NATIONAL UNIVERSITY OF SINGAPORE
FACULTY OF SCIENCE
SEMESTER 2 EXAMINATION 2019–2020
MA1512
DIFFERENTIAL EQUATIONS FOR ENGINEERING
April 2020 Time allowed: 2 hours
INSTRUCTIONS TO CANDIDATES
1. Different students may be given different questions.
2. Type the numerical answers into the answer boxes provided.
Working must be handwritten using blue- or black-ink pen on
paper. On every page of working, write down clearly
• your Student Number near the top-left corner; and
• the question number and the page number near the top-right
corner, e.g., Q1P1, Q1P2, . . . , Q2P1, Q2P2, . . . .
Do not write your name anywhere. Use different pages for different
questions. Write on only one side of the paper.
3. Take photos of your working and embed them in the corresponding
rationale boxes. The following are the steps to embed these photos.
(a) Click the “Image” button at the rich-text editor.
(b) Upload your photos in the “Upload” tab.
(c) Select the right photo to be embedded in the “Image Info” tab.
(d) Click the “OK” button to finish the embedding.
Make sure the photos you embed do not miss any page, are
legible, and are in the correct orientation.
4. The examination closes at 19:05. You must type the numerical
answers and embed photos of your working within the 120-
minute examination time.
MA1512 Examination
5. Do not click the “Finish Quiz” button except when you finish the
examination. You can use the “Save For Later” button to temporarily
save your progress. All saved attempts will be submitted after the end
of the examination.
6. After the examination closes, examination answers and working that
are neither saved nor submitted will not be accepted, unless there is
a valid reason.
7. This examination consists of six (6) questions, each of which consists
of two (2) parts. Answer all questions.
8. The marks allocated for each question are indicated at the beginning of
the question. The maximum possible total marks for this examination
is seventy (70).
9. This is an open book examination, but there should be no com-
munication (neither face-to-face nor via communication devices) with
anyone except the invigilator during the examination.
10. You may use any calculators. However, you should lay out systemat-
ically the various steps in the calculations.
11. Join the Zoom meeting assigned to you with your video enabled but
your microphone muted at all times during the examination. Adjust
your camera such that your face and your upper body, including your
hands, are captured on Zoom. Use the private chat function in Zoom
if you want to communicate with the invigilator. Do not silence
your device during this Zoom meeting.
12. You may go for a short toilet-break (for not more than 5 minutes)
during the examination.
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MA1512 Examination
Question 1 [10 marks]
1. Let y(x) be the solution of the initial value problem
y′ +
6x5 + 20x3
x6 + 5x4 + 1512
y = −5x5 + 3, y(0) = −2.
What is the value of y(0.9)? Give your answer correct to two decimal
places.
2. Professor Dreistein leaves 40 mg of mendelevium-256 in a closed box,
and he wants to let 10 mg of it decay away radioactively. How long does
this take? Give your answer in minutes correct to two decimal places. Do
not enter the unit in your answer. For the calculations, use 76.6667 minutes
as the half-life of mendelevium-256.
Outline of solution
1. exp
(∫ 6x5 + 20x3
x6 + 5x4 + 1512
dx
)
= exp
(
ln(x6 + 5x4 + 1512) + C
)
= (x6 + 5x4 + 1512) eC
∴ y = 1
x6 + 5x4 + 1512
∫
(−5x5 + 3)(x6 + 5x4 + 1512) dx
=
1
x6 + 5x4 + 1512
∫
−5x11 − 25x9 − 7560x5 + 3x6 + 15x4 + 4536 dx
=
1
x6 + 5x4 + 1512
(−5x12
12
− 25x
10
10
− 7560x
6
6
+
3x7
7
+
15x5
5
+ 4536x+D
)
.
Substituting in the initial condition gives D = 1512 y(0) = −3024. Thus
y(0.9) = 0.2571 . . . ≈ 0.26.
Numerical answers for other versions of this question: 0.20, 0.40, 0.49
2. N = N0 exp(−kt) = N0 exp
(
− ln 2
t1/2
t
)
.
The number of minutes needed is
t1/2
ln 2
ln
N0
N0 − 10 = 31.8195 . . . ≈ 31.82.
Numerical answers for other versions of this question: 56.50, 44.85, 37.22
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MA1512 Examination
Question 2 [12 marks]
1. Let y(x) be the solution of the initial value problem
y
dy
dx
= 3xy2 − 2y2 + 6x− 4, y > 0, y(0) = 5.
What is the value of y(1)? Give your answer correct to two decimal places.
2. A bowl of sauerkraut soup is served at 90 ◦C, but Alice can only drink
soup that is at most 70 ◦C. So she waits. After 3 minutes, the soup is
still at 82 ◦C. Suppose the rate at which the soup cools is proportional to
the difference between the temperature of the soup and the temperature
of the surrounding air, which is maintained at 25.5 ◦C. How much longer
does Alice need to wait before she can start drinking the soup? Give your
answer in minutes correct to two decimal places. Do not enter the unit in
your answer.
Outline of solution
1. y
dy
dx
= (3x− 2)(y2 + 2)
⇔
∫
y
y2 + 2
dy =
∫
3x− 2 dx
⇔ ln(y
2 + 2)
2
=
3x2
2
− 2x+ C
⇔ ln(y2 + 2) = 3x2 − 4x+D
⇔ y2 + 2 = Ae3x2−4x.
Substituting into the initial condition gives A = y(0)2 + 2 = 27. So
y(1) =
√
27e−1 − 2 = 2.8165 . . . ≈ 2.82.
Numerical answers for other versions of this question: 2.15, 4.09, 4.72
2. T = Ae−kt + 25.5, A, k > 0.
Substituting into T (0) = 90 gives A = 90− 25.5 = 64.5. Then
64.5 e−k·3 + 25.5 = 82
∴ k = −1
3
(
ln(56.5)− ln(64.5)).
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MA1512 Examination
Let s be such that T (3 + s) = 70. Then
Ae−k(3+s) + 25.5 = 70
∴ s = −1
k
ln
70− 25.5
A
− 3 = ln(44.5)− ln(64.5)
ln(56.5)− ln(64.5) × 3− 3
= 5.4087 . . . ≈ 5.41.
This is the number of minutes Alice still needs to wait.
Numerical answers for other versions of this question: 5.88, 5.91, 5.62
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MA1512 Examination
Question 3 [12 marks]
1. Let y(x) be the solution of the initial value problem
y′′ + 2y′ − 3y = 4x+ 5e6x, y(0) = 0, y′(0) = 0.
What is the value of y(1)? Give your answer correct to two decimal places.
2. Let γ be a positive constant such that the solutions of the differential
equation
x¨+ 2345x = cos(γt) + sin(γt)
exhibit resonant behaviour. What is the value of γ? Give your answer
correct to two decimal places.
Outline of solution
1. Characteristic equation: λ2 + 2λ− 3 = (λ+ 3)(λ− 1) = 0.
So y = ce−3x + dex is a general solution of y′′ + 2y′ − 3y = 0.
Try y = Ax+B + Ce6x. Then
y′ = A+ 6Ce6x and y′′ = 36Ce6x.
So 36Ce6x + 2(A+ 6Ce6x)− 3(Ax+B + Ce6x) = 4x+ 5e6x. Equating coefficients,
−3A = 4 ⇒ A = −4/3
2A− 3B = 0 ⇒ B = 2A/3 = −8/9
45C = 5 ⇒ C = 1/9.
Thus y = −4
3
x− 8
9
+ 1
9
e6x is a particular solution.
A general solution is y = ce−3x + dex − 4
3
x− 8
9
+ 1
9
e6x. Then
y′ = −3ce−3x + dex − 4
3
+
2
3
e6x.
Substituting into y(0) = 0 and y′(0) = 0 gives c + d = 7/9 and −3c + d = 2/3
respectively. Solving gives c = 1/36 and d = 3/4. Thus
y =
1
36
e−3x +
3
4
ex − 4
3
x− 8
9
+
1
9
e6x.
Hence
y(1) = 44.6432 . . . ≈ 44.64.
Numerical answers for other versions of this question: 43.98, 42.59, 41.93
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MA1512 Examination
2. The differential equation can be written as
x¨+ ω2x = A cos(γt− δ),
where
• ω is the natural frequency;
• γ is the input frequency;
• A and δ are constants.
Since resonance occurs, we know
γ = ω =
√
2345 = 48.4252 . . . ≈ 48.43.
Numerical answers for other versions of this question: 38.88, 43.82, 44.94
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MA1512 Examination
Question 4 [12 marks]
1. Let c be a constant such that the differential equation
(x2 + y4 + 3)
dy
dx
= y2 + 4y − c
has exactly one equilibrium solution. What is the value of c? Give the
exact answer.
2. The sheep population in a certain mountain is known to satisfy the
logistic model. The population size is stable at 4000 with a per capita
birth rate of 10% per year. One year, a village is built in the area. The
villagers start hunting 40 sheep per year for food. Assuming the birth
rate and the hunting rate stay the same, what will the size of the sheep
population settle down to eventually? Give your answer correct to the
nearest integer.
Outline of solution
1. dy
dx
=
y2 + 4y − c
x2 + y4 + 3
= 0 ⇔ y2 + 4y − c = 0.
The differential equation has exactly one equilibrium solution exactly when
v2 + 4v − c = 0 has exactly one solution ⇔ 42 − 4(−c) = 0 ⇔ c = −4.
Numerical answers for other versions of this question: 1, −1, 4
2. logistic equilibrium without harvesting = N∞ = B/s, B = 0.1.
So s = B/N∞. In the long run, the sheep population settles down to
B +
√
B2 − 4(B/N∞)E
2(B/N∞)
= 3549.1933 . . . ≈ 3549.
Numerical answers for other versions of this question: 2662, 4436, 5324
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MA1512 Examination
Question 5 [12 marks]
1. Let F (s) be the Laplace transform of
f(t) = e2t δ(t− 1) + t3,
where δ denotes the Dirac delta function. What is the value of F (1)? Give
your answer correct to two decimal places.
2. Let y(t) be the solution of the initial value problem
y¨ + y = 9
(
u(t− pi)− u(t− 3pi)), y(0) = 0, y˙(0) = 1,
where u denotes the unit step function. What is the value of y(2pi)? Give
the exact answer.
Outline of solution
1. By s-shifting,
F (s) = e−(s−2) +
6
s4
.
So
F (1) = e+ 6 = 8.71828 . . . ≈ 8.72.
Numerical answers for other versions of this question: 4.72, 9.39, 13.39
2. s2Y − s y(0)− y˙(0) + Y = 9
(e−pis
s
− e
−3pis
s
)
⇔ (s2 + 1)Y − 1 = 9
s
(e−pis − e−3pis)
⇔ Y = 9(e
−pis − e−3pis) + s
s(s2 + 1)
= 9(e−pis − e−3pis)
(1
s
− s
s2 + 1
)
+
1
s2 + 1
⇔ y = 9u(t− pi) · (1− cos(t− pi))
− 9u(t− 3pi) · (1− cos(t− 3pi))+ sin(t)
by t-shifting. Thus
y(2pi) = 9× (1− (−1)) + 0 = 18.
Numerical answers for other versions of this question: 12, 14, 16
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MA1512 Examination
Question 6 [12 marks]
1. Let w(x, y) be the solution of the partial differential equation
∂w
∂x
= xy2
∂w
∂y
+
9
5
xw, x > 0, y > 0
together with the conditions
w(
√
2, 1) = e9/5 and w(2, 2) = e16/13
obtained using the method of separation of variables. What is the value of
w(3, 3)? Give your answer correct to two decimal places.
2. Let y(t, x) be the solution of the wave equation
∂2y
∂t2
= 4
∂2y
∂x2
together with the boundary and initial conditions
y(t, 0) = 0, y(t, pi) = 0, y(0, x) = 2 sin(4x),
∂y
∂t
(0, x) = 0
obtained using d’Alembert’s formula. What is the value of y(2pi9 ,
2pi
9 )? Give
your answer correct to two decimal places.
Outline of solution
1. Let w = XY . Then
X ′Y = xy2XY ′ +
9
5
xXY
∴ X
′
xX
= k =
y2Y ′ + 9
5
Y
Y
.
From the LHS equation,
X ′
X
= kx
∴ ln |X| = k · x
2
2
+ C0
∴ X = A0 exp
(k
2
x2
)
.
From the RHS equation,
Y ′
Y
=
(
k − 9
5
)
y−2
∴ ln |Y | =
(
k − 9
5
) y−1
−1 + C1
∴ Y = A1 exp
( 9
5
− k
y
)
.
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MA1512 Examination
Thus w = A exp
(
k
2
x2 +
9
5
−k
y
)
. Substituting this into w(
√
2, 1) = e9/5 gives
e9/5 = A exp
(
k +
9
5
− k
)
= Ae9/5.
So A = 1. Using w(2, 2) = e16/13, we get
e16/13 = exp
(
2k +
9
5
− k
2
)
= exp
( 9
5
+ 3k
2
)
.
So k = (2× 16
13
− 9
5
)/3 = 43/195. Hence
w = exp
( 43
390
x2 +
308
195y
)
and
w(3, 3) = 4.5667 . . . ≈ 4.57.
Numerical answers for other versions of this question: 3.39, 4.84, 8.26
2. According to d’Alembert,
y =
1
2
(
2 sin(4x+ 8t) + 2 sin(4x− 8t)) = sin(4x+ 8t) + sin(4x− 8t).
So
y(2pi/9, 2pi/9) = 0.5240 . . . ≈ 0.52.
Numerical answers for other versions of this question: 0.22, 0.35, 0.54
END OF PAPER
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