > 0.
Note that δ(t + 3π) means that the impulse function occurs outside the
domain of integration. Therefore
L[f(t)] = w
∞
0
δ(t + 3π)e
s t dt = 0
An alternative correction would be to change the sign in the argument of f so
that f(t) = δ(t 3π).
(ii) [2 marks] Let the Laplace Transform of g(t) be g (s) = e
2s
(s + 1)2 + 4
.
Copyright 2023 University of Southampton Page 2 of 19
Solutions iii MATH2048W1
Therefore, the original function, g(t), is given as
g(t) = H(t 2)e
t
sin(2t)
The e
2s
is a result of the 2nd shift theorem so t is shifted by the amount 2.
The inverse of 2
s
2+4 is sin(2t) with the s + 1 a result of the first shift theorem.
Therefore
g(t) = 1
2
H(t 2)e
(t 2) sin(2(t 2))
(b) [16 marks] Use the method of Laplace transforms to solve the following
initial-value problem:
y¨ ˙y 2y = δ(t 2π) y(0) = 0, ˙y(0) = 2,
where δ(t) is the Dirac delta function.
Taking Laplace Transforms
s
2
y sy 2 y 2 = e
2πs (10)
= y ((s + 1)(s 2)) = e
2πs + 2 (11)
= y =
e
2πs + 2
(s + 1)(s 2) (12)
=
e
2πs + 2
3
_x0012_
1
s 2

1
s + 1
. (13)
Note that L

1
s 2

1
= e
2t
, L

1
s+1
1
= e
t
therefore
y =
2
3
_xfffe_
e
2t e
t
_x0001_ +
1
3
H(t 2π)
_x0010_ e
2(t 2π) e
(t 2π)

.
6 marks obtaining Y (s);
2 marks partial fractions on Y (s);
Copyright 2023 University of Southampton
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Solutions iv MATH2048W1
4 marks explaining how to go from Y (s) to y(t);
4 marks correct y (t).
Note that there was a typo on the original paper, where the final line of the
question read “where H(x) is the Heaviside function”. This confused some.
There are two potential confusions that might (in context) be marked as correct:
(i) writing H(t 2π) on the RHS in the initial-value problem;
(ii) treating δ as an unknown constant.
In the first case we have
y¨ ˙y 2y = H(t 2π) y(0) = 0, ˙y(0) = 2, (14)
so Laplace transforms gives
s
2
y sy 2 y 2 =
e
2πs
s
(15)
= y ((s + 1)(s 2)) = e
2πs
s
+ 2 (16)
= y =
e
2πs
s(s + 1)(s 2) +
2
(s + 1)(s 2) (17)
=
e
2πs
6

1
s 2
+
2
s + 1

3
s

+
2
3

1
s 2

1
s + 1_x0013_
.
(18)
Inverting this gives
y(t) = 2
3
_xfffe_
e
2t e
t
+
1
6
H(t 2π)
e
2(t 2π) + 2e
(t 2π) 3
. (19)
In the second case where δ is considered a constant we have
y¨ ˙y 2y = δ(t 2π) y(0) = 0, ˙y(0) = 2, (20)
Copyright 2023 University of Southampton Page 4 of 19
Solutions v MATH2048W1
so Laplace transforms gives
s
2
y sy 2 y 2 = δ

s
1
2

2π
s

(21)
= y ((s + 1)(s 2)) = δ
_x0012_
s
1
2

2π
s

+ 2 (22)
= y =
δ
s
2
(s + 1)(s 2) +
2 2πδ
s(s + 1)(s 2) (23)
=
2 2πδ
6

1
s 2
+
2
s + 1

3
s
_x0013_
+
δ
12
1
s 2

4
s + 1
+
10
s

6
s
2

. (24)
Inverting this gives
y(t) = 2 2πδ
6

e
2t + 2e
t 3
+
δ
12

e
2t 4e
t + 10 6t
. (25)
Copyright 2023 University of Southampton
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Solutions vi MATH2048W1
A2. [Total for this question: 25 marks]
LO1 Demonstrate knowledge and understanding of ODE eigenvalue problems,
Fourier series and transformations, Laplace transforms and partial differential
equations
LO2 Demonstrate organisational and time-management skills
LO3 Critically analyse and solve some mathematical problems relevant to
engineering
LO4 Perform calculations in simple situations and work through some longer
examples
Unseen but similar to examples seen in class
The forced heat equation with unit diffusion constant is
y
t =

2
y
x2
+ F(x, t). (1)
The domain is x ∈ [0, 1]. The boundary conditions are
y
x(0, t) = 0 = y
x(1, t). (2)
The forcing term is
F(x, t) = exp( t)(1 x).
(a) [10 marks] Show that the solution to the homogeneous problem
y
t =

2
y
x2
subject to the boundary condition in equation (2) is
y(x, t) = T0(t) +
∞
X
n=1
Tn(t) cos(nπx). (5)
You do not need to compute the explicit form of Tn(t) or T0(t).
Use the standard separation of variables ansatz
y(x, t) = X(x)T(t). (26)
[1 mark]
Copyright 2023 University of Southampton Page 6 of 19
Solutions vii MATH2048W1
The homogeneous PDE gives
˙T
T
=
X′′
X
= λ (27)
where λ is the separation constant, as both sides are functions of different
independent variables.
[2 marks]
Re-arrange to get the coupled ODEs
X
′′ λX = 0 (28)
˙T λT = 0. (29)
[1 mark]
Substitute the ansatz into the boundary conditions to find
X
′
(0) = 0 = X
′
(1). (30)
[1 mark]
Solve the eigenvalue problem for X(x) in three cases.
For λ = 0 we have X′′ = 0 and hence X = Ax + B, giving X′ = A. The
boundary conditions give A = 0. So X0 = B is a solution, and λ0 = 0.
[1 mark]
For λ = k
2 with k > 0 we have X = Cekx + De kx giving
X
′
(x) = k
Cekx De kx
. (31)
The boundary condition at x = 0 gives D = C and the boundary condition at
x = 1 gives C = 0. Hence the only solution is trivial.
[1 mark]
For λ = k
2 with k > 0 we have X = E cos(kx) + F sin(kx) giving
X
′
(x) = k ( E sin(kx) + F cos(kx)). (32)
The boundary condition at x = 0 gives F = 0. The boundary condition at x = 1
gives k = nπ. So Xn = En cos(nπx) is a solution, and λn = (nπ)
2
.
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Solutions viii MATH2048W1
[2 marks]
Combining these results together, whilst leaving the time dependent behaviour
unsolved, we get
y(x, t) = T0(t) +
∞
X
n=1
Tn(t) cos(nπx). (33)
[1 mark]
(b) [10 marks] Using the information from part (a), and that the Fourier cosine
series for 1 x is
1 x =
1
2 +
∞
X
n=1
2
(nπ)
2
(1 ( 1)n
) cos(nπx),
show that the general solution to the forced heat equation in equation (1) subject
to the boundary condition in equation (2) is
y(x, t) = C0
1
2
e
t+
∞
X
n=1

2
(nπ)
2
(1 ( 1)n
)
e
t
(nπ)
2 1
+ Cne
(nπ)
2
t

cos(nπx).
(7)
From part (a) and the result for the Fourier series we write
y(x, t) = T0(t) +
∞
X
n=1
Tn(t) cos(nπx), F(x, t) = F0(t) +
∞
X
n=1
Fn(t) cos(nπx),
(34)
where F0, Fn are given by the Fourier series expression multiplied by e
t
. Using
the PDE we find
∞
X
n=0
h
˙Tn + (nπ)
2Tn Fne
t
i
cos(nπx) = 0. (35)
Here we have F0 =
1
2
and
Fn =
2
(nπ)
2
(1 ( 1)n
). (36)
Note that the sum runs from zero to include the constant term: any consistent
way of writing this is fine.
Copyright 2023 University of Southampton Page 8 of 19
Solutions ix MATH2048W1
[4 marks]
From this we deduce
˙Tn + (nπ)
2Tn = Fne
t
. (37)
[1 mark]
We immediately solve the n = 0 case to get
T0 =
1
2
e
t + C0. (38)
[1 mark]
Any appropriate method (such as integrating factors) solves the general case to
give
Tn =
2
(nπ)
2
(1 ( 1)n
)
e
t
(nπ)
2 1
+ Cne
(nπ)
2
t
. (39)
[3 marks]
Substituting back into our form for y(x, t) gives the result.
[1 mark]
(c) [5 marks] Set the initial data to be
y(x, 0) = 0.
Show, using (7) from (b), that the solution to the original PDE satisfying the initial
condition is
y(x, t) = 1
2

1 e
t
+
∞
X
n=1

2
(nπ)
2
(1 ( 1)n
)
(nπ)
2 1

e
t e
(nπ)
2
t

cos(nπx).
(9)
Set t = 0 in the result for part (b) and use the initial data to get
0 = C0
1
2 +
∞
X
n=1

2
(nπ)
2
(1 ( 1)n
)
(nπ)
1
2 1
+ Cn
cos(nπx). (40)
[2 marks]
We read off
C0 =
1
2
(41)
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[1 mark]
and
Cn =
2
(nπ)
2
(1 ( 1)n
)
(nπ)
1
2 1
. (42)
[1 mark]
Substituting back into our form for y(x, t) gives the result.
[1 mark]
Section B: All students except Civil Engineering
Solutions xii MATH2048W1
B. LO5 Demonstrate knowledge and understanding of vector calculus (Mech,
Ship, Aero, ISVR).
Unseen but similar to problem sheet question. [Total for this question: 35 marks]
(a) [16 marks] Define the following:
f(x, y, z) = ze1 xy
,
g(x, y, z) = sin x
2
y z ,
v(x, y, z) = y(z 1)i x(1 z)j + (1 + xy 2z) k
u(x, y, z) = sin(yz)i + cos(yz)j + e
xzk
where i, j, k are the three unit vectors of the Cartesian axes.
Showing all your working, evaluate the following or explain why it is not possible.
(i) Find the Gradient ( ) of f(x, y, z)
yze1 xyi xze1 xyj + e
1 xyk.
(ii) Find the Gradient ( ) of v(x, y, z)
The gradient acts on scalar fields only, so this is not possible.
(iii) Find the Divergence ( ·) of g(x, y, z)
The divergence acts on vector fields only, so this is not possible.
(iv) Find the Divergence ( ·) of u(x, y, z)
·u = z sin(yz) + xexz
.
(v) Find the Curl ( ×) of g(x, y, z)
The curl acts on vector fields only, so this is not possible.
(vi) Find the Curl ( ×) of v(x, y, z)
×v = (0) i + (0) j + (0) k = 0.
(vii) Find the Laplacian ( 2
) of f(x, y, z)
Copyright 2023 University of Southampton Page 12 of 19
Solutions xiii MATH2048W1
2
f = · f = · yze1 xy
, xze1 xy, e1 xy =
y
2
z + x
2
z
e
1 xy
.
(viii) Find the Laplacian ( 2
) of u(x, y, z) The Laplacian acts on each
component. We have
2u = (y
2 + z
2
) (sin(yz)i + cos(yz)j) + (x
2 + z
2
)e
xzk
(b) [5 marks] Define the conservative vector field u as
u = ze1 z
i e
1 z
j + e
1 z
(x + y xz) k
Find a scalar potential for u.
We are looking for where u = = ( x , y , z ). Do each component in
order. First
x = ze1 z
.
Direct integration gives
= xze1 z + f(y, z).
Differentiate to find, using the next component,
y = yf = e
1 z
.
Direct integration gives
f(y, z) = ye1 z + g(z)
= (xz y)e
1 z + g(z).
Differentiate to find, using the final component,
z = (x + y xz)e
1 z + zg = (x + y xz)e
1 z
.
Hence a solution is zg = 0, giving
= (xz y)e
1 z + C.
Copyright 2023 University of Southampton
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Solutions xiv MATH2048W1
(c) [4 marks] Find the work done by a particle subject to the force
F = (2x 3y)i + (y z)j x
2
yk
when moving from t = 0 to t = 1 along a paramaterized curve, C, defined as
r = (t 1, t, 1 t).
We have
x = t 1
y = t
z = 1 t
and hence
F = ( t 2)i + (2t 1)j t(t 1)2k
and also
dr
dt
= i + j k.
Hence the work done is
w
1
0

( t 2) + (2t 1) + t(t 1)2

dt =
1
w
0

t
3 2t
2 + 2t 3
dt
=

t
3
3
3t

1
0
=
29
12
.
(d) [5 marks] A closed curve C has coordinates
sin(t)i + cos(t)j + 2 cos(t)k
where t ∈ [0, 2π] is a parameter. A vector field F is given by
F = y sin(z)i + x sin(z)j + xy cos(z)k.
Using Stoke’s Theorem compute the line integral
I
C
F·dr.
Copyright 2023 University of Southampton Page 14 of 19
Stoke’s theorem says that if a simply-connected volume V has boundary V
given by C then
I
C
F·dr =
Z Z
S
( ×F)·dS.
We compute the curl as
×F = (x cos(z) x cos(z)) i (y cos(z) y cos(z)) j+(sin(z) sin(z)) k = 0.
Therefore the line integral is zero.
(e) [5 marks] A cylinder encloses the region V where x
2 + y
2 ≤ 1 and 0 ≤ z ≤ 1.
A vector field F is given by
F = yzi + xzj + (xy + z)k.
Use the divergence theorem to find the flux of the vector field through the surface
of the cylinder,
Z Z
V
F·dS.
The divergence theorem gives that the flux is
Z Z Z
V
·F.
The divergence is
·F = 1.
Therefore the flux is the volume of the cylinder, which is π.
Solutions xvi MATH2048W1
C1. LO6 Demonstrate knowledge and understanding of statistics (Civil
Engineering).
Unseen but similar to problem sheet question.
(a) (i)
P (B2B takes off) = 1 P (B2B fails to take off)
= 1 P (both engines fail)
= 1 p
2
.
(ii) Find p such that P (B2B takes off) > P (B4B takes off), i.e.
1 p
2 > 1 4p
3 + 3p
4
,
i.e.
p
2

3p
2 4p + 1 < 0. Solving the quadratic equation p1,2 = 4 ± √ 16 12 6 = 1 or 1 3 . Therefore P (B2B takes off) > P (B4B takes off) for p ∈

1
3
, 1
.
(iii) Engines failures are unlikely to be independent since there may be factors
that could impact the operation of both engines, e.g. birdstrikes, weather
conditions, fuel contamination, etc.
(b) Let denote the event of failure. Then P ( |A) = 0.2, P ( |B) = 0.1,
P (A) = 0.7 and P (B) = 0.3.
By Bayes’ theorem
P (A| ) = P ( |A) P (A)
P ( )
=
P ( |A) P (A)
P ( |A) P (A) + P ( |B) P (B)
=
0.2 × 0.7
0.2 × 0.7 + 0.1 × 0.3
=
14
17
= 0.8235.
Copyright 2023 University of Southampton Page 16 of 19
(c) (i) Pdf is given by
f(y) = dF(y)
dy
=
d
dy

1 exp y
2

= 2y exp y
2

for y > 0.
(ii) Since Y is measured in 100s of hours, the probability of a component
operates for at least 200 hours is
P(Y ≥ 2) = 1 P(Y < 2) = 1 F(2) = exp 2 2 = 0.0183. Solutions xviii MATH2048W1 C2. LO6 Demonstrate knowledge and understanding of statistics (Civil Engineering). Unseen but similar to problem sheet question. (a) (i) The combined sample variance is s 2 c = (n1 1)s 2 1 + (n2 1)s 2 2 n1 + n2 2 = 6 × 2102 + 9 × 1902 7 + 10 2 = 39300. (ii) The test statistic is t = yˉ1 yˉ2 sc q 1 n1 + 1 n2 = 3250 3240 √ 39300 × q 1 7 + 1 10 = 0.1024. Since |t| < t15,0.975 = 2.131, do not reject H0 : μ1 = μ2. No evidence of a difference between the two drying methods. (b) (i) The estimated slope is negative meaning that it is estimated that as engine volume increases, fuel efficiency decreases. (ii) 95% CI for β1 is given by β 1 ± tn 2,0.975se β 1 , i.e. 5.518 ± 2.364 × 1.211, i.e. ( 8.381, 2.655). Because the 95% CI does not include 0, reject H0 : β1 = 0 at 5% level. Hence, evidence that fuel efficiency decreases as volume increases. (iii) s 2 = 1.362 = 1.850. Copyright 2023 University of Southampton Page 18 of 19 (iv) R2 = 0.7478 means that about 75% of the variability in the data is explained by the model. (v) (1) Anscombe plot of residuals against fitted values and (2) QQ-plots of residuals. (vi) y = 36.730 5.518 × 2 = 25.69.