ADVANCED GCE
4758/01
MATHEMATICS (MEI)
Differential Equations
Candidates answer on the answer booklet.
OCR supplied materials:
• 8 page answer booklet
(sent with general stationery)
• MEI Examination Formulae and Tables (MF2)
Other materials required:
• Scientific or graphical calculator
Wednesday 18 May 2011
Morning
Duration: 1 hour 30 minutes
*475801*
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4
7
5
8
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1
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Read each question carefully. Make sure you know what you have to do before starting
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Answer any three questions.
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You are permitted to use a scientific or graphical calculator in this paper.
Final answers should be given to a degree of accuracy appropriate to the context.
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INFORMATION FOR CANDIDATES
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The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised that an answer may receive no marks unless you show sufficient detail
of the working to indicate that a correct method is being used.
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© OCR 2011 [R/102/2661]
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Turn over
2
1
The differential equation
d2 y
dy
+4
+ 3y = 13 cos 2t
2
dt
dt
(∗)
is to be solved.
(i) Find the general solution.
[9]
(ii) Find the particular solution, given that when t = 0, y and
dy
are both zero.
dt
[6]
Now consider the differential equation
d2 ß
dß
d3 ß
+
4
+3
= −26 sin 2t.
dt
dt3
dt2
(iii) Show that the general solution may be expressed as ß = y + c where y is the general solution of
[2]
(∗) and c is a constant.
(iv) When t = 0, ß = 2,
2
dß
d2 ß
= 0 and 2 = 13. Use these conditions to find the particular solution. [7]
dt
dt
(a) A curve in the x-y plane satisfies the differential equation
for x > 0.
dy 2y p
−
= x
dx
x
(i) Find the general solution for y in terms of x.
[8]
The curve passes through (1, 0).
(ii) Find the equation of this curve.
[2]
(iii) Find the coordinates of the stationary point of this curve and find the values to which y and
dy
tend as x → 0. Sketch the curve.
[6]
dx
(b) The differential equation
dy p 2
= x + y2
dx
is to be solved approximately by using a tangent field.
(i) Describe the shape of the isocline for which
dy
= 1.
dx
[2]
dy
dy
dy
(ii) Sketch, on the same axes, the isoclines for the cases
= 1,
= 2,
= 3. Use these
dx
dx
dx
isoclines to draw a tangent field.
[3]
(iii) Sketch the solution curve through (0, 1).
(iv) Sketch the solution curve through the origin.
© OCR 2011
4758/01 Jun11
[1]
[2]
3
3
(a) A particle of mass 2 kg moves on a horizontal straight line containing the origin O. When its
displacement is x m from O, it is subject to a force of magnitude 2k2 x N directed towards O,
where k is a positive constant.
(i) Show that the velocity, v m s−1 , of the particle satisfies the differential equation
v
dv
= −k2 x.
dx
[3]
The particle is at rest when x = a, where a is a positive constant.
(ii) Solve the differential equation, subject to this condition. Hence show that, while the particle
moves in the negative direction,
p
dx
[6]
= −k a2 − x2 .
dt
Initially the particle is at x = a.
(iii) Use the standard integral
ä p
x
dx = arcsin + c
a
a −x
1
2
2
to find x in terms of t, k and a.
[5]
(b) At time t s, the angle, θ rad, that a pendulum makes with the vertical satisfies the differential
equation
dω
ω
= −9 sin θ
dθ
dθ
where ω =
.
dt
(i) Solve the differential equation for ω in terms of θ subject to the condition ω = 0 when
θ = 13 π . Hence show that, while θ is decreasing,
p
dθ
= −3 2 cos θ − 1.
dt
[6]
(ii) Starting from θ = 13 π when t = 0, use Euler’s method with a step length of 0.1 to estimate
θ when t = 0.1. The algorithm is given by tr+1 = tr + h, θr+1 = θr + hθ˙ r . State whether this
algorithm can usefully be continued, justifying your answer.
[4]
[Question 4 is printed overleaf.]
© OCR 2011
4758/01 Jun11
Turn over
4
4
The quantities x and y at time t are modelled by the simultaneous differential equations
dx
= −3x − 2y + 3t,
dt
dy
= 2x + y + t + 2.
dt
(i) Show that
dx
d2 x
+2
+ x = −5t − 1.
2
dt
dt
[5]
(ii) Find the general solution for x.
[8]
(iii) Find the corresponding general solution for y.
[4]
When t = 0, x = 9 and y = 0.
(iv) Find the particular solutions.
[4]
(v) Find approximate expressions for x and y in terms of t, valid for large positive values of t.
[3]
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© OCR 2011
4758/01 Jun11
GCE
Mathematics (MEI)
Advanced GCE
Unit 4758: Differential Equations
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
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on the Examination.
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4758
1(i)
Mark Scheme
June 2011
 2  4  3  0
  1 or -3
CF Ae  t  Be 3t
PI y  a cos 2t  b sin 2t
M1
A1
F1
B1
y  2a sin 2t  2b cos 2t
M1
Differentiate twice and
substitute
M1
Compare coefficients
Auxiliary equation
CF for their roots

y  4a cos 2t  4b sin 2t
4a cos 2t  4b sin 2t  8a sin 2t  8b cos 2t  3a cos 2t  3b sin 3t  13cos 2t
8b  a  13
b  8a  0
1
8
a , b
5
5
y  15 (8sin 2t  cos 2t )  Ae  t  Be 3t
GS
A1
A1
F1
PI + CF with two arbitrary
constants
M1
Use condition
M1
F1
M1
Differentiate
9
(ii)
t  0, y  0  0   15  A  B
y  15 (16 cos 2t  2 sin 2t )  Ae  t  3Be 3t
t  0, y  0  0  165  A  3B
13
3
, B=
10
2
13  t
1
y  5 (8sin 2t  cos 2t )  10
e  32 e 3t
 A
(iii)
(iv)
Use condition
A1
If z  y  c , differentiating (*) gives new DE
and has 3 arbitrary constants so must be GS
or
Integrating gives (*) with  k on RHS
PI will be previous PI  13 k , CF as before, so GS y  c
A1
Cao
M1
A1
Recognise derivative
6
M1
A1
SC1 for showing that correct y from (i) + c satisfies new DE
z  15 (8sin 2t  cos 2t )  De  t  Ee 3t  c
t  0, z  2  2   15  D  E  c
t
z  (16 cos 2t  2sin 2t )  De  3Ee
1
5
t  0, z  0  0 
16
5
3 t
 D  3E
2
M1
Use condition
F1
Derivative
M1
Use condition
Second derivative:
condone, for this mark
only, +c appearing
Use condition

z  15 (32sin 2t  4 cos 2t )  De  t  9 Ee 3t
F1
t  0, 
z  13  13  54  D  9 E
M1
13
3
,E ,c2
10
2
13  t
z  15 (8sin 2t  cos 2t )  10
e  32 e 3t  2
D
B1
A1
Cao
7
1
4758
2(a)(i)
Mark Scheme
I  exp(   2x dx)
M1
 exp(2 ln x )
M1
1
M1
A1
F1
y  2 x 2  Ax 2
3
(iii)
Multiply both sides by IF
M1
x 2 y  2 x  2  A
Integrate both sides
Must divide constant
8
0  2  A
y  2 x2  2 x
Attempt integrating factor
A1
A1
 x 2
dy
3
x 2
 2 x 3 y  x  2
dx
d 2
3
( x y)  x  2
dx
(ii)
June 2011
3
M1
A1
2
2
x  0, y  0
F1
dy
9
1
 4 x  3x 2  0  x 
(as x  0)
dx
16
dy
x  0,
0
dx
M1
F1
1
B1
B1
Behaviour at origin
and shape for x  1
Through
B1
Stationary point at
6
(b)(i)
Circle centre origin
Radius 1
B1
B1
2
(ii)
B1
B1
B1
One isocline correct
All three isoclines correct
Reasonably complete and accurate
direction indicators
3
(iii)
B1
Solution curve
(iv)
B1
B1
Solution curve
Zero gradient at origin
1
2
2
4758
3(a)(i)
Mark Scheme
N2L: ma  2k 2 x
dv
2v  2k 2 x
dx
dv
v
 k 2 x
dx
June 2011
M1
M1
Acceleration
E1
3
(ii)
 vdv   k
2
xdx
M1
Separate and integrate
x  a, v  0  A  12 k 2 a 2
A1
A1
M1
LHS
RHS
Use condition
v 2  k 2 (a 2  x 2 )
A1
1
2
v  k x A
2
1
2
2
So for v  0,
2
dx
 k a 2  x 2
dt
E1
6
(iii)
1

dx   kdt
a  x2
arcsin ax  B  kt
M1
Separate and integrate
x  a, t  0  B   12 
A1
A1
M1
LHS
RHS
Use condition
x  a sin( 12   kt )  a cos kt
A1
Either form
2
5
(b)(i)
 d   9sin  d
M1
Separate and integrate
  13  ,   0  C   92
A1
A1
M1
LHS
RHS
Use condition
So  2  9(2 cos   1)
A1
d
 3 2 cos   1 (decreasing)
dt
E1
1
2
  9 cos   C
2
6
(ii)
      0
1
3
M1
So estimate    0  
1
3
1
3
A1
The algorithm will keep giving   
1
3
B1
B1
but  is not constant so not useful
4
3
4758
4(i)
Mark Scheme
y   12 x  32 x  32 t
M1
y   
x  x 
M1
1
2
3
2
3
2
June 2011
 12 
x  32 x  32  2 x  ( 12 x  32 x  32 t )  t  2
M1
Eliminate
M1
E1
Eliminate

x  2 x  x  5t  1
M1
A1
F1
B1
Auxiliary equation
Root
CF for their root(s) (with two constants)
M1
Differentiate and substitute
M1
A1
F1
Compare and solve
5
(ii)
 2  2  1  0
  1 (repeated)
CF: ( A  Bt )e t
PI: x  at  b
x  a, 
x0
In DE: 0  2a  at  b  5t  1
a  5
2 a  b  1
a  5, b  9
GS: x  9  5t  ( A  Bt )e t
GS = PI + CF with two arbitrary constants
8
(iii)
y   12 x  32 x  32 t
M1
t
t
  [5  Be  ( A  Bt )e ]
1
2
t
 [9  5t  ( A  Bt )e ]  t
3
2
 9t  11  ( A  B  Bt )e
1
2
(iv)
3
2
t
M1
Differentiate (product rule)
M1
Substitute
A1
4
t  0, x  9  A  0
t  0, y  0  0  11  12 B  B  22
M1
M1
x  9  5t  22te  t
y  9t  11  (11  22t )e t
A1
A1
Use condition
Use condition
4
(v)
t
e 0
x  9  5t
y  9t  11
M1
F1
F1
3
4
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© OCR 2011
Examiners’ Reports – June 2011
4758/01: Differential Equations
(Written Examination)
General Comments
The overall performance on this paper was very good. Many candidates scored high
marks and very few scored less than half of the available marks. As usual, the familiar
topics tested in Questions 1 and 4 were attempted by almost all of the candidates, with
Question 3 the least popular choice. Most candidates have a very good working
knowledge of the topics on this syllabus, the exception being, on this occasion, an
understanding of the terminology of isoclines and tangent fields.
A high standard of algebraic and arithmetical accuracy of solutions is expected on this
paper, and it is pleasing to note an improvement in this aspect.
Comments on Individual Questions
1
This question was attempted by all candidates and many earned the majority of the
available marks.
(i)
The method was well-understood by all, but a minority of candidates made arithmetical
or algebraic errors in solving the linear simultaneous equations in finding the particular
integral.
(ii)
As in part (i), there were some algebraic errors.
(iii)
Most candidates scored one out of the two marks available here, by recognising that
one differential equation was the integral/differential of the other. Few candidates were
able to go on to give a convincing argument to show that z was equal to y + c.
(iv)
Apart from arithmetical and algebraic errors, a minority of candidates worked with the
particular solution to the original differential equation, rather than the general solution.
2
Candidates usually answered part (a) well, but many seemed unclear of the
terminology and/or methods involved in part (b).
(a)
(i)
Candidates showed a good understanding of the integrating factor method of solving
this first order differential equation, and they applied it with accuracy.
(ii)
Again, this use of an initial condition was well-executed.
(iii)
Most candidates were able to find the stationary point of the curve and the values of y
and its derivative as x approaches zero, but they did not always go on to use this
information to help them sketch the curve.
(b)
There were a few excellent solutions to this part of the question, but the work of many
candidates suggested that they were not familiar with the words “isocline” and “tangent
field.”
(i)
A statement that the isocline is a circle with centre at the origin and with unit radius was
required here.
26
Examiners’ Reports – June 2011
(ii)
Many candidates showed confusion in their attempted solutions here, indicating that
they were unsure of what was required in the requests for three isoclines and
corresponding tangent fields. The isoclines, which were concentric circles in this case,
were often not drawn.
(iii)
The majority of candidates were able to recover here and sketch the solution curve
through (0,1).
(iv)
Again, candidates recognised the general shape of the solution curve through the
origin, but relatively few indicated its gradient at the origin to be zero.
3
This was the least popular choice of question, but those who selected it were usually
successful in scoring the majority of the marks.
(a)
(i)
This was invariably answered well.
(ii)
The separation of variables and integration was done well, but the justification for
taking the negative sign in the final expression was not always present.
(iii)
Again, the separation of variables and integration was handled correctly by the
candidates.
(b)
(i)
There were many fully correct solutions here, though a minority of candidates made a
sign error in the trigonometric integration. Again, the justification for the negative sign
in the given expression was omitted by some candidates.
(ii)
Almost all candidates applied the Euler method accurately to obtain the requested
estimation. Very few candidates were able to explain that it was not helpful to continue
using the given algorithm, because of the non-constancy of θ.
4
The vast majority of candidates attempted this question and many scored full marks.
(i)
Solutions were almost always convincing and correct.
(ii)
Candidates were clearly very familiar with the method and worked through it accurately.
(iii)
Occasionally there were arithmetical errors in finding the general solution for y. It was
particularly pleasing that only a handful of candidates failed to use their general
solution for x as the starting-point.
(iv)
Again, there were a few arithmetical slips, but the method was well-known by all.
(v)
This request presented no problems to candidates and with follow through from
previous incorrect answers, almost all candidates scored the three marks available.
27
For a description of how UMS marks are calculated see:
www.ocr.org.uk/learners/ums_results.html
GCE Mathematics (MEI)
4751/01 (C1) MEI Introduction to Advanced Mathematics
4752/01 (C2) MEI Concepts for Advanced Mathematics
4753/01
4753/02
4753/82
4753
4754/01
(C3) MEI Methods for Advanced Mathematics with Coursework: Written Paper
(C3) MEI Methods for Advanced Mathematics with Coursework: Coursework
(C3) MEI Methods for Advanced Mathematics with Coursework: Carried Forward Coursework Mark
(C3) MEI Methods for Advanced Mathematics with Coursework
(C4) MEI Applications of Advanced Mathematics
4755/01 (FP1) MEI Further Concepts for Advanced Mathematics
4756/01 (FP2) MEI Further Methods for Advanced Mathematics
4757/01 (FP3) MEI Further Applications of Advanced Mathematics
4758/01
4758/02
4758/82
4758
4761/01
(DE) MEI Differential Equations with Coursework: Written Paper
(DE) MEI Differential Equations with Coursework: Coursework
(DE) MEI Differential Equations with Coursework: Carried Forward Coursework Mark
(DE) MEI Differential Equations with Coursework
(M1) MEI Mechanics 1
4762/01 (M2) MEI Mechanics 2
4763/01 (M3) MEI Mechanics 3
4764/01 (M4) MEI Mechanics 4
4766/01 (S1) MEI Statistics 1
4767/01 (S2) MEI Statistics 2
4768/01 (S3) MEI Statistics 3
4769/01 (S4) MEI Statistics 4
4771/01 (D1) MEI Decision Mathematics 1
4772/01 (D2) MEI Decision Mathematics 2
4773/01 (DC) MEI Decision Mathematics Computation
4776/01
4776/02
4776/82
4776
4777/01
(NM) MEI Numerical Methods with Coursework: Written Paper
(NM) MEI Numerical Methods with Coursework: Coursework
(NM) MEI Numerical Methods with Coursework: Carried Forward Coursework Mark
(NM) MEI Numerical Methods with Coursework
(NC) MEI Numerical Computation
Unit level raw mark and UMS grade boundaries June 2011 series: GCE
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
UMS
Raw
Raw
Raw
UMS
Raw
UMS
Max Mark
72
100
72
100
72
18
18
100
90
100
72
100
72
100
72
100
72
18
18
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
100
72
18
18
100
72
100
a
55
80
53
80
54
15
15
80
63
80
59
80
55
80
55
80
63
15
15
80
60
80
64
80
59
80
54
80
53
80
60
80
56
80
56
80
51
80
58
80
46
80
62
14
14
80
55
80
b
49
70
46
70
48
13
13
70
56
70
52
70
48
70
48
70
57
13
13
70
52
70
57
70
51
70
47
70
45
70
53
70
49
70
49
70
45
70
53
70
40
70
55
12
12
70
47
70
c
43
60
39
60
42
11
11
60
50
60
45
60
41
60
42
60
51
11
11
60
44
60
51
60
43
60
40
60
38
60
46
60
42
60
42
60
39
60
48
60
34
60
49
10
10
60
39
60
d
37
50
33
50
36
9
9
50
44
50
39
50
34
50
36
50
45
9
9
50
36
50
45
50
35
50
33
50
31
50
39
50
35
50
35
50
33
50
43
50
29
50
43
8
8
50
32
50
e
32
40
27
40
29
8
8
40
38
40
33
40
27
40
30
40
39
8
8
40
28
40
39
40
27
40
26
40
24
40
33
40
28
40
28
40
27
40
39
40
24
40
36
7
7
40
25
40
u
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
19