Morning Time: 1 hour 30 minutes Additional materials (enclosed):

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1 ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph paper MEI Examinaion Formulae and Tables (MF) INSTRUCTIONS TO CANDIDATES Wrie your name in capial leers, your Cenre Number and Candidae Number in he spaces provided on he Answer Bookle. Read each quesion carefully and make sure you know wha you have o do before saring your answer. Answer any hree quesions. You are permied o use a graphical calculaor in his paper. Final answers should be given o a degree of accuracy appropriae o he conex. The acceleraion due o graviy is denoed by g ms. Unless oherwise insruced, when a numerical value is needed, use g = 9.8. INFORMATION FOR CANDIDATES The number of marks is given in brackes [ ] a he end of each quesion or par quesion. The oal number of marks for his paper is 7. You are advised ha an answer may receive no marks unless you show sufficien deail of he working o indicae ha a correc mehod is being used. This documen consiss of prined pages. OCR 008 [R/0/66] OCR is an exemp Chariy [Turn over

2 The differenial equaion d y + dy + y = f() is o be solved for 0 subjec o he condiions ha dy = 0andy = 0when = 0. Firsly consider he case f() =. (i) Find he soluion for y in erms of. [0] Now consider he case f() =e. Explain briefly why a paricular inegral canno be of he form ae or ae. Find a paricular inegral and hence solve he differenial equaion, subjec o he given condiions. [8] (iii) For > 0, show ha y > 0 and find he maximum value of y. Hence skech he soluion for 0. [You may assume ha k e 0as for any k.] [6] A raindrop falls from res hrough mis. Is velociy, v ms verically downwards, a ime seconds afer i sars o fall is modelled by he differenial equaion ( + ) dv + 3v =( + )g 3. (i) Solve he differenial equaion o show ha v = g( + ) +( g)( + )3. [0] The model is refined and he erm 3 is replaced by he erm v, giving he differenial equaion ( + ) dv + 3v =( + )g v. Find he soluion subjec o he same iniial condiions as before. [9] (iii) For each model, describe wha happens o he acceleraion of he raindrop as. [] OCR /0 Jan08

3 3 3 The populaion, P, of a species a ime years is o be modelled by a differenial equaion. The iniial populaion is 000. A firs he model dp = 0.P is used. (i) Find P in erms of. [3] To ake accoun of observed flucuaions, he model is refined o give dp = 0.P + 70 sin. Sae he complemenary funcion for his differenial equaion. Find a paricular inegral and hence sae he general soluion. [8] (iii) Find he soluion subjec o he given iniial condiion. [] The model is furher refined o give dp = 0.P + P 3 sin. This is o be solved using Euler s mehod. The algorihm is given by r+ = r + h, P r+ = P r + hṗr. (iv) Using a sep lengh of 0. and he given iniial condiions, perform wo ieraions of he algorihm o esimae he populaion when = 0.. [] The populaion is observed o end o a non-zero finie limi as, so a furher model is proposed, given by dp = 0.P( P 000 ). (v) Wihou solving he differenial equaion, (A) find he limiing value of P as, [3] (B) find he value of P for which he rae of populaion growh is greaes. [] The simulaneous differenial equaions areobesolvedfor 0. dx =3x + y + 9, dy =x + y +, (i) Show ha d x + dx + x = 6. [] Find he general soluion for x. [7] (iii) Hence find he corresponding general soluion for y. [3] (iv) Find he soluions subjec o he condiions ha x = y = 0when = 0. [] (v) Skech, on separae axes, graphs of he soluions for 0. [] OCR /0 Jan08

4 Permission o reproduce iems where hird-pary owned maerial proeced by copyrigh is included has been sough and cleared where possible. Every reasonable effor has been made by he publisher (OCR) o race copyrigh holders, bu if any iems requiring clearance have unwiingly been included, he publisher will be pleased o make amends a he earlies possible opporuniy. OCR is par of he Cambridge Assessmen Group. Cambridge Assessmen is he brand name of Universiy of Cambridge Local Examinaions Syndicae (UCLES), which is iself a deparmen of he Universiy of Cambridge. OCR /0 Jan08

5 78 Mark Scheme January Differenial Equaions (i ) α + α + = 0 M Auxiliary equaion α = (repeaed) A CF y = ( A+ B)e F CF for heir roos PI y = a B Consan PI in DE y = B PI correc y = + ( A+ B) e Their PI + CF (wih wo F arbirary consans) = 0, y = 0 0 = + A A= M Condiion on y y = ( B AB)e M Differeniae (produc rule) = 0, y = 0 0 = B A B = M Condiion on y y = + e A Boh erms in CF hence will give zero if subsiued in LHS E PI y = b e B y ( b b ) e =, y = ( b b+ b ) e in DE ( b b b ( b b ) b ) e = e Differeniae wice and M subsiue b = A PI correc y = ( C+ D+ ) e Their PI + CF (wih wo F arbirary consans) = 0, y = 0 0 = C M Condiion on y y = ( D+ CD ) e = 0, y = 0 0 = DC D = 0 M Condiion on y e y = A (iii) > 0 > 0 and e > 0 y > 0 E e y = so y = 0 = 0 = 0 or M Solve y = 0 Maximum a =, y = e A Maximum value of y B Sars a origin B Maximum a heir value of y B y >

6 78 Mark Scheme January 008 (i ) (iii) dv v = g M Rearrange M Aemp inegraing facor 3 + = = = + + A Correc A Simplified dv + + 3( + ) v = g( + ) 3 3( + ) F Muliply DE by heir I d (( + ) 3 v) = g( + ) 3 3( + ) 3 3 ( + ) v = ( g( + ) 3( + ) ) dx M Inegrae 3 3 = g A A RHS v = g + + A + F Divide by heir I (mus also divide consan) = 0, v = 0 0 = g + A M Use condiion v = g + + g + E Convincingly shown 3 dv ( + ) + v = ( + ) g M Rearrange dv + v = g + M Aemp inegraing facor + I = = = + + A Simplified dv + + ( + ) v = g( + ) F Muliply DE by heir I d (( + ) v) = g( + ) ( + ) v = g( + ) dx M Inegrae = g + + B A RHS v g ( ) ( ) v = g + + B + F Divide by heir I (mus also divide consan) = 0, v = 0 0 = g+ B M Use condiion ( ) = F Follow a non-rivial GS dv = + M Find acceleraion,+ 0 B Firs model: g 3 ( g)( ) As Hence acceleraion ends o g A dv 6 Second model = g 6 ( + ( + ) ) M Find acceleraion Hence acceleraion ends o 6 g A Idenify erm(s) 0 in heir soluion for eiher model 0 9 8

7 78 Mark Scheme January 008 3(i) 0. P = Ae M Any valid mehod = 0, P = 000 A= 000 M Use condiion 0. P = 000e A 0. CF P = Ae F Correc or follows (i) PI P = acos + bsin B P = asin+ bcos M Differeniae asin + bcos = 0.( acos + bsin ) + 70sin M Subsiue a = 0.b+ 70 M Compare coefficiens b = 0.a M Solve solving a = 80, b = 0 A 0. GS P = Ae 80cos 0sin Their PI + CF (wih one arbirary F consan) (iii) = 0, P = 000 A = 080 M Use condiion 0. P = 080e 80cos 0sin F Follow a non-rivial GS (iv) P P M Use of algorihm A A A 08 (v) (A) Limiing value P = 0 M Se P = 0 P P = (as limi non-zero) limiing value = 000 (B) Growh rae max when P f( P) = P max 000 M A M P P f ( P) = P M P f ( P) = 0 P = M P = 8000 A Solve Recognise expression o maximise Reasonable aemp a derivaive Se derivaive o zero

8 78 Mark Scheme January 008 (i) x = 3x + y M Differeniae firs equaion = 3 x + ( x+ y+ ) M Subsiue for y y = 3x 9+ x M y in erms of x, x x =3x x+ 3x 9+ x + M Subsiue for y x+ x + x = 6 E λ + λ+ = 0 M Auxiliary equaion λ = ± j A CF x e ( Acos Bsin) = + M CF for complex roos F CF for heir roos PI x = a B Consan PI a = 6 a = 3 B PI correc GS x = 3+ e ( Acos+ Bsin) Their CF + PI (wih wo arbirary F consans) (iii) y = 3x 9+ x M y in erms of x, x ( A B ) ( A B ) ( A B ) ( ) = 9+ 3e cos + sin 9 e cos + sin + e sin + cos M y = e A+ B cos+ B A sin A Differeniae x and subsiue (iv) 0= 3+ A A= 3 M Condiion on x 0= A+ B B = 6 M Condiion on y x = 3+ 3e sin cos F Follow heir GS y = e sin F Follow heir GS Consans mus correspond wih hose in x (v) B Skech of x sars a origin B Asympoe x = B Skech of y sars a origin B Decaying oscillaions (may decay rapidly) B Asympoe y = 0 30

9 Repor on he Unis aken in January : Differenial Equaions (Wrien paper) General Commens Many candidaes demonsraed a good undersanding of he specificaion and high levels of algebraic compeency. Quesions and proved o be he mos popular choices. When skeching graphs, if candidaes use a graphic calculaor, merely copying he screen may no be enough if i does no idenify he key feaures of he soluion. However, deailed analysis is no required in skech graphs unless specifically requesed in he quesion. When a differenial equaion and condiions are given, a reques o find he soluion implies ha he condiions should be used. (ie in hese circumsances, find he paricular soluion unless he specific erm general soluion is used.) Commens on Individual Quesions Secion A ) (i) This was generally done very well, alhough a few candidaes did no use he given condiions o calculae he arbirary consans. (iii) Alhough here were some excellen answers, mos candidaes sruggled o explain why he given expressions could no form a paricular inegral. They were required o say more han jus idenify hem as erms in he complemenary funcion, bu o remark ha hey would give zero in he lef hand side of he differenial equaion. This was ofen done well, alhough some candidaes found he value of a he maximum, raher han he value of y. I was expeced ha he maximum value was marked on he skech graph. ) (i) This was ofen done very well, alhough some candidaes made errors when dividing or muliplying he equaion hrough by an appropriae expression. For example, when dividing hrough by ( + ), some candidaes did no divide he 3 erm. This was also ofen done well, alhough some candidaes wrongly used he same inegraing facor as before. (iii) Compleely correc answers were no common. Many candidaes made errors in differeniaing heir velociy expressions. A few described he velociy raher han he acceleraion. 3) (i) A surprising number of candidaes did no use he condiion o calculae he arbirary consan. Noe ha he quesion did no ask for he general soluion and so he condiion should have been used. (iii) This was ofen done well, alhough algebraic errors when calculaing he coefficiens were common. The paricular soluion was usually well answered.

10 Repor on he Unis aken in January 008 (iv) (v) The numerical soluion was ofen done well, alhough a sizeable minoriy made numerical errors in he second sep. Finding he limiing value was someimes well done, alhough many did no realise ha his mus correspond wih a zero derivaive. Finding he populaion when he growh rae is greaes was rarely compleed. A number of candidaes differeniaed he expression, bu few were able o solve he resuling equaion. ) (i) Mos candidaes compleed his correcly, bu a few did no seem o know how o do he eliminaion. (iii) (iv) (v) This was ofen correc, bu some candidaes assumed ha he paricular inegral was 6. Many candidaes gave correc answers by using heir soluion for x in he firs of he displayed differenial equaions. Pleasingly few candidaes aemped o consruc a differenial equaion for y. Many correc soluions were seen. The skeches were ofen done well, bu some candidaes omied o idenify he key feaures, in paricular he iniial condiions and he asympoes. 3