* * MATHEMATICS (MEI) 4758/01 Differential Equations ADVANCED GCE. Wednesday 18 May 2011 Morning

Transcription

1 ADVANCED GCE MATHEMATICS (MEI) 4758/0 Differential Equations Candidates answer on the answer booklet. OCR supplied materials: 8 page answer booklet (sent with general stationery) MEI Examination Formulae and Tables (MF) Other materials required: Scientific or graphical calculator Wednesday 8 May 0 Morning Duration: hour 0 minutes * * * * INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces provided on the answer booklet. Please write clearly and in capital letters. Use black ink. 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 three questions. Do not write 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 g m s. Unless otherwise instructed, when a numerical value is needed, use g = 9.8. INFORMATION FOR CANDIDATES 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. The total number of marks for this paper is 7. This document consists of 4 pages. Any blank pages are indicated. OCR 0 [R/0/66] OCR is an exempt Charity RP 0I Turn over

2 The differential equation is to be solved. d y dt + 4 dy + y = cos t dt ( ) (i) Find the general solution. [9] Find the particular solution, given that when t = 0, y and dy dt Now consider the differential equation d dt + 4 d dt + d = 6 sin t. dt are both zero. [6] (iii) Show that the general solution may be expressed as = y + c where y is the general solution of ( ) and c is a constant. [] (iv) When t = 0, =, d dt = 0 and d =. Use these conditions to find the particular solution. [7] dt (a) A curve in the x-y plane satisfies the differential equation for x > 0. dy y x = x (i) Find the general solution for y in terms of x. [8] The curve passes through (, 0). Find the equation of this curve. [] (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] (b) The differential equation dy = x + y is to be solved approximately by using a tangent field. (i) Describe the shape of the isocline for which dy =. [] Sketch, on the same axes, the isoclines for the cases dy dy dy =, =, =. Use these isoclines to draw a tangent field. [] (iii) Sketch the solution curve through (0, ). [] (iv) Sketch the solution curve through the origin. [] OCR /0 Jun

3 (a) A particle of mass 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 k x N directed towards O, where k is a positive constant. (i) Show that the velocity, v m s, of the particle satisfies the differential equation The particle is at rest when x = a, where a is a positive constant. v dv = k x. [] Solve the differential equation, subject to this condition. Hence show that, while the particle moves in the negative direction, Initially the particle is at x = a. (iii) Use the standard integral dt = k a x. [6] a x = arcsin( x a ) + c 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 ω dω = 9 sin θ dθ where ω = dθ dt. (i) Solve the differential equation for ω in terms of θ subject to the condition ω = 0 when θ = π. Hence show that, while θ is decreasing, dθ dt = cos θ. [6] Starting from θ = π when t = 0, use Euler s method with a step length of 0. to estimate θ when t = 0.. The algorithm is given by t r+ = t r + h, θ r+ = θ r + h θ r. State whether this algorithm can usefully be continued, justifying your answer. [4] [Question 4 is printed overleaf.] OCR /0 Jun Turn over

4 4 The quantities x and y at time t are modelled by the simultaneous differential equations 4 = x y + t, dt dy = x + y + t +. dt (i) Show that d x dt + + x = 5t. [5] dt 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. [] 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 ( 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 opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB GE. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR /0 Jun

5 GCE Mathematics (MEI) Advanced GCE Unit 4758: Differential Equations Mark Scheme for June 0 Oxford Cambridge and RSA Examinations

6 OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners meeting before marking commenced. 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 skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 0 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG5 0DL Telephone: Facsimile: publications@ocr.org.uk

7 4758 Mark Scheme June 0 (i) 4 0 M Auxiliary equation or - A CF t t Ae Be F CF for their roots PI y acos t bsin t B y asint bcost Differentiate twice and M substitute y 4acost 4bsint 4acos t 4bsin t 8asin t 8bcos t acos t bsin t cos t M Compare coefficients 8b a b 8a 0 A 8 a, b 5 5 A t t GS y (8sin t cos t) Ae Be PI + CF with two arbitrary 5 F constants t 0, y 0 0 A B 5 M Use condition y t t (6 cos sin t) Ae B e 5 M Differentiate F 6 t 0, y 0 0 A B 5 M Use condition A, B= 0 A t t y (8sin t cos t) e e 5 0 A Cao 6 (iii) If z y c, differentiating (*) gives new DE M Recognise derivative and has arbitrary constants so must be GS A or Integrating gives (*) with k on RHS M PI will be previous PI k, CF as before, so GS y c A SC for showing that correct y from (i) + c satisfies new DE (iv) (8sin cos ) e t z t t D Ee c 5 t 0, z D E c 5 M Use condition (6 cos sin ) e t z t t D E e 5 F Derivative 6 t 0, z 0 0 D E 5 M Use condition Second derivative: t t z ( sint 4cos t) De 9Ee 5 F condone, for this mark only, +c appearing 4 t 0, z D 9E 5 M Use condition D, E, c 0 B t t z (8sin t cos t) e e A Cao

8 4758 Mark Scheme June 0 (a)(i) I exp( d x) x M Attempt integrating factor exp( ln x) A x A dy x x y x M Multiply both sides by IF d ( ) x y x M x y x A M Integrate both sides A y x Ax F Must divide constant 0 A M y x x A (iii) x 0, y 0 F dy 9 4x x 0 x (as x 0) 6 M dy x 0, 0 F B Behaviour at origin B Through and shape for x 8 (b)(i) Circle centre origin B Radius B B Stationary point at B One isocline correct B All three isoclines correct B Reasonably complete and accurate direction indicators (iii) B Solution curve (iv) B Solution curve B Zero gradient at origin 6

9 4758 Mark Scheme June 0 (a)(i) NL: ma k x M dv v k M Acceleration dv v k E vdv k x M Separate and integrate v k x A A LHS A RHS x a, v 0 A k a M Use condition v k ( a x ) A So for v 0, k a x d t E (iii) a x kt d arcsin x B kt a A LHS A RHS x a, t 0 B M Use condition x asin( kt) acos kt A Either form M Separate and integrate (b)(i) d 9sin d M Separate and integrate 9cos C A LHS A RHS 9, 0 C M Use condition So 9( cos ) A d cos (decreasing) dt E 0 M So estimate 0 A The algorithm will keep giving B but is not constant so not useful B

10 4758 Mark Scheme June 0 4(i) y x x t M y x x M x x x ( x x t) t M Eliminate M x x x 5t E Eliminate 0 M Auxiliary equation (repeated) A Root CF: ( A Bt)e t F CF for their root(s) (with two constants) (iii) PI: x at b B x a, x 0 In DE: 0 a at b 5t M Differentiate and substitute a 5 a b M Compare and solve a 5, b 9 A GS: x 9 5 t ( A Bt)e t F GS = PI + CF with two arbitrary constants y x x t M t [ 5 e ( )e t B A Bt ] [9 5 t ( A Bt)e ] t M Substitute t t M 9t ( A B Bt)e A Differentiate (product rule) (iv) t 0, x 9 A 0 M Use condition t 0, y 0 0 B B M Use condition x 9 5t te t A y 9t ( t)e t A (v) e t 0 M x 9 5t F y 9t F

11 OCR (Oxford Cambridge and RSA Examinations) Hills Road Cambridge CB EU OCR Customer Contact Centre 4 9 Qualifications (General) Telephone: Facsimile: general.qualifications@ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; Hills Road, Cambridge, CB EU Registered Company Number: OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: Facsimile: OCR 0

12 Examiners Reports June /0: 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 and 4 were attempted by almost all of the candidates, with Question 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 (i) (iii) (iv) This question was attempted by all candidates and many earned the majority of the available marks. 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. As in part (i), there were some algebraic errors. 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. 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. Candidates usually answered part (a) well, but many seemed unclear of the terminology and/or methods involved in part (b). (a) (i) (iii) (b) (i) Candidates showed a good understanding of the integrating factor method of solving this first order differential equation, and they applied it with accuracy. Again, this use of an initial condition was well-executed. 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. 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. A statement that the isocline is a circle with centre at the origin and with unit radius was required here. 6

13 Examiners Reports June 0 (iii) (iv) (a) (i) (iii) (b) (i) 4 (i) (iii) (iv) (v) 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. The majority of candidates were able to recover here and sketch the solution curve through (0,). 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. This was the least popular choice of question, but those who selected it were usually successful in scoring the majority of the marks. This was invariably answered well. 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. Again, the separation of variables and integration was handled correctly by the candidates. 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. 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. The vast majority of candidates attempted this question and many scored full marks. Solutions were almost always convincing and correct. Candidates were clearly very familiar with the method and worked through it accurately. 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. Again, there were a few arithmetical slips, but the method was well-known by all. This request presented no problems to candidates and with follow through from previous incorrect answers, almost all candidates scored the three marks available. 7

14 For a description of how UMS marks are calculated see: GCE Mathematics (MEI) Max Mark a b c d e u 475/0 (C) MEI Introduction to Advanced Mathematics Raw UMS /0 (C) MEI Concepts for Advanced Mathematics Raw UMS /0 (C) MEI Methods for Advanced Mathematics with Coursework: Written Paper Raw /0 (C) MEI Methods for Advanced Mathematics with Coursework: Coursework Raw /8 (C) MEI Methods for Advanced Mathematics with Coursework: Carried Forward Coursework Mark Raw (C) MEI Methods for Advanced Mathematics with Coursework UMS /0 (C4) MEI Applications of Advanced Mathematics Raw UMS /0 (FP) MEI Further Concepts for Advanced Mathematics Raw UMS /0 (FP) MEI Further Methods for Advanced Mathematics Raw UMS /0 (FP) MEI Further Applications of Advanced Mathematics Raw UMS /0 (DE) MEI Differential Equations with Coursework: Written Paper Raw /0 (DE) MEI Differential Equations with Coursework: Coursework Raw /8 (DE) MEI Differential Equations with Coursework: Carried Forward Coursework Mark Raw (DE) MEI Differential Equations with Coursework UMS /0 (M) MEI Mechanics Raw UMS /0 (M) MEI Mechanics Raw UMS /0 (M) MEI Mechanics Raw UMS /0 (M4) MEI Mechanics 4 Raw UMS /0 (S) MEI Statistics Raw UMS /0 (S) MEI Statistics Raw UMS /0 (S) MEI Statistics Raw UMS /0 (S4) MEI Statistics 4 Raw UMS /0 (D) MEI Decision Mathematics Raw UMS /0 (D) MEI Decision Mathematics Raw UMS /0 (DC) MEI Decision Mathematics Computation Raw UMS /0 (NM) MEI Numerical Methods with Coursework: Written Paper Raw /0 (NM) MEI Numerical Methods with Coursework: Coursework Raw /8 (NM) MEI Numerical Methods with Coursework: Carried Forward Coursework Mark Raw (NM) MEI Numerical Methods with Coursework UMS /0 (NC) MEI Numerical Computation Raw UMS Unit level raw mark and UMS grade boundaries June 0 series: GCE 9