de paper jan13
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Monday 28 January 2013 Morning
A2 GCE MATHEMATICS (MEI)4758/01 Differential Equations
QUESTION PAPER
INSTRUCTIONS TO CANDIDATES
These instructions are the same on the Printed Answer Book and the Question Paper.
The Question Paper will be found in the centre of the Printed Answer Book. Write your name, centre number and candidate number in the spaces provided on the
Printed Answer Book. Please write clearly and in capital letters. Write your answer to each question in the space provided in the Printed Answer
Book. Additional paper may be used if necessary but you must clearly show yourcandidate number, centre number and question number(s).
Use black ink. HB pencil may be used for graphs and diagrams only. Read each question carefully. Make sure you know what you have to do before starting
your answer. Answer any threequestions. Do notwrite in the bar codes. 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. The acceleration due to gravity is denoted by gm s
2. Unless otherwise instructed, when
a numerical value is needed, use g= 9.8.
INFORMATION FOR CANDIDATES
This information is the same on the Printed Answer Book and the Question Paper.
The number of marks is given in brackets [ ]at the end of each question or part questionon the Question Paper.
You are advised that an answer may receive no marksunless you show sufficient detailof the working to indicate that a correct method is being used.
The total number of marks for this paper is 72. The Printed Answer Book consists of 16pages. The Question Paper consists of 4pages.
Any blank pages are indicated.
INSTRUCTION TO EXAMS OFFICER / INVIGILATOR
Do not send this Question Paper for marking; it should be retained in the centre orrecycled. Please contact OCR Copyright should you wish to re-use this document.
OCR is an exempt Charity
Turn over OCR 2013 [R/102/2661]
DC (SLM) 63805/3
Candidates answer on the Printed Answer Book.
OCR supplied materials: Printed Answer Book 4758/01 MEI Examination Formulae and Tables (MF2)
Other materials required:
Scientific or graphical calculator
Duration: 1 hour 30 minutes
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2
4758/01 Jan13 OCR 2013
1 The differential equation
sinx
y
x
y
x
yy x2 5 6
d
d
d
d
d
d
3
3
2
2
+ - - =
is to be solved.
(i) Show that 2 is a root of the auxiliary equation. Find the other two roots and hence find the generalsolution of the differential equation. [10]
When x 0= , y 1= andx
y0
d
d= . Also, y is bounded as x " 3.
(ii) Find the particular solution. [6]
(iii) Write down an approximate solution for large positive values of x . Calculate the amplitude of this
approximate solution and sketch the solution curve for large positive x . [4]
Suppose instead that a solution is required that is bounded as x " 3- .
(iv) Determine whether there is a solution for which y 1= andx
y 0d
d= when x 0= . [4]
2 A ball of mass mkg falls vertically from rest through a liquid. At time t s, the velocity of the ball is vm s1
and the ball has fallen a distancexm. The forces on the ball are its weight and a total upwards force ofRN.
A student investigates three models forR.
In the first model R mkv= , where kis a positive constant.
(i) Show that .t
vkv9 8
d
d= - and hence find vin terms of tand k. [7]
The terminal velocity of the ball is observed to be 7 m s1
.
(ii) Find k. [1]
In the second model, .R mv0 2 2
= .
(iii) Find vin terms of t. Show that your solution is consistent with a terminal velocity of 7 m s1
. [10]
In the third model, .R mv0 529 23
= . Eulers method is to be used to solve for vnumerically.
The algorithm is given by t t hr r1
= ++
, v v hvr r r1
= ++
o with , ,t v 0 00 0
=_ _i i.
(iv) Show that . .t
v v9 8 0 529dd 2
3
= - and find vwhen .t 0 2= using Eulers method with a step length of 0.1.
[5]
(v) Show that this model is consistent with a terminal velocity of approximately 7 m s1
. [1]
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4758/01 Jan13 OCR 2013
3 (a) Solve the differential equation
tan sinx
yy x x
d
d- =
to findyin terms ofxsubject to the condition y 1= when x 0= . [9]
(b) Consider the differential equations
( ) ( )fx
yx y x
d
dg+ = , (1)
( )fx
yx y 0
d
d+ = . (2)
Show that if ( )py x= satisfies (1) and ( )cy x= satisfies (2), then ( ) ( )py x A xc= + satisfies (1),
whereAis an arbitrary constant. [5]
(c) The differential equation
xy
xy
xx
22 1
d
de
x
22
+ = +d n (3) is to be solved.
(i) Verify that y ex2
= satisfies (3). [3]
(ii) Find the general solution ofx
y
x
y20
d
d+ = , givingyin terms ofx. [4]
(iii) Use the result of part (b)to find a solution of (3) for which y 1= when x 1= . [3]
4 The simultaneous differential equations
tx y t
t
yx y t
x
2
1
2
3
2
3
2
12
d
d
d
d= - - +
= - +
are to be solved.
(i) Eliminateyto obtain a second order differential equation for xin terms of t. Hence find the general
solution forx. [13]
(ii) Find the corresponding general solution fory. [4]
When t 0= , x 1= and y 0= .
(iii) Find the particular solutions. [3]
(iv) Show that in this case x y+ tends to a finite limit as t" 3and state its value. Determine whether
x y+ is equal to this limit for any values of t. [4]
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4758/01 Jan13 OCR 2013
Copyright Information
OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders
whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright
Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.
If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible
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THERE ARE NO QUESTIONS PRINTED ON THIS PAGE.
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Monday 28 January 2013 Morning
A2 GCE MATHEMATICS (MEI)4758/01 Differential Equations
PRINTED ANSWER BOOK
INSTRUCTIONS TO CANDIDATESThese instructions are the same on the Printed Answer Book and the Question Paper.
The Question Paper will be found in the centre of the Printed Answer Book. Write your name, centre number and candidate number in the spaces provided on the Printed
Answer Book. Please write clearly and in capital letters. Write your answer to each question in the space provided in the Printed Answer Book.
Additional paper may be used if necessary but you must clearly show your candidate number,centre number and question number(s).
Use black ink. HB pencil may be used for graphs and diagrams only. Read each question carefully. Make sure you know what you have to do before starting your
answer. Answer any threequestions. Do notwrite in the bar codes.
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. The acceleration due to gravity is denoted by gm s2. Unless otherwise instructed, when a
numerical value is needed, use g= 9.8.
INFORMATION FOR CANDIDATES
This information is the same on the Printed Answer Book and the Question Paper.
The number of marks is given in brackets [ ]at the end of each question or part question onthe Question Paper.
You are advised that an answer may receive no marksunless you show sufficient detail of theworking to indicate that a correct method is being used.
The total number of marks for this paper is 72. The Printed Answer Book consists of 16pages. The Question Paper consists of 4pages. Any
blank pages are indicated.
* 4 7 5 8 0 1 *
OCR is an exempt Charity
Turn over OCR 2013 [R/102/2661]
DC (SLM) 63807/3
Candidates answer on this Printed Answer Book.
OCR supplied materials: Question Paper 4758/01 (inserted) MEI Examination Formulae and Tables (MF2)
Other materials required:
Scientific or graphical calculator
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whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright
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If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible
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Oxford Cambridge and RSA Examinations
GCE
Mathematics (MEI)Advanced GCE
Unit 4758:Differential Equations
Mark Scheme for January 2013
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OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range ofqualifications to meet the needs of candidates of all ages and abilities. OCR qualificationsinclude AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals,Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications inareas such as IT, business, languages, teaching/training, administration and secretarial skills.
It is also responsible for developing new specifications to meet national requirements and theneeds of students and teachers. OCR is a not-for-profit organisation; any surplus made isinvested back into the establishment to help towards the development of qualifications andsupport, which keep pace with the changing needs of todays society.
This mark scheme is published as an aid to teachers and students, to indicate the requirementsof the examination. It shows the basis on which marks were awarded by examiners. It does notindicate the details of the discussions which took place at an examiners meeting before markingcommenced.
All examiners are instructed that alternative correct answers and unexpected approaches in
candidates scripts must be given marks that fairly reflect the relevant knowledge and skillsdemonstrated.
Mark schemes should be read in conjunction with the published question papers and the reporton the examination.
OCR will not enter into any discussion or correspondence in connection with this mark scheme.
OCR 2013
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4758 Mark Scheme
1
Annotat ions
Annotat ion in scoris Meaning
and
BOD Benefit of doubt
FT Follow through
ISW Ignore subsequent workingM0, M1 Method mark awarded 0, 1
A0, A1 Accuracy mark awarded 0, 1
B0, B1 Independent mark awarded 0, 1
SC Special case
^ Omission sign
MR Misread
Highlighting
Other abbreviations inmark scheme
Meaning
E1 Mark for explainingU1 Mark for correct units
G1 Mark for a correct feature on a graph
M1 dep* Method mark dependent on a previous mark, indicated by *
cao Correct answer only
oe Or equivalent
rot Rounded or truncated
soi Seen or implied
www Without wrong working
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4758 Mark Scheme
4
g Rules for replaced work
If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be mthe candidate requests.
If there are two or more attempts at a question which have not been crossed out, examiners should ma
(complete) attempt and ignore the others.
NB Follow these maths-specific instructions rather than those in the assessor handbook.
h For a genuinemisreading (of numbers or symbols) which is such that the object and the difficulty of theaccording to the scheme but following through from the candidates data. A penalty is then applied; 1 mthis may differ for some units. This is achieved by withholding one A mark in the question.
Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unlesestablished by equivalent working.
Fresh starts will not affect an earlier decision about a misread.
Note that a miscopy of the candidates own working is not a misread but an accuracy error.
i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow f(provided, of course, that there is nothing in the wording of the question specifying that analytical methis wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with in doubt, consult your Team Leader.
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4758 Mark Scheme
5
Question Answer Marks Gu
1 (i) 3 22 5 6 0 M13 22 2 2 5 2 6 0 E1 Allow if implicit in factorisation of cu
( 2)( 1)( 3) 0 M1 Attempt roots (any method)(2), 1, 3 A1
CF 2 3e e ex x xA B C F1PI sin cosy a x b x B1
cos siny a x b x sin cosy a x b x
cos siny a x b x
( cos sin ) 2( sin cos )a x b x a x b x M1 Differentiate and substitute5( cos sin ) 6( sin cos ) sina x b x a x b x x
3225 50
6 8 0,
8 6 1
a ba b
a b
M1 Compare coefficients and solve
A1
GS2 33 2
50 25cos sin e e e
x x xy x x A B C
F1
[10]1 (ii) bounded so 0A F1
350
0, 1 1x y B C F1 Use condition33 2
50 25sin cos e 3 ex xy x x B C M1 Differentiate
225
0, 0 0 3x y B C M1 Use condition29 5120 100
,B C A133 29 512
50 25 20 100cos sin e ex xy x x F1
[6]
1 (iii) 3 250 25
cos siny x x F1
amplitude2 21 1
50 103 4
M1
A1
B1
Sketch showing oscillations with thei
amplitude. More than one oscillation,
origin
[4]
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4758 Mark Scheme
6
Question Answer Marks Gu
1 (iv) bounded so 0B C B13 47
50 500, 1 1x y A A M1
47225 50d
0 2 0d
yx
x M1 Or
1
25A
So no such solution A1 www[4]
2 (i)N2L:
d9.8
d
vm m mkv
t
d9.8
d
vkv
t E1
EITHER1
d d9.8
v tkv
M1 Separate and integrate
1ln 9.8 kv t c
k
A1
A1
LHS
RHS (including constant on one side)
9.8 e ktkv A M1 Rearrange, dealing properly with con0, 0 9.8t v A M1 Use condition
9.8 1 e ktv k A1 caoOR Integrating factor ekt M1
e 9.8e d kt kt
v t A1Multiply both sides by IF and recogn
derivative on LHS
9.8e ekt kt v A
k M1 Integrate both sides
A1 Must include constant
9.80, 0t v A
k M1 Use condition
9.81 e
ktv
k
A1 cao
OR Auxiliary equation 0k M1
CF e ktv A A1
PI9.8
, 0v b v bk
M1
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4758 Mark Scheme
7
Question Answer Marks Gu
GS9.8
e ktv Ak
A1
9.80, 0t v A
k M1 Use condition
9.8 1 e ktv k A1 cao[7]
2 (ii) 9.81.4
7k B1
[1]
2 (iii) 2d9.8 0.2
d
vm m mv
t M1
2
1d d
9.8 0.2v t
v
M1
22
5
d49 v t cv A1 RHS (including constant on one side)
2
5 1 1d
14 7 7v t c
v v
M1 Integrate
5 214 ln 7 ln 7v v t c A1 LHS
14 257
ln7
vt c
v
14 /57e
7
tvB
v
M1
Rearrange into a form without ln, dea
properly with constant
0, 0 1t v B M1 Use condition
14 /57 e 7tv v M1 Rearrange to get vin terms of t14 /5 14 /5
14 /5 14 /5
e 1 1 e7 7
e 1 1 e
t t
t tv
A1 oe
as 1 01 0, 7 7t v
E1
[10]
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4758 Mark Scheme
8
Question Answer Marks Gu
2 (iv) 3/20.529v g v E1
t v v hv M1 Use algorithm
0 0 9.8 0.98 A1 (0.1)v
0.1 0.98 9.2868 0.92868 A1 (0.1)v at least 3d.p.
0.2 1.9087 A1 (0.2)v =1.91 to 3s.f.[5]
2 (v) 3/20 0.529 7.00v g v v (3 sf) E1 Or3
29.8 0.529 7 0 [1]
3 (a) exp tan d I x x M1 exp ln secx or exp ln cosx A1
cosx A1d
cos sin sin cosd
yx y x x x
x
d cos sin cosd
y x x xx
M1 Multiply and recognise derivative
cos sin cos d y x x x x M1 Attempt integral12
sin 2 d x x M1 Use identity, substitution or inspectio1
14cos2x c (or 21
2sin x k ) A1 oe (but must include constant)
114
0, 1 1x y c M1 Use condition
5 cos 2
4cos
xy
x
or
2sin 2
2cos
xy
x
or
23 cos
2cos
xy
x
A1 oe
[9]
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4758 Mark Scheme
9
Question Answer Marks Gu
3 (b) p '( ) f ( )p( ) g( )x x x x M1 Must be ( )p x oe
c '( ) f ( )c( ) 0x x x M1 Must be ( )c x oe
d
f ( ) p '( ) c '( ) f ( ) p( ) c( )d
yx y x A x x x A x
x M1 Substitute in DE
p '( ) f ( )p( ) c '( ) f ( )c( )x x x A x x x M1 Separate p and c termsg( ) 0 g( )x A x E1 Complete argument
[5]
3 (c) (i) 2 2de 2 e
d
x xyy x
x B1
so LHS of DE2 22
2 e ex xxx
M1
2 221 1
2e 2ex x x
xx x
E1
[3]
3 (c) (ii) d 2 1 2d d
d
y yy x
x x y x M1
2ln 2lny x c A1 LHS
A1 RHS including constant2
y Ax A1 cao
OR Integrating factor =2
x B1
2d
0d
yxx
M1
2yx A A12y Ax A1 cao
[4]
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4758 Mark Scheme
10
Question Answer Marks Gu
3 (c) (iii) 2 2ex
y Ax B1HereAcombines the arbitrary consta
and (c) (ii) into a single arbitrary con1
1 e A M1 Use condition
2 2e 1 exy x A1
[3]4 (i) 2 13 2y x x t M1
2 13 2 1y x x M1 Differentiate
32 1 1 2 13 2 2 2 3 21 2x x x x x t t M1 Substitute2 2 5 2 5x x x t A1 oe
AE 22 2 5 0 M131
2 2j A1
CF 12 3 3
2 2e cos sin
tA t B t
M1 Correct form
F1 FT wrong rootsPI x a bt B1
, 0 2 5( ) 2 5x b x b a bt t M1 Differentiate and substitute
45
2 5 2, 1
5 5
b aa b
b
M1
A1Equate coefficients and solve
GS 12 3 34
5 2 2e cos sin
tx t A t B t
F1
[13]
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4758 Mark Scheme
11
Question Answer Marks Gu
4 (ii) 2 13 2y x x t M1
12
12
3 312 2 2
3 3 3 32 2 2 2
1 e cos sin
e sin cos
t
t
x A t B t
A t B t
M1
F1
Must be using product rule
Must be GS from(i)
12 3 32
5 2 2e sin cos
ty t A t B t
A1
[4]
4 (iii) 4 15 5
1, 0 1x t A A M12 25 5
0, 0 0y t B B M1
12 3 34 1 2
5 5 2 5 2e cos sin
tx t t t
12 3 32 1 2
5 5 2 5 2e sin cos
ty t t t
A1 Both
[3]
4 (iv)
1
26 3 3 315 5 2 5 2e sin cost
x y t t
M1 Adding and attempting the limit12 6
5e 0
tt x y
E1 FT for finite limit
6 3 3 3 31 15 5 2 5 2 2 3
sin cos 0 tanx y t t t M1 Establish equation and indicate metho
which occurs (infinitely often) E1Correctly investigate the existence of
solution, but explicit solution for t no
[4]
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Oxford Cambridge and RSA Examinationsis a Company Limi ted by GuaranteeRegistered in EnglandRegistered Office; 1 Hills Road, Cambridge, CB1 2EURegistered Company Number: 3484466OCR is an exempt Charity
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Oxford Cambridge and RSA Examinations
GCE
Mathematics (MEI)
Advanced GCE A27895-8
Advanced Subsidiary GCE AS 3895-8
OCR Report to Centres
January 2013
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OCR Report to CentresJanuary 2013
24
4758 Differential Equations (Written Examination)
General Comments
Candidates showed a sound understanding of the basic methods of solution of the types ofdifferential equations covered by this specification. The presentation of solutions was usually ofa good quality with clear explanations. As in previous series, Questions 1 and 4, on second andhigher order differential equations, were chosen by almost all candidates. There was evidencethat Question 3 was the next question of preference, but many candidates struggled with the firsttwo parts and then opted for Question 2 instead. Others appeared to answer any parts ofQuestions 2 and 3 that they could, leaving the examiner to determine which of the two questionscounted towards their total.
There were fewer fully correct solutions to questions than in previous series. There were morearithmetical errors than usual, particularly in the first parts of Questions 1 and 4. Candidatesshould be encouraged to work accurately even with work that is familiar. Although incorrect work
is followed through and given credit, it is inevitable that errors early in a solution will lead tosome loss of marks. In the case of second order linear differential equations an incorrectauxiliary equation, or the incorrect solution of an auxiliary equation, can lead to a different formof general solution and this may impact on the ease or feasibility of the requests in the later partsof a question.
Comments on Individual Questions
1 Third order linear differential equation
(i) Almost all candidates were able to use the information in the question to make a goodattempt at solving the given third order differential equation. The complementaryfunction was usually found accurately, but there were a surprising number of numericalerrors when finding the particular integral.
(ii) The majority of candidates were successful in using the initial conditions to find aparticular solution; it was pleasing to see that the less familiar form of condition, that ywas bounded for very large values ofx, was interpreted correctly.
(iii) Candidates seemed less confident in answering this part. A common error was toassume that the coefficient of either the sine or the cosine term was equal to theamplitude of the motion.
(iv) The majority of candidates were able to use the condition that ywas bounded for verylarge negative values ofxto deduce that two of the unknown coefficients in their
particular solution from part (i) had to be zero. Many candidates then went on to applythe other two initial conditions to obtain two different values for the remainingcoefficient, indicating that there was no consistent solution. A common error was toapply just one of the initial conditions to find the unknown coefficient and then state thatthis was a solution.
2 First order differential equations
(i) Almost all candidates applied Newton's second law to establish the given differentialequation. They then proceeded to find the general solution either by the integratingfactor method or by separating the variables or by finding the complementary functionand particular integral. The three methods were equally popular and successful. A
significant minority of candidates did not seem to realise that they needed to apply theinitial condition, that the ball fell from rest, to find the value of their constant ofintegration.
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OCR Report to CentresJanuary 2013
25
(ii) This was almost always correct.
(iii) There were some excellent solutions to this part of the question with very clear accuratework and gaining full marks. However, the majority of candidates struggled to solve thedifferential equation. Separating the variables leads to an integral of the form
2 2
1dx
a xand evaluating this proved to be a stumbling block for the majority. The
most common attempts resulted in inverse tangents or in incorrect logarithmic terms of
the form2 2ln( )a x . A denominator that is the difference of two squares is a common
occurrence in problems of this type and candidates should be encouraged to be vigilant.
(iv) Euler's method is familiar territory for candidates and the majority of candidates weresuccessful, presenting their working clearly.
(v) This was almost always correct.
3 First order differential equation
A significant number of candidates were not able to complete their attempts at the firsttwo parts of this question. They chose to omit the remainder of the question and attemptanother question instead.
(a) Almost all candidates recognised the method of solution as the integrating factormethod and made a good start. Most candidates evaluated the integrating factor
correctly as cosx, but they did not recognise a method of evaluating sin cos dx x x , the
integral that appeared on the right hand side of the differential equation. There aremany valid approaches, either by direct integration or by using a double angle formulaor using a substitution and it was surprisingly that none of these came to mind for manycandidates.
(b) Only a handful of candidates made any meaningful progress with this request and themajority offered very little as an attempt at a solution. A common error was to substitute
p( )y x into equation (1) only partially, withd
d
y
xretained. A similar procedure for
equation (2) led to inevitable confusion between y, p and c.
(c)(i) Many candidates missed the obvious approach here, expressed clearly in theinstruction to verify. All that was necessary was to differentiate the given function andsubstitute into equation (3). A surprising number of candidates attempted to solve
equation (3), not realising that this was not within their scope, and that the structure ofthe question was to lead them through a method of solution.
(ii) Many candidates had abandoned their attempts at this question by this stage, but tothose who proceeded, this was familiar territory.
(iii) This part brought together parts (b) and (c) but only a few candidates grasped thestructure of the question and pursued it to the end.
4 Simultaneous linear differential equations
(i) Almost all candidates gained the majority of the marks in this part. The method ofapproach was understood and any marks lost were due to arithmetic slips inmanipulating the functions or in solving the auxiliary equation.
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(ii) Again, candidates knew what they had to do. Answers to part (i) were followed throughfor the majority of the marks.
(iii) The method was applied successfully, although by this stage most candidates hadmade at least one arithmetical slip. Fully correct expressions forxand ywere rarelyseen.
(iv) Candidates who had made errors in the previous parts of this question were at adisadvantage because their expression for x y did not usually tend to a finite limit.
Credit was given for the correct interpretation of the behaviour of the candidate'sincorrect expression.