The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 72.

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1 ADVANCED GCE UNIT 758/0 MATHEMATICS (MEI) Differential Equations THURSDAY 5 JANUARY 007 Additional materials: Answer booklet (8 pages) Graph paper MEI Examination Formulae and Tables (MF) Morning Time: hour 0 minutes INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces provided on the answer booklet. Answer any three questions. You are permitted to use a 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 ms. 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. The total number of marks for this paper is 7. ADVICE TO CANDIDATES Read each question carefully and make sure you know what you have to do before starting your answer. 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. This document consists of printed pages. HN/ OCR 007 [R/0/66] OCR is an exempt Charity [Turn over

2 The differential equation d y dy y e kt dt dt is to be solved for t 0. Consider the case k. (i) Find the general solution. [9] (ii) Find the particular solution subject to the conditions y 0 when t 0, and y tends to zero as t tends to infinity. Show that this solution is zero only when t 0 and sketch a graph of the solution. [7] Consider now the case k. (iii) Find the general solution. Find also the particular solution subject to the same conditions as in part (ii). [You may assume that te t Æ 0 as t Æ. ] [8] The differential equation dy y cot x k, dx (*) where k is a constant, is to be solved for 0 x p. d (i) Show that [] dx ( ln sin x) cot x. (ii) In the case k 0, solve the differential equation by separating the variables to find the general solution for y in terms of x. [6] Now assume that k 0. (iii) Solve the differential equation to show that the general solution is y A cosec x k cot x where A is an arbitrary constant. [7] (iv) Find the particular solution subject to the condition y 0 when x p. Sketch the graph of the solution for 0 x p, showing the behaviour of y as x tends to zero. [] (v) By using the double angle formulae for sin x and cos x, or otherwise, show that there is a solution to (*) for which y tends to a finite limit as x tends to zero. State the solution and its limiting value. [6] OCR /0 Jan 07

3 In an experiment, a ball-bearing of mass m kg falls through a liquid. The ball-bearing is released from rest and t seconds later its displacement is x m and its velocity is v ms. The forces acting on the ball-bearing are its weight and a resistance force RN. Three models for R are to be considered. In the first model, R mk v, where is a positive constant. k dv (i) Show that Hence show that v g ( e k t ). [7] dt g k v. k (ii) Find an expression for x in terms of t. [] In the second model, R mk v, where is a positive constant. k v (iii) Show that dv and hence find v in terms of x. [7] g k v dx In the third model, R mk v, where is a positive constant. Euler s method is used to solve the = dv resulting differential equation = g -k v dt. The algorithm is given by t r t r h, v r v r hv.. r k (iv) Given k.5 and using a step length of 0., perform two iterations of the algorithm to estimate v when t 0.. [5] The terminal velocity of the ball-bearing is m s. (v) Verify the value of given in part (iv). [] k [Question is printed overleaf.] OCR /0 Jan 07 [Turn over

4 The following simultaneous differential equations are to be solved. dx =-x - y + 0, dt dy = x - y + 5. dt dx dy (i) Find the values of x and y for which 0. [] dt dt d x dx (ii) Show that 5x 5. [5] dt dt (iii) Find the general solution for x in terms of t. Hence obtain the corresponding general solution for y. [0] (iv) Given that x y 0 when t 0, find the particular solutions. [] (v) Sketch the graph of the solution for x, making clear the behaviour of the curve initially and for large values of t. [] Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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 Jan 07

5 9 Mark Scheme 758 January 007

6 758 Mark Scheme Jan 007 (i) λ λ = 0 M Auxiliary equation λ = or A t t CF y = Ae + Be F CF for their roots PI ae t y = B t t y = ae, y = ae M Differentiate twice ( ) t t t t ae ae ae = e M Substitute a = M Compare and solve a = A t t t y = Ae + Be + e F Their CF with constants + their PI (ii) 0 = A+ B+ M Use initial condition t t t t e 0,e 0,e so y 0 B = 0 M Use asymptotic condition t t ( ) y = e e A cao t t t y = 0 e = e e = t = 0 M Valid method to establish 0 is only root E Complete argument B Curve satisfies both conditions B y 0 for t > 0 and consistent with their solution 9 (iii) CF PI t t y = Ce + De F Correct or same as in (i) e t y = bt B t t t t y = be bte, y = be + bte ( ) t t t t t t be + bte be bte be = e M Differentiate (product) and substitute b b= b= A cao GS t t t y = Ce + De te F Their CF + their non-zero PI y = 0, t = 0 C+ D = 0 M Use condition y 0 D = 0 M Use condition e t t y = A cao 7 8 0

7 758 Mark Scheme Jan 007 (i) (ii) d ( ln sin ) cos cot d = x sinx = x E Differentiate (chain rule) dy cotx y dx M Rearrange (iii) dy = cotxdx y M Integrate ln y = ln sin x + c A One side correct (ignore constant) A All correct, including constant y = Acosec x M Rearrange, dealing properly with constant A dy ycotx k dx + = ( cot xdx) I = exp M Attempt integrating factor = exp( ln sin x) M Integrate = sin x A Simplified form of IF dy sin x + y cos x = k sin x dx M Multiply by their IF ysin x = ksin xdx M Integrate both sides = k x+ A A cao cos y = Acosec x kcot x E (iv) x = π, y = 0 0= A M Use condition y = kcot x A B Increasing and through ( π ) B Asymptote x = 0,0 6 7 (v) ( ) Both double angle formulae correct (or small A kcos x A k sin x B y = = angle approximations or series expansion) sin x sin xcos x M Use expressions in general solution A ksin x M Identify value of A A= k y = cos x E Correct solution, fully justified which tends to zero as x 0 B Must be from correct solution 6

8 758 Mark Scheme Jan 007 (i) dv m = mg R dt B NL equation (accept ma, allow sign errors) dv = g kv dt E Must follow from correct NL dv = dt g k v M Separate and integrate ln g kv = t + c A LHS k g k v = A M Rearrange (dealing properly with constant) e kt Alternatively M Attempt integrating factor d A ( e kt ) e kt v = g dt M Integrate Alternatively M Auxiliary equation A CF Ae kt M Constant PI ( g / k ) t = 0, v = 0 A= g M Use condition ( e kt ) g v = E k (ii) g M Integrate v x v dt t e kt = = + + B k k A cao (including constant) t = 0, x = 0 B = M Use condition k g e kt x = t+ k k k (iii) dv mv mg mkv dx dv mkv (accept ma or m ) d t v dv = g k dx v E Must follow from correct NL v dv = dx g kv M Integrate ln g kv k = x+ c A LHS g kv = Ce kx M Rearrange (dealing properly with constant) x = 0, v = 0 C = g M Use condition v = g ( e kx ) k A cao (iv) t v v B First line M Use algorithm A M Use algorithm A.86 (accept sf or better) A cao 7 7 5

9 758 Mark Scheme Jan 007 (v) g g kv = 0 when v = k = =.5 E Deduce or verify value (must relate to resultant force or acceleration being zero)

10 758 Mark Scheme Jan 007 (i) subtracting 5x + 5= 0 M Solve simultaneously x = A y = 7 A (ii) x = x y M Differentiate = x (x y+ 5) M Substitute for y = x x+ ( x x+ 0) 5 M y in terms of x, x M Substitute x+ x + 5x = 5 E (iii) λ + λ+ 5= 0 M Auxiliary equation M Solve to get complex roots λ = ± j A t CF x e ( Acost Bsint) = + F CF for their roots 5 PI x = = 5 B t GS x = e ( Acost+ Bsint) + F Their CF with constants + their PI y = x x+ 0 M y in terms of x, x t t ( A t B t) ( A t B t) t ( A t B t) (( A B) t ( A B) t) = e sin + cos + e cos + sin M Differentiate their x e cos + sin + 0 M Substitute t = e cos + sin + 7 A cao (iv) t = 0, x = 0 A+ = 0 M Use condition on x t = 0, y = 0 A+ B+ 7 = 0 M Use condition on y A=, B = 8 t x = e 8sint cost + ( ) t ( ) y = e 7cost+ 9sint + 7 A Both correct (v) B Through origin B Positive gradient at t = 0 B Asymptote x =, or their non-zero constant PI (accept oscillatory or non-oscillatory) NB Oscillates about x =, but not apparent at this scale due to small amplitude 5 0

11 Report on the units taken in January 007 General Comments Differential Equations The standard of work was generally very good, demonstrating a clear understanding of the techniques required. Almost all candidates answered questions and ; questions and were equally popular as a third choice. Candidates often produced accurate work when solving second order differential equations. However, when solving first order equations, it was common to see errors in integration, omission of the constant of integration, or not dealing properly with the constant. With regard to graph sketching, I would like to emphasise the following advice given in last summer s report. Candidates should note that generally the expectation of sketches in this unit is that any known information (such as initial conditions or relevant results found earlier in the question) should be indicated on the sketch. Also any obvious shape and features of the graph (e.g. oscillating, increasing, decreasing, unbounded, asymptotes), should be shown. Detailed calculations are not required, unless specifically requested. Comments on Individual Questions Second order differential equations (i) This was often completely correct. (ii) (iii) Many candidates correctly found the particular solution, but a few candidates were unable to use the asymptotic condition. Only a minority of candidates were able to demonstrate that the solution is zero only when t = 0. Commonly candidates simply showed that t = 0 implies y = 0, or gave a vague argument why it is the only root. Sketch graphs were often good, except they often showed y positive rather than negative. The solution was often correct, but some did not realise the correct form of the particular integral and some made slips with their arbitrary constants. First order differential equations (i) This derivative (using the chain rule) was intended to help with the integral later in the question, but unfortunately many candidates misunderstood the request and approached it as an integral or a differential equation. (ii) (iii) This was often done well, but slips in integration were common, in particular not dealing with the x properly and omitting the constant. Another common error was not dealing with the constant properly when rearranging. Most candidates attempted to use an integrating factor, but slips were common. I = exp cot xdx = exp ln sin x = sin x, which gave a A common error was ( ) ( ) valid integrating factor but contained two errors, i.e. omitting the in the integral and then wrongly dealing with the. 8

12 Report on the units taken in January 007 (iv) (v) Many candidates successfully found the particular solution, but some made slips in the calculation. The sketch graphs varied widely in quality. Some candidates simply drew a sketch of cotx. Many sketches did not show the given condition, indeed some wrongly had a break in the curve around the point ( π,0 ), presumably due to entering y = /tanx into a graphical calculator. If using a graphical calculator, candidates must use it intelligently and not just copy the screen without thinking about what the graph should look like. Very few candidates successfully completed this part. Many tried to solve the differential equation again, which was rarely successful, rather than use their general solution. Modelling the motion of a ball-bearing falling through a liquid (i) Candidates usually formulated the differential equation correctly from Newton s second law, and many solved it correctly. However, errors did occur, in particular with the integral. Also, some candidates omitted the constant or made mistakes with the constant when rearranging their solution. (ii) (iii) (iv) (v) Most candidates integrated the previous solution, but many omitted the constant or did not calculate it. This part was often done well, but candidates were not as successful as in part (i). Common errors with formulating the equation were: not starting from Newton s second law or not recalling the alternative form of acceleration. Common errors when solving the differential equation were: not integrating correctly, omitting the constant, or including the constant but wrongly stating that it was the initial displacement and hence zero. This was usually correct, although as usual with an Euler s method calculation, a few candidates produced unrecognisable figures with no method shown. Some candidates mistakenly gave the value of v at t = 0. by tabulating tv, r, v& r and in each row and then stating the final value of v in their final row. v r + Virtually all attempts at this calculation were correct. Simultaneous differential equations (i) This calculation was usually correct, although some made slips. (ii) (iii) (iv) (v) The elimination of y was often done well, although a few differentiated the first equation with respect to x rather than t. The solution for x was often done very well. When finding y, many candidates correctly used the first equation. As usually occurs, a few tried instead to set up and solve a differential equation for y. Such attempts were very lengthy and time-consuming and never attempted to relate the arbitrary constants in the two solutions, which is a vital feature of the solutions. The particular solutions were often done well. The sketch was often correct, but some candidates seemed to ignore the request to make clear the initial and the long-term behaviour of the solution on the sketch. 9

Concepts for Advanced Mathematics (C2) WEDNESDAY 9 JANUARY 2008

ADVANCED SUBSIDIARY GCE 4752/01 MATHEMATICS (MEI) Concepts for Advanced Mathematics (C2) WEDNESDAY 9 JANUARY 2008 Additional materials: Answer Booklet (8 pages) MEI Examination Formulae and Tables (MF2)