2610 MEI Differential Equations January 2005 Mark Scheme

Transcription

1 60 MEI Differenial Equaions January 00 Mark Scheme (i) I ep( ) = M = e A d ( ) e y e e y + = M muliply d ( e y) = e F follow heir inegraing facor e y = e M inegrae = + A e A ( e A ) y = e + F divide by heir I (mus divide consan also) (ii) 0 e ( A) A e ( e y ) and = + = M condiion on y = A cao > > 0 e > M aemp inequaliy for y e > 0 so y > 0 E fully jusified B hrough (,0) and y > 0 for > 7 (iii) dy maimum = 0 M ( ) B asympoic o y = 0 y = e M mus use DE = y = y = E ( ) = e e + A M subsiue ino GS for y ( ) A= 0 y = e A cao ¼ B general shape consisen wih heir soluion B maimum labelled a (, ) 6 7

2 60 MEI Differenial Equaions January 00 Mark Scheme (i) for < 0, au.eq. α + 9= 0 α = ± j M imaginary roo or recognise SHM equaion y = Acos+ Bsin A d y for > 0, 6y = 0 B may be implied α 6 = 0 M α =± A y = Ce + De F accep A,B again here bu no in (ii) (ii) () A = y0 F C+ D = y 0 F dy () < 0, = Asin + B cos M differeniae dy > 0, = Ce + De M differeniae B = C+ D A () only e is unbounded so D = 0 B hence C = y0 B B = y B 0 (iii) B curve for < 0 B curve for > 0 B coninuous gradien a = 0 (mus have reasonable aemp a each of < 0 and > 0) 6 8 (iv) d y y 0 y Ecos Fsin + = = + B d bounded so () provides no equaion hence insufficien ( equaions, unknowns) M A

3 60 MEI Differenial Equaions January 00 Mark Scheme (i) (ii) (iii) dv rae of flow = area speed = g M d dv M g d = E dv = = g B d = g d M separae 0 = g+ A M inegrae 0 accep eiher in words or symbols for M bu need boh for E use of chain rule complee argumen = 0, = A= M condiion on ( g) = A cao = A ( ) + = g M subsiue for d V d ( + ) = g d M separae + = g + B M inegrae A = 0, = B =.7 M condiion on 6 = 0 0 A (iv) (0) = 0.000( g + 0) / = E mus show working (0.) = M use of algorihm =.9997 E mus show working (0.) = 0.00 B (0.) =.9990 B 6 6

4 60 MEI Differenial Equaions January 00 Mark Scheme (i) d dy M differeniae = 8 d d d A = 8 ( + 7 y) M subsiue for d y d d = d d M subsiue for y d + 0 = 0 d d E (ii) α α + 0 = 0 M auiliary equaion α = or 0 A 0 = Ae + Be F CF for heir roos y ( ) M rearrange equaion () 8 ( 0 ( 0 A B A B )) = 8 e + 8 e e + 0 e M subsiue for and 0 Be = Ae A cao (iii) ae = 8ae be + e M subsiue in () be = ae + 7be + e M subsiue in () 9a+ b =, a 8b = M compare coefficiens and solve a = 6 A b = 6 A (iv) subsiuing f(), g() ino LHS gives zero B saed (or subsiuion) subsiuing soluions in (iii) gives e B as equaions linear, subsiuing sums gives sum 0+ e = e E (or verified by subsiuion) General soluions because wo arbirary consans as epeced for wo firs order equaions. E 6

5 60 Mark Scheme June 00 (i) d dy = + 6 M differeniae d d d = + 6( + y+ 6) M subsiue for d y d d = M subsiue for y d 6 d d = 00 E d d (ii) α + α + 0 = 0 M auiliary equaion α = ± j A CF = e ( Acos+ Bsin ) F CF for heir soluions 00 PI = = 0 B 0 GS = 0 + e ( Acos+ Bsin ) F heir CF + heir PI = e ( Acos Bsin Asin + Bcos ) d M differeniae y 6 ( 8) M rearrange and subsiue y = + e (( A+ B)cos + ( B A)sin) A cao (iii) 0 + A = 0 A = 0 M use = 0 wih heir = 0 + ( A + B) = 0 B = 6 use = 0 wih heir y = 0 (or heir M = 8 ) (iv) = 0 + e (6sin 0cos ) y = + e (8sin cos ) A B B B boh (cao) iniial condiion and asympoe for one graph boh generally correc (mus sar a origin) long-erm values 8 long-erm values can be found by seing = y = 0 in DE s M and solving he resuling equaions A

6 60 Mark Scheme June 00 (i) B B r sars a 6 and decreases, ending o I sars a 0, gradien is posiive bu decreases (ii) (iii) dr M kr ( ) d A dr = k r d ln r = k + c A all correc r = + Ae k M rearranging = 0, r = 6 A = M use iniial condiion r = + e E M separae and inegrae ( ) I r d e k = = + c k M inegrae r = 0, I = 0 c = k M use iniial condiion I = + ( e k ) k A cao k 60k ( ) 000 = 60 + e M condiion on heir I 60k 0k = e E mus follow correc I 60k unless k very small, e 0 0k k 0. E independen of oher marks (iv) 0. α + 0. = 0 α = 0. so CF r = Be B for given PI d r bsin ccos d 0.a + (0.c b) sin + (0.b + c) cos = cos M differeniae and subsiue 0.a = 0. 0.c b = 0 compare a leas wo coefficiens and M solve 0.b + c = 0. a =, b =, c = A condiions B =.99, so 0. ( ) r =.99e + + cos + 0sin A 0 7 6

7 60 Mark Scheme June 00 (a)(i) dy cos sin y = + M may be implied M use of algorihm A y(0.8) M use of algorihm A y(0.0) (a)(ii) dy/ decreases B For each sep, gradien used is greaer han dy/ over inerval, hence overesimaes y. B E skech showing curve and sep by sep soluion convincing argumen (b)(i) dy cos + y = + + M rearrange M aemp inegraing facor I = ep d = ( + ) + A dy ( + ) + ( + ) y = ( + )cos M muliply ( + ) y = ( + )cosd = ( + )sin sin M aemp inegraion by pars = ( + ) sin + cos + A A = 0, y = 0 A = M ( + ) sin + cos y = ( + ) E = 0. y = 0.7 B (b)(ii) dy cos sin y Since sin y < y, replacing sin y by y in = + gives an underesimae for dy/. Hence (while y > 0), he appro. DE will underesimae y. M A E consider effec on dy/ 9

8 60 Mark Scheme June 00 (i) α + α α = 0 M auiliary equaion ( α )( α + α + ) = 0 M facorise or demonsrae is a roo α =,, A CF y = Ae + Be + Ce CF for heir roos (mus have F consans) PI y = ae B correc form y ae =, y = 9ae, y = 7ae M differeniae and subsiue 7 a + 8a + a a = M compare coefficiens a = A e e e e y = A + B + C F heir CF + heir PI (ii) decays C = 0 B = 0, y = = A + B M condiion on y = A y = Ae Be + e M differeniae = 0, y = B + M condiion A =, B = so y = e + e e (iii) le u e = so y = u + u u = u( u u + ) e u = 0 and u u + = ( u ) + > 0 M consider quadraic (any valid mehod) hence y 0 for all E if discriminan used, value or working mus be shown mus indicae u non-zero ( u u ) u y = e e + e = + (( u ) ) 8 u 0 9 = + > M consider quadraic (any valid mehod) hence no urning poins E E 9 B sars a and asympoe y = 0 B Shape (increasing) 6

9 (i) λ + 6λ+ 8= 0 M λ = or A CF I = Ae + Be F PI I = ae B ae 6ae + 8ae = 6e M differeniae and subsiue a = 6 dm compare a = A I = e + Ae + Be F CF + PI (ii) λ + 6λ+ 9= 0 M λ = (repeaed) A CF I = ( C+ D)e F PI I = be B be 6be + 8be = 6e M subsiue and compare b = A ( )e I = + C+ D F CF + PI. = + C C = 0 M condiion on I (iii ) I = e ( C+ D)e + De M differeniae 0= C+ D D = M condiion on I ( ) I = e + e A cao as, I 0 F recognise e 0 λ + 6λ+ k = 0 λ = ± 9 k M 0< k < 9 9 k < wo negaive roos λ λ hence I = Ae + Be 0 E k > 9 λ = ± β j M e ( Acosβ+ Bsin β) 0 E comple roos wih negaive real par (or CF) 8

10 (i) (ii) (iii ) dy = y M separae dy = y M inegrae A ± LHS ln y = + c A ± RHS y = Ae M rearrange = 0, = = e M condiion y A y = ( ep( ) ) A = y 0.9 A from correc soluion dy + y = cos M divide I = ep( d ) M aemp inegraing facor = A d ( ) y = cos F follow heir I M inegrae (by pars) y = cos = sin sin = sin + cos + B RHS (or muliple) consan no A required here y = sin + (cos + B) M divide o ge y =, y = 0 B = cos sin M use condiion y = sin + (cos cos sin) A = y M subsiue = A cao ( cos ( 0. )) y = y+ y B y y M use algorihm A y(.9) ( ) Using smaller h would give greaer accuracy A B seen or implied by correc numerical value y(.0) 8

11 (i) dv F = ma mv = mg 0.00mv M NL (accep jus ma for M) dv v = g 0.00v E down posiive so weigh posiive and resisance negaive as i opposes moion B (ii) 0.00v M separae dv = 0.00 g 0.00v M inegrae (iii) ln g 0.00v = c A LHS (or muliple) ( ) 0.00 v = 000 g Ae M rearrange = 0, v = 0 A= g M use condiion 0.00 ( ) v = 000g e A cao = 0 v = 0. E mus follow correc work dv mv mg mv M dv v g v A v dv = g v M separae A + c g dv c + = + g v M aemp o inegrae LHS v gln g v = + c A = 0, v = 0. c = M use condiion (correc value of v a leas) v = = 76.6 M so 6. m deep A (iv) erminal velociy when acceleraion zero M v =.9 F follow heir DE 7 9 B increasing from (0,0) B decreasing o asympoe a.9 (or follow heir value) B cusp/ma a (0, 0.) (boh coordinaes shown).9

12 (i) = y + cos M differeniae = ( y+ cos ) + cos M subsiue = 8+ 6y cos ( sin ) M y in erms of, = 8+ ( + sin ) cos M subsiue + + = sin cos E (ii) λ + λ+ = 0 M auiliary equaion λ = ± j M solve o ge comple roos A CF = e ( Acos+ Bsin) CF for heir roos (for comple roos F mus be in ep/rig form, no comple eponenials) PI = asin + bcos B = acos bsin, = asin bcos M differeniae wice and subsiue a b+ a = M compare b+ a+ b = M solve a =, b = A = e ( Acos + Bsin ) + ( sin cos) F CF + PI (iii) y = ( + ) M y in erms of, sin = e ( Acos + Bsin ) + e ( Asin + Bcos + cos+ sin M ( ) ( ) ( ) y = e ( A B)cos + ( A+ B)sin + sin cos (iv) ~ ( sin cos) F y ~ ( sin cos) F hence for large, y B bu unless A = B = 0, A B A or A+ B B so y differeniae M subsiue for, A CF par A PI par B mus follow correcly from heir soluions mus follow correcly from heir soluions 0

13 78 Mark Scheme June 006 (i) λ = 0 B = Acos + Bsin M A cos or sin or Acosω+ Bsinω seen or GS for heir λ (ii) ( λ) < 0 M Use of discriminan A Correc inequaliy 0< λ < A Accep lower limi omied or (iii) α + α + = 0 M Auiliary equaion α = ± j A = e Ccos+ Dsin F CF for heir roos (iv) 0 ( ) = C M Condiion on ( ) ( ) & = e Ccos + Dsin + e Csin + Dcos M Differeniae (produc rule) 0= C+ D M Condiion on & D = 0 0 ( ) = e cos+ sin A cao (v) cos + sin = 0 M an = M =.07 A cao (vi) α + 6α + M Auiliary equaion α =, A = Ee + Fe F CF for heir roos 0 = E+ F M Condiion on & = Ee Fe 0= E F M Condiion on & E =, F = ( ) ( ) = e e A cao 0 = e e M Aemp complee mehod > 0 > e, > 0,e > 0 > 0 i.e. never zero E Fully jusified (only 0 required) 0 8

14 78 Mark Scheme June 006 (i) λ + = 0 λ = M CF = Ae A PI = a + b B a+ ( a+ b) = + M Differeniae and subsiue a =, a+ b= M Compare a =, = A b e 0, e = + + A F CF + PI = = = + A M Condiion on = + + F Follow a non-rivial GS Alernaively: I = ep M ( d) = e A Inegraing facor d e + e = e ( + ) d B Muliply DE by heir I e = e ( + ) d M Aemp inegral = e + e d M Inegraion by pars ( ) ( ) e 0, e e = e + e + A A = + + A F Divide by heir I (mus also divide consan) = = = + A M Condiion on = + + F Follow a non-rivial GS (ii) dy = y M Separae dy = y M Inegrae ln y = ln+ c y = B M Make y subjec, dealing properly wih consan ( ) = 0, =, y = y = M Condiion e y = + + F y = (heir in erms of ) (iii) dz + z = 6 M Divide DE by I = ep M Aemp inegraing facor ( ) = A Simplified ( z) d = 6 F Follow heir inegraing facor z = + C A z = + C F Divide by heir I (mus also divide consan) ( = 0, ) =, z = C = M Condiion on z z = + A cao (in erms of ) = = 0.8 y =.69 B Any values (a leas sf) z =.08 B All correc (and sf) 9 0

15 78 Mark Scheme June 006 (i) dv dv = f( ) so ( unless f( ) = 0 ), v 0 ± Consider d v v when v = 0, bu no if dv M dv 0 i.e. gradien parallel o v-ais (verical) E Mus conclude abou direcion dv = 000 v = = 0 M Consider d v when = 000 d so if v 0 hen gradien parallel o -ais (horizonal) E Mus conclude abou direcion M Add o angen field A Several verical direcion indicaors on -ais (ii) M Aemp one curve A M Aemp second curve A 6 V 0 = 0.0 probe reaches B B Mus be consisen wih heir curve V 0 = 0.0 probe reurns o A B Mus be consisen wih heir curve N.B. Canno score hese if curve no drawn (iii) ( = + ) vd v (9000 ) (000 ) M Separae M Inegrae v = c B LHS A RHS V = + + c M Condiion v = + + V0 0 (iv) minimum when = 000 B Clearly saed vmin = + + V0 M Subsiue heir ino v or v F Their v or v when = 000 need v min > 0 M For v min > 0 v > 0 if V > M Aemp inequaliy for min V 0 > A cao A V

16 78 Mark Scheme June 006 (i) && = & y& M Differeniae firs equaion = & ( y+ 8) M Subsiue for &y y = + & M y in erms of, & && = & + (+ & ) 8 M Subsiue for y && + & = 6 E LHS E RHS (ii) λ + λ = 0 M Auiliary equaion λ = or A CF = Ae + Be F CF for heir roos PI = a B Consan PI a = 6 a = B PI correc = + Ae + Be F Their CF + PI y = + & M y in erms of, & = + Ae + Be + ( Ae + Be ) M Differeniae and subsiue y = 7+ Ae + Be A Consans mus correspond wih hose in (iii) = + A + B M Condiion on 7 = 7 + A + B M Condiion on y A=, B = 0 M Solve = + e F Follow heir GS y = 7+ 0e F Follow heir GS 6 9 B Skech of sars a and decreases B Asympoe = B Skech of y sars a 7 and decreases B Asympoe y =7 7 9

17 78 Mark Scheme Jan 007 (i) λ λ = 0 M Auiliary equaion λ = or A CF y = Ae + Be F CF for heir roos PI ae y = B y = ae, y = ae M Differeniae wice ( ) ae ae ae = e M Subsiue a = M Compare and solve a = A y = Ae + Be + e F Their CF wih consans + heir PI (ii) 0 = A+ B+ M Use iniial condiion e 0,e 0,e so y 0 B = 0 M Use asympoic condiion ( ) y = e e A cao y = 0 e = e e = = 0 M Valid mehod o esablish 0 is only roo E Complee argumen B Curve saisfies boh condiions B y 0 for > 0 and consisen wih heir soluion 9 (iii) CF PI y = Ce + De F Correc or same as in (i) e y = b B y = be be, y = be + be ( ) be + be be be be = e M Differeniae (produc) and subsiue b b= b= A cao GS y = Ce + De e F Their CF + heir non-zero PI y = 0, = 0 C+ D = 0 M Use condiion y 0 D = 0 M Use condiion e y = A cao 7 8 0

18 78 Mark Scheme Jan 007 (i) (ii) d ( ln sin ) cos co d = sin = E Differeniae (chain rule) dy co y M Rearrange (iii) dy = co y M Inegrae ln y = ln sin + c A One side correc (ignore consan) A All correc, including consan y = Acosec M Rearrange, dealing properly wih consan A dy yco k + = ( co ) I = ep M Aemp inegraing facor = ep( ln sin ) M Inegrae = sin A Simplified form of IF dy sin + y cos = k sin M Muliply by heir IF ysin = ksin M Inegrae boh sides = k + A A cao cos y = Acosec kco E (iv) = π, y = 0 0= A M Use condiion y = kco A B Increasing and hrough ( π ) B Asympoe = 0,0 6 7 (v) ( ) Boh double angle formulae correc (or small A kcos A k sin B y = = angle approimaions or series epansion) sin sin cos M Use epressions in general soluion A ksin M Idenify value of A A= k y = cos E Correc soluion, fully jusified which ends o zero as 0 B Mus be from correc soluion 6

19 78 Mark Scheme Jan 007 (i) dv m = mg R d B NL equaion (accep ma, allow sign errors) dv = g kv d E Mus follow from correc NL dv = d g k v M Separae and inegrae ln g kv = + c A LHS k g k v = A M Rearrange (dealing properly wih consan) e k Alernaively M Aemp inegraing facor d A ( e k ) e k v = g d M Inegrae Alernaively M Auiliary equaion A CF Ae k M Consan PI ( g / k ) = 0, v = 0 A= g M Use condiion ( e k ) g v = E k (ii) g M Inegrae v v d e k = = + + B k k A cao (including consan) = 0, = 0 B = M Use condiion k g e k = + k k k (iii) dv mv mg mkv dv mkv (accep ma or m ) d v dv = g k v E Mus follow from correc NL v dv = g kv M Inegrae ln g kv k = + c A LHS g kv = Ce k M Rearrange (dealing properly wih consan) = 0, v = 0 C = g M Use condiion v = g ( e k ) k A cao (iv) v v B Firs line M Use algorihm A M Use algorihm A.86 (accep sf or beer) A cao 7 7

20 78 Mark Scheme Jan 007 (v) g g kv = 0 when v = k = =. E Deduce or verify value (mus relae o resulan force or acceleraion being zero)

21 78 Mark Scheme Jan 007 (i) subracing + = 0 M Solve simulaneously = A y = 7 A (ii) = y M Differeniae = ( y+ ) M Subsiue for y = + ( + 0) M y in erms of, M Subsiue + + = E (iii) λ + λ+ = 0 M Auiliary equaion M Solve o ge comple roos λ = ± j A CF e ( Acos Bsin) = + F CF for heir roos PI = = B GS = e ( Acos+ Bsin) + F Their CF wih consans + heir PI y = + 0 M y in erms of, ( A B ) ( A B ) ( A B ) (( A B) ( A B) ) = e sin + cos + e cos + sin M Differeniae heir e cos + sin + 0 M Subsiue = e cos + sin + 7 A cao (iv) = 0, = 0 A+ = 0 M Use condiion on = 0, y = 0 A+ B+ 7 = 0 M Use condiion on y A=, B = 8 = e 8sin cos + ( ) ( ) y = e 7cos+ 9sin + 7 A Boh correc (v) B Through origin B Posiive gradien a = 0 B Asympoe =, or heir non-zero consan PI (accep oscillaory or non-oscillaory) NB Oscillaes abou =, bu no apparen a his scale due o small ampliude 0

22 78 Mark Scheme June 007 (i) λ + λ+ 9= 0 M Auiliary equaion M Solve for comple roos λ = ± j A CF y = e ( Acos+ Bsin) CF for heir roos (if comple, mus F be ep/rig form) PI y = acos+ bsin B Correc form for PI y = asin + bcos, y = acos bsin M Differeniae wice acos bsin + ( asin + bcos) + 9( acos+ bsin ) = cos M Subsiue b+ 8a = Compare coefficiens (boh sin and M a+ 8b =0 cos) a = 0.0 M Solve for wo coefficiens b = 0.0 A Boh y = e ( Acos + B sin ) + 0.0cos+ 0.0sin GS = PI + CF (wih wo arbirary F consans) (ii) = 0, y = 0 0 = A+ 0.0 M Use condiion on y A = 0.0 F y = e ( Acos+ Bsin) + e ( Asin + Bcos) 0.0sin + 0.0cos M Differeniae (produc rule) = 0, y = 0 0 = A+ B+ 0.0 M Use condiion on y B = 0.0 y = e ( 0.0cos + 0.0sin ) + 0.0cos+ 0.0sin A cao For large, y 0.0cos+ 0.0sin M Ignore decaying erms ampliude M Calculae ampliude from soluion of his form A cao (iii) y(0 π ) 0.0 y (0 π ) 0.0 ( ) B B Their a from PI, provided GS of correc form Their b from PI, provided GS of correc form (iv) y = e Ccos+ Dsin F Correc or follows previous CF Mus no use same arbirary consans as before oscillaions B wih decaying ampliude (or ends o zero) B Mus indicae ha y approaches zero, no ha y 0 for > 0π 8 6

23 78 Mark Scheme June 007 (i) dy n y = + M Rearrange I = ep M Aemp IF = ep ln M Inegrae o ge kln ( ) = A Simplified form of IF d n Muliply boh sides by IF and recognise ( y ) = + M derivaive y n M Inegrae n = + +A A RHS including consan n y = + +A F Their inegral (wih consan) divided by IF n (ii) From soluion, 0 y B Limi consisen wih heir soluion From DE, = 0 0 y = M Use DE wih = 0 y = E Correcly deduced (iii) y =, = = + +A M Use condiion A = n n n n ( ) y = + F n =, y = + Consisen wih heir GS and given condiion 8 B Shape for > 0 consisen wih heir soluion (provided no y = consan), or (0, heir value from (ii)) B Through ( ) (iv) ½ (, ½) d Use resul from (i) or aemp o solve from ( y ) = + M scrach F Follow work in (i) y = + ln + B M Inegrae A RHS (accep repeaed error in firs erm from (i)) y = + ln + B Divide by IF, including consan (here or M laer) y() = + B M Use condiion a = y() = + ln + B M Use condiion a = y() = y() B = ln B = ln M Equae and solve ( ln ln ) y = + A cao 9 7

24 78 Mark Scheme June 007 (i) (ii) y dy = k( + 0.cos ) d M Separae y = k( sin) +c M Inegrae = 0, y = c= ( ( 0.00sin )) y = k + ( ) ( ) A A M F M A =, y = = k sin + M Subsiue LHS RHS (condone no consan) Use condiion (mus have consan) Rearrange, dealing properly wih consan cao k 0.86 E Calculae k (mus be from correc soluion) M Subsiue = y = ( 0.86( sin 0) ) 0.7 A cao (iii) soluion curve on inser M Reasonable aemp a curve A From (0,) and decreasing A Curve broadly in line wih angen field ank empy afer.0 minues F Answer mus be consisen wih heir curve ( ) (iv) (0.) = M A 0.66 = 0.9 E Mus be clearly shown (0.) = ( 0.98) M A 0.98 = A awr 0.88 (v) y < 0.0 y < 0. y + 0.cos < 0 for some M Consider size of y + 0.cos dy > 0 for some values of E Complee argumen d y and sign of 8 6 8

25 78 Mark Scheme June 007 (i) = + y e M Differeniae ( ) 8 ( ) M Subsiue for y = y+ e e = 6+ + e + 0e M y in erms of, M Subsiue for y + = e E (ii) λ λ+ =0 M Auiliary equaion λ = (repeaed) A CF = ( A+ B)e F CF for heir roos PI = ae B Correc form for PI ae =, = ae M Differeniae wice ( ) ae e + ae = e M a = A GS e ( ) = + A+ B F (iii) y = ( + e ) ( ( ) = e + Be + A+ B e + e + ( A+ B) e e ) Subsiue and compare GS = PI + CF (wih wo arbirary consans) M y in erms of, ( ) y = e 6 A+ B+ 6 B A cao M F Differeniae follows heir (bu mus use produc rule) (iv) + A = 0 M Condiion on ( + B ) = 0 M Condiion on y A= = e + ( ) e y = e A Boh soluions correc = 0 =, y = B Boh values correc 8 B B B hrough origin and consisen wih heir soluion for large (bu no linear) y hrough origin and consisen wih heir soluion for large (bu no linear) Gradien of boh curves a origin consisen wih heir values of, y 7 9

26 78 Mark Scheme January Differenial Equaions (i ) (ii) α + α + = 0 M Auiliary 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 A B)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+ C D ) e = 0, y = 0 0 = D C 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 Maimum a =, y = e A Maimum value of y B Sars a origin B Maimum a heir value of y B y >

27 78 Mark Scheme January 008 (i ) (ii) (iii) dv + v = g M Rearrange d + + ( ) M Aemp inegraing facor ( ) ( ) + = = = + + A Correc A Simplified ( ) dv + + ( + ) v = g( + ) ( + ) d F Muliply DE by heir I d (( + ) v) = g( + ) ( + ) d ( + ) v = ( g( + ) ( + ) ) M Inegrae ( ) ( ) ( ) ( ) = 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 ( ) ( )( ) dv ( + ) + v = ( + ) g d M Rearrange dv + v = g d + ( ) M Aemp inegraing facor ( ) ( ) + I = = = + + A Simplified ( ) dv + + ( + ) v = g( + ) d F Muliply DE by heir I d (( + ) v) = g( + ) d ( + ) v = g( + ) 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 d,+ 0 B Firs model: g ( g)( ) As ( ) Hence acceleraion ends o g A dv 6 Second model = g 6 ( + ( + ) ) d M Find acceleraion Hence acceleraion ends o 6 g A Idenify erm(s) 0 in heir soluion for eiher model 0 9 8

28 78 Mark Scheme January 008 (i) 0. P = Ae M Any valid mehod = 0, P = 000 A= 000 M Use condiion 0. P = 000e A (ii) 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 ma when P f( P) = P ma 000 M A M P P f ( P) = P M P f ( P) = 0 P = M P = 8000 A Solve Recognise epression o maimise Reasonable aemp a derivaive Se derivaive o zero 8 9

29 78 Mark Scheme January 008 (i) = + y M Differeniae firs equaion = + ( + y+ ) M Subsiue for y y = 9+ M y in erms of, ( ) = M Subsiue for y + + = 6 E (ii) λ + λ+ = 0 M Auiliary equaion λ = ± j A CF e ( Acos Bsin) = + M CF for comple roos F CF for heir roos PI = a B Consan PI a = 6 a = B PI correc GS = + e ( Acos+ Bsin) Their CF + PI (wih wo arbirary F consans) (iii) y = 9+ M y in erms of, ( A B ) ( A B ) ( A B ) (( ) ( ) ) = 9+ e cos + sin 9 e cos + sin + e sin + cos M y = e A+ B cos+ B A sin A Differeniae and subsiue (iv) 0= + A A= M Condiion on 0= A+ B B = 6 M Condiion on y = + e sin cos F Follow heir GS ( ) y = e sin F Follow heir GS Consans mus correspond wih hose in (v) B Skech of sars a origin B Asympoe = 7 B Skech of y sars a origin B Decaying oscillaions (may decay rapidly) B Asympoe y = 0 0

30 78 Mark Scheme June Differenial Equaions (i) ( ) = g g kv NL equaion wih all forces using given M epressions for ension and resisance Weigh posiive as down, ension negaive as up. B Resisance negaive as opposes moion. B + k + = 0 E Mus follow correc NL equaion (ii) = Acos + Bsin B = 0, = 0. A= 0. M Find he coefficien of cos = Asin+ Bcos so = 0, = 0 B = 0 M Find he coefficien of sin = 0.cos A cao (iii) α + α + = 0 M Auiliary equaion α = ± j A M CF for comple roos = e ( Ccos + Dsin ) F CF for heir roos = 0, = 0. C = 0. M Condiion on = e ( Ccos + Dsin ) + e Csin + Dcos M Differeniae (produc rule) ( ) 0= C+ D M Condiion on 0. D = ( ) = 0.e cos + sin A cao B Curve hrough (0,0.) wih zero gradien B Oscillaing B Asympoe = 0 (iv) k > 0 M Use of discriminan A Correc inequaliy (As k is posiive) k > A Accep k < in addiion (bu no k > ) B Curve hrough (0,0.) B Decays wihou oscillaing (a mos one inercep wih posiive ais) 8

31 78 Mark Scheme June 008 (i) Ae = M Any valid mehod = 0, = 8 A= 8 M Condiion on = 8e A (ii) y + y = 6e M Subsiue for α + = 0 α = M Auiliary equaion CF y = Be A PI y = ae B e a ae + = 6e M Differeniae and subsiue a = 6 A cao GS y = 6e + Be Their PI + CF (wih one arbirary F consan) = 0, y = 0 B = 6 M Condiion on y y = 6( e e ) F Follow a non-rivial GS Alernaive mark scheme for firs 7 marks: M Subsiue for I = e M Aemp inegraing facor A IF correc d(y e )/d = 6e B M Inegrae y e = 6e + B A cao y = 6e + Be F Divide by heir I (mus divide consan) (iii) y 6e ( e ) = M Or equivalen (NB e > e needs jusifying) 6e 0 > < hence y > 0 E Complee argumen B Sars a origin B General shape consisen wih heir soluion and y > 0 B Tends o zero 9 (iv) (v) d ( ) ( ) ( ) ( ) 0 d + y + z = + y + y = M Consider sum of DE s + y+ z = c E Hence iniial condiions + y+ z = 8 E z = 8 y M Subsiue for and y and find z ( ) ( ) z = 8 e + e = 8 e E Convincingly shown (, y mus be correc) ( ) = 8 e B Correc equaion (any form) = or.98 99% is Z afer.0 hours B Accep value in [.9,.] 9

32 78 Mark Scheme June 008 k (i) y + y = M Divide by (condone LHS only) k k I = ep( d) = ep( kln ) = M Aemp inegraing facor A Inegraing facor k k k y + k y = F Muliply DE by heir I d k k ( y ) = d M LHS k k y = d M Inegrae k + = k + + A A cao (including consan) k y = + A k + F Divide by heir I (mus divide consan) =, y = 0 0 = k + A A= + + M Use condiion y = F Follow a non-rivial GS k + k ( ) (ii) y = ( ) B Shape consisen wih heir soluion for > B Passes hrough (, 0) B Behaviour for large 0 (iii) y = d M Follow heir (i) = ln + B A cao y = ( ln + B) F Divide by heir I (mus divide consan) =, y = 0 B = 0 y = ln A cao (iv) dy = + sin y d M Rearrange DE (may be implied) y dy/d 0 M Use algorihm A y(.) A y(.) (v) 0.8 as smaller sep size B Mus give reason Decreasing sep lengh has increased esimae. Assuming his esimae is more accurae, decreasing sep lengh furher will M Idenify effec of decreasing sep lengh increase esimae furher, so rue value likely o be greaer. Hence underesimaes. A Convincing argumen Alernaive mark scheme for las marks: dy/d seems o be increasing, hence Euler s mehod will underesimae rue value + skech (or eplanaion). M A Idenify derivaive increasing Convincing argumen 0

33 78 Mark Scheme June 008 (i) = 6y 9cos M Differeniae firs equaion = 6( y 7sin ) 9cos M Subsiue for y y 6 ( ) M y in erms of, = 8+ ( 9sin ) + sin 9cos M Subsiue for y + = sin 9cos E LHS E RHS (ii) α + α = 0 M Auiliary equaion α = or A CF = Ae + Be F CF for heir roos PI = acos+ bsin B PI of his form ( ac bs) + ( as + bc) ( ac + bs) = s 9c M Differeniae wice and subsiue a+ b a = 9 Compare coefficiens ( M equaions) b a b = M Solve ( equaions) a =, b = 0 A cos e = + A + Be Their PI + CF (wih wo arbirary F consans) (iii) y = ( ) 6 9sin M y in erms of, cos e e sin e e = + A + B + A + B 9sin M Differeniae and subsiue 6 ( ) cos sin e y = + A + Be A (iv) bounded A = 0 Consans mus correspond wih hose in Idenify coefficien of eponenially growing erm mus be zero M y bounded E Complee argumen (v) = 0, y = 0 0 = B+ B = M Condiion on y = cos e, y = cos sin e Follow heir (non-rivial) general F soluions = cos A cao y = cos sin A cao 6 9

34 76 Mark Scheme January Differenial Equaions (i) α + α α = 0 B ( ) + ( ) ( ) = 0 E Or facorise ( α + )( α ) = 0 M Solve α =, ± A y = Ae + Be + Ce M Aemp CF F CF for heir hree roos (ii) PI y = = M Consan PI A Correc PI GS y = + Ae + Be + Ce F GS = PI + CF (iii) e as M Consider as so finie limi C = 0 F Mus be shown, no jus saed = 0, y = 0 0 = -+ A + B M Use condiion = ln, y = 0 0 = + A + B M Use condiion Solving gives A = -, B = M y = e + e E Convincingly shown (iv) y = ( e )( e ) y = 0 e = or M Solve = ln or 0 E Convincingly show no oher roos dy = e e = e (e ) dy 0 = e = as e 0 M Solve = ln E Show only one roo Saionary poin a (ln, 8) A (v) y B Through (0, 0) (ln(/),/8) B Through (ln, 0) ln Saionary poin a heir answer o B (iv) B y as 6 6 9

35 76 Mark Scheme January 009 (i) dy + y an = cos M Rearrange I = ep an M Aemp IF = eplnsec A Correc IF = sec A Simplified d ( ysec ) = M Muliply and recognise derivaive y sec = + A M Inegrae A RHS Divide by heir IF (mus divide y = ( + A) cos F consan) = 0, y = A = M Use condiion y = ( + ) cos F Follow heir non-rivial GS y (ii) B Shape correc for π < < π B Through (0,) 0 (iii) (iv) cos sin ysin y = cos M Rearrange y ( 0) = 0 B y ( 0.) = B y ( 0.) = B y ( 0.) = = M Use of algorihm for second sep A sf or beer I = sec M d π/ π/ ( ysec ) = an A = 0. [ y sec ] = = 0. Same IF as in (i) or aemp from scrach 0 an 0 M Inegrae A Accep no limis y ( 0.)sec(0.) sec M Subsiue limis (boh sides) y ( 0.) A Awr

36 76 Mark Scheme January 009 (i) (ii) (iii) dv 60v = 60g v M NL A Correc NL equaion v dv = 0g v 0 E Convincingly shown v dv = 0g v 0 M Inegrae ln 0g v = + c A ln 0g v seen 0 A RHS Rearrange, dealing properly 0 0g v = Ae M wih consan = 0, v = 0 A = 0g M Use condiion v 0 = 0g( e ) A Cao 0 = 0 v = 0 0g( e ).7 E Convincingly shown dv 60 = 60g 60v 90g d M NL dv dv = g v or + v = g d d A Correc DE 9 Solving DE (hree alernaive mehods): dv + = d v g M Separae ln v + g = + k M Inegrae A LHS Rearrange, dealing properly v + g = Ae M wih consan or α + = 0 α = M Solve auiliary equaion CF Ae M CF for heir roo PI g M Aemp o find consan PI v = Ae g A All correc or I = e M Aemp inegraing facor d d e v ( e v) = ge M Muliply = ge + A M Inegrae v = Ae g A All correc v =.7, = 0.7 = A g A = 8.6 M Use condiion v = 8.6e.9 E Complee argumen 8

37 76 Mark Scheme January 009 (iv) A greaes deph, v = 0 M e = =. Deph = ( 8.6e.9) d M Inegrae = 0 [ 8.6e.9 ] 0. A A Ignore limis Se velociy o zero and aemp o solve M Use limis (or evaluae consan and subsiue for ) = 7.7 m A All correc 6 (i) y + 7 = 0 = B y + = 0 y = B (ii) = y M Differeniae = ( y + ) M Subsiue for y y = + 7 M y in erms of, = + 7 M Subsiue for y + + = E Complee argumen (iii) α + α + = 0 M Auiliary equaion α = ± i A M CF for comple roos CF e ( Acos + Bsin ) F CF for heir roos PI = = B GS = PI + CF wih wo GS = + e ( Acos + Bsin ) F arbirary consans 6 (iv) y = + 7 M y in erms of, Differeniae heir (produc = e ( Acos + Bsin ) + e ( Asin + B cos) M rule) y = + e (( A B)sin ( A + B)cos) A Consans mus correspond (v) + A = M Use condiion on A B = 0 M Use condiion on y A =, B = = + e (cos + sin ) y = + e (sin cos) A Boh soluions (vi) B (0, ) B B (0, 0) B As he soluions approach he asympoes, he gradiens approach zero. B Mus refer o gradiens

38 78 Mark Scheme June Differenial Equaions (i) α + = 0 M Auiliary equaion α =± j A CF y= Acos + Bsin F CF for heir roos PI y= acos + bsin B y = acos asin + bsin + bcos y = 0asin acos + 0bcos bsin M Differeniae wice In DE 0bcos 0asin = 0cos M Subsiue and compare coefficiens b=, a= 0 A PI y= sin GS y= sin + Acos + Bsin F 8 (ii) = 0, y= A= B From correc GS y = sin + 0cos Asin + Bcos M Differeniae = 0, y = 0 B= 0 M Use condiion on y y= sin + cos A (iii) Curve hrough (0,) B Curve wih zero gradien a (0,) B Oscillaions B Oscillaions wih increasing ampliude B (iv) y= sin, y= 0cos, y = 0sin y + y + y= 0sin + 0cos + 0sin M Subsiue ino DE = 0cos E α + α + = 0 M Auiliary equaion α = ± j A CF e ( Ccos + Dsin ) F CF for heir comple roos GS y= sin + e ( Ccos + Dsin ) F Their PI + heir CF wih wo arbirary consans 6 (v) Oscillaions of ampliude B or bounded oscillaions; or boh oscillae Compared o unbounded oscillaions in firs model B or equivalen; or one bounded, one unbounded

39 78 Mark Scheme June 009 dy sin (i) + y = M Rearrange I = ep M Aemping inegraing facor = ep(ln) A = A Correc and simplified ( y ) d = sin M Muliply and recognise derivaive y= sin = cos + cos M Inegrae A = cos + sin + A A All correc y= ( cos+ sin + A)/ F Mus include consan 9 (ii) y + + A / 6 M Subsiue given approimaions F A = + M Use finie limi o deduce A A = 0 A y= (sin = cos ) / B Correc paricular soluion limy 0 = B Correc limi 6 (iii) y= 0 sin cos= 0 M Equae o zero and aemp o ge an an = E Convincingly shown (iv) dy, muliply by + y= I= 6 M Rearrange and muliply by IF B Same IF as in (i) or correc IF d ( y) = 6 A Recognise derivaive and RHS correc y= + B 0 M Inegrae B y= + 0 A cao Finie limi B = 0 M Use condiion o find consan

40 78 Mark Scheme June 009 as (ii) y = E Show correc limi (or same limi lim 0 7 (a)(i) α + = 0 α = M Find roo of auiliary equaion Ae CF A PI I= acos + bsin B I = asin + bcos M Differeniae asin + bcos + acos + bsin = cos M Subsiue a+ b= 0,b+ a= a= b= 8 M Compare coefficiens and solve PI I= (cos + sin ) 8 A GS I= Ae + (cos + sin ) 8 F 8 Their PI + heir CF wih one arbirary consan (ii) = 0, I= 0 0 = A+ A= 8 8 M Use condiion (cos sin e I= + ) 8 A cao (iii) For large, I (cos + sin ) 8 M Consider behaviour for large (may be implied) Ampliude = + = 8 8 A Curve wih oscillaions wih consan ampliude B Their ampliude clearly indicaed B (b)(i) dy 0 (A) = 0, y = 0 = (0) + e d M Subsiue ino DE Gradien = A (B) A saionary poin, d y = 0, y = 9 d 8 M Subsiue ino DE 9 0= ( ) + e e = 8 M Solve for = ln A (C) d y 0,e 0 d M Subsiue ino DE Giving 0= y+ 0, so y A 7 (ii) Curve hrough origin wih posiive gradien B Wih maimum a (ln, 9/8) B Follow heir ln Wih y as B Follow heir (C) (i) = 7 + 6y 6e M Differeniae = 7 + 6( 0y+ sin ) 6e M Subsiue for y ( 7 e ) y= M y in erms of,, 6 6

41 78 Mark Scheme June 009 = 7 7 0( 7 e ) + 0sin 6e M Subsiue for y E Complee argumen + + = e + 0sin (ii) = ae 9cos + sin = e a + 9sin + cos = 9e a + 9cos sin M Differeniae wice In DE gives M Subsiue 9e a + 9cos sin +( e a + 9sin + cos ) +( ae 9cos + sin ) = e a + 0sin So PI wih a = E Correc form shown a = 7 A AE α + α + = 0 M Auiliary equaion α =, A CF Ae + Be F CF for heir roos GS = Ae + Be + 7e 9cos + sin F Their PI + heir CF wih wo arbirary 8 consans ( 7 e ) (iii) = 6 M y in erms of,, = Ae Be e + 9sin + cos M Differeniae GS for F Follow heir GS y= Ae Be e + cos sin A cao (iv) sin 9cos B Follow heir y cos sin B Follow heir y (v) = y cos sin sin 9cos M Equae 0cos sin an A Complee argumen Ampliude of + 9 = 0 M Aemp boh ampliudes Ampliude of y + = A One correc Raio is 6 A cao (accep reciprocal) 7

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43 78 Mark Scheme January Differenial Equaions (i) α + 6α+ 9= 0 M Auiliary equaion α= (repeaed) A - y = e ( A+ B) F CF for heir roos PI y = asin + bcos B y = acos bsin y = asin bcos ( ) ( ) asin bcos + 6 acos bsin GS ( ) + 9 asin+ bcos = 0.sin M Differeniae wice and subsiue 8a 6b= 0. M Compare coefficiens 8b+ 6a= 0 M Solve Solving gives a= 0.0, b= 0.0 A y = e A+ B sin 0.0 cos F PI + CF wih wo arbirary consans (ii) = 0, y= 0 A= 0.0 M Use condiion y = e ( B A B) cos sin M Differeniae F Follows heir GS = 0, y = 0 0 = B A+ 0.0 M Use condiion ( ( ) ) y = 0.0 e + + sin cos A Cao (iii) For large, he paricle oscillaes B Oscillaes Wih ampliude consan ( 0.0) B Ampliude approimaely consan (iv) π y 0.0 A y 0.0 A 60π = 0 e very small M (v) y e ( C D) = + M CF of correc ype or same ype as in (i) A Mus use new arbirary consans π B y 0.0 a = 0π B Gradien a 0π consisen wih (iv) B Shape consisen 9

44 78 Mark Scheme January 00 (a)(i) I = ep an M Aemp IF = ep ln sec A Correc IF ( ) ( sec ) = = cos A Simplified dy cos ysin = sin M Muliply by IF d ( ycos ) = sin M Recognise derivaive M Inegrae ycos = cos + A A RHS (including consan) ( y = Asec ) A LHS 8 dy ( + ) an d = y ln(+y)=lnsec + A (ii) 0, y 0 0 A y sec M A M A M A M A Rearrange equaion Separae variables RHS LHS = = = M Use condiion = A B Shape and hrough origin B Behaviour a ±½π y 8 π/ π/ (b)(i) M Aemp one curve A Reasonable aemp a one curve M Aemp second curve A Reasonable aemp a boh curves y' = + y an M Rearrange (ii) ( ) (iii) = 0, y = y' = 0 y ( 0.) = = M Use of algorihm = 0., y= y' = 0.0 A y ( 0.) = M Use of algorihm for second sep =.00 E an π undefined so canno go pas π M π So approimaion canno coninue o.6 > A (iv) Reduce sep lengh B 0

45 78 Mark Scheme January 00 (i) = Ae k M A = 0, = v A= v M Use condiion = v e A e = v k d M Inegrae Any valid mehod (or no mehod shown) v = e k + B k A v = 0, = 0 B= k M Use condiion (ii) (iii) v = ( e k ) E k dy d y + g/ k = k M Separae and inegrae g In y+ = k+ C k A LHS A RHS g y + = De k Rearrange, dealing properly wih M k consan g = 0, y = v D= v + k M Use condiion g k g y = v + e k k A g k g y = v + e d k k M Inegrae g g = v + e k + E k k k A g = 0, y = 0 0 = v + + E k k M Use condiion g y = ( kv + g)( e k ) k k E e k k = v = ln k y k A v kv + g g k M Subsiue kv k v E Convincingly shown = + ln M (iv) = 8 y =.686 M Hence will no clear wall A 8 0

46 78 Mark Scheme January 00 (i) y = + M y or y in erms of, y = M Differeniae ( ) = + ( + ) 7 M Subsiue for y = M Subsiue for y + + = E (ii) α + α + = 0 M Auiliary equaion α = ± i A M CF for comple roos e Acos+ Bsin F CF for heir roos CF ( ) PI = = B Consan PI B Correc PI = + e Acos+ Bsin PI + CF wih wo arbirary F consans GS ( ) = + M = e ( Acos+ Bsin) + M + e ( Acos+ Bsin) F e ( Asin+ Bcos) y = e (( A+ B) cos+ ( B A) sin) A (iii) y ( ) (iv) = 0, = 8 + A= 8 A= 7 M Use condiion = 0, y = 0 ( A+ B) = 0 B= M Use condiion = + e ( 7cos+ sin) A y = e cos sin A ( ) Differeniae and subsiue Epression for follows heir GS (v) For large, e - ends o 0 M y B B y> E Complee argumen 7

47 GCE Mahemaics (MEI) Advanced GCE 78 Differenial Equaions Mark Scheme for June 00 Oford Cambridge and RSA Eaminaions

48 78 Mark Scheme June 00 (i) α α 8 0 M Auiliary equaion α j A CF e ( Acos Bsin ) M CF for comple roos F CF for heir roos PI y a b c B y ab, ya Differeniae wice and a( ab) 8( a bc) M subsiue 8a M Compare coefficiens 8a8b 0 ab8c 0 M Solve a, b, c A GS PI + CF wih wo arbirary y e ( AcosBsin ) F consans (ii) 0, y 0 A M Use condiion y 8e ( AsinBcos Acos Bsin ) M Differeniae (produc rule) 0, y 0 0 (B A) B M Use condiion y e (sincos ) A Cao (iii) y oscillaes B Oscillaes Wih (eponenially) growing ampliude B Ampliude growing (iv) y ( ) or B (v) B Minimum poin a origin B Oscillaes for <0 wih growing ampliude B Approimaely parabolic for >0 (vi) A saionary poin d y 0 Se firs derivaive (only) o M zero in DE d y So 8y A d 0 y y 0 Deduce sign of second M derivaive minimum E Complee argumen 0

49 78 Mark Scheme June 00 (a)(i) IF ep d M Aemp IF e A dy d M* Muliply by IF d (e y ) A e y A *MA Inegrae boh sides 6 e e y [ y e ( A)] Alernaive mehod: CF e y E PI e y F In DE: e ( F F) Fe e F = y e ( E ) B B M MA F (ii) dz z e ( A) d I e B Correc or follows (i) d (e z ) A d M Muliply by IF and inegrae e z A B A ze ( A B) 0, z B M Use condiion z e ( AB) e ( A) M Differeniae (produc rule) 0, z 0 0 B A A M Use condiion ze ( ) A 7 Alernaive mehod: PI ( P Q )e P = A and Q = 0. ze ( A B) Then as above B MA Correc form of PI Complee mehod (b)(i) α0α CF Ce B CF correc PI asin bcos B Correc form of PI acos bsin In DE: acos bsinasin bcos sin M Differeniae and subsiue ab0, ba M Compare and solve a, b A GS (sincos ) Ce F Their PI + CF 6 (ii) 0, 0 0 (from DE) M Or differeniae 0 C M Use condiion (sincos e A (iii) For large, (sin cos ) sin (-) M Complee mehod So varies beween and A Accep

50 78 Mark Scheme June 00 (i) y dy kd M Separae and inegrae y k B A LHS A RHS 0, y B M Use condiion, y0.8.8 k M Use condiion k 0. A y 0.0 y Valid for 0.0 0, i.e. 0 ( 0.0 ) A y B on arihmeical error in k B B Shape Inerceps (ii) πy dy 0.d M Separae and inegrae 0 πy 0. C A LHS A RHS 0, yc π M Use condiion y A 0. y (iii) y M π( y y ) Rearrange (implied by correc values) y y hy M Use algorihm A y (0.) (awr 0.987) M Use algorihm A y (0.) (0.97 o 0.97) 0 (iv) If V volume, v velociy, A horizonal cross-secional area, dv hen kv d M Rae of change of volume v k y A dy dv d d M Relae raes of change of y and volume dy Eliminae volume and/or A kk y M d velociy dy k y E Complee argumen d

51 78 Mark Scheme June 00 (i) y M y or y in erms of, 9e y 8e M Differeniae (8e ) M Subsiue for y ( 9e ) e M Subsiue for y e E (ii) α α 0 M Auiliary equaion α, A CF Ae Be F CF for heir roos PI ae B PI of correc form ae, ae M Differeniae and subsiue a a a e e Compare coefficiens and M solve (iii) a GS A B A e e e F y M ( 9e ) PI + CF wih wo arbirary consans (Ae B e e 9e M Differeniae and subsiue Epression for follows ( Ae Be e )) F heir GS y Ae B e e A (iv) 0, 0 0 AB M Use condiion 0, y AB M Use condiion A0, B e e A y e e A (v) As, 0, y 0 B y(0) A 0 Consider coefficien(s) of e M and menion of y <, y as E Complee argumen 8

52 GCE Mahemaics (MEI) Advanced GCE Uni 78: Differenial Equaions Mark Scheme for January 0 Oford Cambridge and RSA Eaminaions

53 78 Mark Scheme January 0 Quesion Answer Marks Guidance (a) (i) 0 M Auiliary equaion j A CF e ( Acos Bsin ) M CF for comple roos F CF for heir roos PI ae B ae, ae M Differeniae wice and subsiue ae ae ae e M Compare coefficiens and solve a A GS e e ( AcosBsin ) F PI + CF wih wo arbirary consans [9] (a) (ii) 0, 0 A 0 M Use condiion e-e ( AcosBsin) M Differeniae (produc rule) e ( AsinBcos ) 0, 0 0 AB M Use condiion e - e (cos sin ) A cao [] (b) (i) AE 6 0 B 6 0 E ( )( 6) 0 M Facorise (or solve by oher means), or A GS y Ae Be Ce F GS = CF wih hree arbirary consans [] (b) (ii) y0 A 0 B y(0) BC M Use condiion y(0) BC M Use condiion B, C y e e A cao []

54 78 Mark Scheme January 0 Quesion Answer Marks Guidance (b) (iii) y 0e (e ) 0 e M ln A [] (a) (i) I ep M Aemp IF e A Correc IF d y e e y sin M Muliply by IF d ( e y ) sin M Recognise derivaive d ye sind M Inegrae cos A A RHS (including consan) 0, y 0 A M Use condiion A A y e ( cos) F Divide by heir IF, including consan [9] (a) (ii) e 0,cos M cos0 y0 E dy Or use DE e (cos ) e (sin ) dy E 0 0 y 0 B B B Through (0,) Shape consisen wih resuls shown [6]

55 78 Mark Scheme January 0 Quesion Answer Marks Guidance (b) (i) dy y y y 0 B Firs row M Use algorihm A. A.78 [] (b) (ii) I e F d (e ) e M y ye A e y(0.)e 0.07 M y(0.).() A [] (i) k 0 k M Roo of auiliary equaion CF Ae k A PI y acos bsin B dy asin bcos M Differeniae asin bcos M Subsiue and compare ka ( cosbsin ) cos akb0 bka A k A a, b 9 k 9 k k F PI + CF wih one arbirary consan y Ae ( kcossin ) 9 k [8] 6

56 78 Mark Scheme January 0 Quesion Answer Marks Guidance (ii) 0, y y 0 (from DE) OR differeniae y MA k 0 A 9 k M Use condiion k A y ( kcossin ke ) 9 k [] (iii) CF Be k F PI y ce k B k k y ce kce M Differeniae k k k ce ( k) kce e M Subsiue and compare c A k k y Be e F PI + CF wih one arbirary consan [6] (iv) d M Recognise relaionship (previous DE) wih k y Be e C F 0, y0 0 B C B Condiion y Be e e M Differeniae 0, y B M Use condiion B, C NEED ALTERNATIVE SOLUTION y e e A [6] OR for firs marks m m 0; CF y Be C and PI y pe giving p GS y Be e C M A Complee mehod 7

57 78 Mark Scheme January 0 Quesion Answer Marks Guidance (i) y 0 M y 0 M M Eliminae y M E Eliminae y [] (ii) M Auiliary equaion 0. 0.j A 0. e ( Acos0. Bsin 0. ) M CF for comple roos F CF for heir roos [] (iii) 0. 0.e ( Acos0.Bsin 0. ) M Differeniae (produc rule) 0. A 0.e ( Asin 0.B cos 0. ) y 0 M Subsiue o find y 0. e (( AB)cos0. ( B A)sin0.) A [] (iv) 0 A B y0 A B M Use condiion 0. e ( 0cos0. ( y0 0)sin0.) A 0. y e ( y0cos0. ( y0 0)sin0.) A [] (v) y M 0 y 0 when an0.. y0 F 0 So (for leas posiive ),. A Or compare values of an 0. 0 M 0 when an0. y0 0 9 F So (for leas posiive ), 0. A Or compare values of an 0. Hence rabbis die ou firs A Complee argumen [7] 8

58 GCE Mahemaics (MEI) Advanced GCE Uni 78: Differenial Equaions Mark Scheme for June 0 Oford Cambridge and RSA Eaminaions

59 78 Mark Scheme June 0 (i) 0 M Auiliary equaion or - A CF Ae Be F CF for heir roos PI y acos bsin B y asinbcos Differeniae wice and M subsiue y acosbsin acos bsin 8asin 8bcos acos bsin cos M Compare coefficiens 8ba b8a 0 A 8 a, b A GS y (8sin cos ) Ae Be PI + CF wih wo arbirary F consans (ii) 0, y 0 0 AB M Use condiion y (6 cos sin ) Ae B e M Differeniae F 6 0, y 0 0 AB M Use condiion A, B= 0 A y (8sin cos ) e e 0 A Cao 6 (iii) If z yc, differeniaing (*) gives new DE M Recognise derivaive and has arbirary consans so mus be GS A or Inegraing gives (*) wih k on RHS M PI will be previous PI k, CF as before, so GS y c A SC for showing ha correc y from (i) + c saisfies new DE (iv) (8sin cos ) e z D Ee c 0, z DEc M Use condiion (6 cos sin ) e z D E e F Derivaive 6 0, z 0 0 DE M Use condiion Second derivaive: z ( sin cos ) De 9Ee F condone, for his mark only, +c appearing 0, z D9E M Use condiion D, E, c 0 B z (8sin cos ) e e A Cao 0 9 7

60 78 Mark Scheme June 0 (a)(i) I ep( d ) M Aemp inegraing facor ep( ln ) A A dy y M Muliply boh sides by IF d ( ) y M y A M Inegrae boh sides A y A F Mus divide consan (ii) 0 A M y A (iii) 0, y 0 F dy 9 0 (as 0) 6 M dy 0, 0 F B Behaviour a origin B Through and shape for 8 (b)(i) Circle cenre origin B Radius B B Saionary poin a (ii) B One isocline correc B All hree isoclines correc B Reasonably complee and accurae direcion indicaors (iii) B Soluion curve (iv) B Soluion curve B Zero gradien a origin 6

61 78 Mark Scheme June 0 (a)(i) NL: ma k M dv v k M Acceleraion dv v k E (ii) vdv k M Separae and inegrae v k A A LHS A RHS a, v 0 A k a M Use condiion v k ( a ) A So for v0, k a d E (iii) a k d arcsin B k a A LHS A RHS a, 0 B M Use condiion asin( k) acos k A Eiher form M Separae and inegrae (b)(i) d 9sind (ii) M Separae and inegrae 9cos C A LHS A RHS 9, 0 C M Use condiion So 9( cos ) A d cos (decreasing) d E 0 M So esimae 0 A The algorihm will keep giving B bu is no consan so no useful B 6 6

62 78 Mark Scheme June 0 (i) y M y M ( ) M Eliminae M E Eliminae (ii) 0 M Auiliary equaion (repeaed) A Roo CF: ( A B)e F CF for heir roo(s) (wih wo consans) (iii) PI: a b B a, 0 In DE: 0aab M Differeniae and subsiue a ab M Compare and solve a, b 9 A GS: 9 ( A B)e F GS = PI + CF wih wo arbirary consans y M [ e ( )e B A B ] [9 ( AB)e ] M Subsiue M 9 ( A B B)e A Differeniae (produc rule) (iv) 0, 9 A 0 M Use condiion 0, y 0 0 B B M Use condiion 9e A y 9 ( )e A (v) e 0 M 9 F y 9 F 8

63 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) 69 0 M (repeaed) A y A Be F y a b c B y a b, y a a 6( a b) 9( a b c) M Differeniae and subsiue 9a a 9b 0 M Compare a leas wo coefficiens a 6b9c 0 M Solve for a leas wo unknowns a, b, c A y ( A B)e F Non-zero PI + CF wih wo arbirary consans [9] (ii) 0, y 0 A 7 M Use condiion dy M Differeniae using produc rule Be ( A B)e 9 7 F FT only c from (i) 0, y 0 0 B A 7 M Use condiion B 7 ( ) e y A cao [] (iii) Boh appear in CF B Or any clear saemen or reason [] (iv) y e B Allow for y e e dy Differeniae wice using produc rule e e M

64 78 Mark Scheme June 0 Quesion Answer Marks Guidance d y e e 9 e A in DE: e e M Subsiue ino DE M Compare coefficiens A Correc PI correc working only e y AB F Non-zero PI + CF wih wo arbirary consans [7] (v) (A) (e 0), quadraic has +ve coefficien of FT for any hree erm quadraic so can have y 0 for all for suiable A and B B Reasonably complee eplanaion or show for a specific eample (v) (B) would need quadraic < 0 for all, bu for large i will FT for any hree erm quadraic be posiive B Reasonably complee eplanaion For boh marks o be awarded in (v), e 0 mus be saed [] (i) dv mv mg mkv M NL erms, allow sign errors and any form for accn, including a dv v g kv E v d v d g kv M* Separaing variables. Inegraing facor aemp ges zero ln g kv k c A LHS A RHS (including consan on one side) gkv Ae k Mdep* Rearrange, dealing properly wih consan 0, v 0 A g Mdep* Use condiion g v e k k E [8] (ii) g g k k 0 B Or 0.00: cao [] 6

65 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iii) dv m mg0.mg v d M NL erms, allow sign errors and any form for accn, including a dv g( 0. v) d A EITHER dv gd 0.v M Separaing variables 0ln 0.v g c A LHS A RHS (including consan on one side) 0.g 0.v Be M Rearrange 0, v B. M Use condiion 0.g v 0 e A [8] OR 0. Inegraing facor e g B FT heir equaion 0.g ve 0.g 0e A M Muliply hrough by IF and inegrae boh sides A FT 0.g v 0 Ae M Divide hrough by IF 0, v A M Use condiion 0.g v 0 e A [8] OR dv 0.gv g d 0.g Auiliary equaion m0.g 0 CF Ae M A PI v =0 B 0.g GS v 0 Ae M 0, v A M Use condiion 0.g v 0 e A [8] 7

66 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iv) 0 ln. s 0.g B FT FT for v obained by a correc mehod [] (v) 0 0.g vd 0 e c M g F FT aemp o inegrae heir answer o par (iii) 0 0, 0 c M g A FT FT for obained by a correc mehod. 7. M A cao [6] (i) dy y sin M Divide boh sides by (NOTE ha a MR (eg sin missing) can earn 7/8) I ep M Allow ln e A A d y sin M Muliply by heir IF and aemp o inegrae boh sides y sin B LHS mus be heir IF y y cos A A + A is needed y cos A F Divide every erm by he muliplier of y (including a consan) [8] 0 cos A M Use condiion A y (cos ) A cao (ii) 8

67 78 Mark Scheme June 0 Quesion Answer Marks Guidance B Oscillaions corresponding o heir epression B Changing ampliude corresponding o heir epression B y 0 wih maima on -ais; cusps ge B0; cao Ma B if MR has made graph simpler [] (iii) dy y M Aemp o separae variables Aemp a IF, M0 for his par dy y M Inegrae boh sides ln B y A LHS A RHS (including consan on one side) y ln B A Any correc form of answer [] (iv) dy sin y B May be implied by correc values y y hy M Use algorihm can ge his even if B A y(.) = Agreemen o sf A 0.08 Agreemen o sf Award for A Agreemen o sf. ONLY award if his is he FINAL answer [] (v) Reduce sep lengh (and carry ou more seps) B No conradicions [] 9

68 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) y 6 M y M 7 M Subsiue for y M Subsiue for y A cao AE 0 M j A cao CF e ( Acos Bsin) M CF for comple roos B CF for heir roos B Appropriae form of PI PI B Correc PI GS e ( AcosBsin ) F Non-zero PI + CF wih arbirary consans (ii) e ( AcosB sin ) e ( Asin B cos ) d M Differeniae using produc rule y 6 e ( AsinBcos ) M A Subsiue cao [] dy Alernaive mehod: e Acos Bsiny 7 d dy y 8e Acos Bsin M Subsiue for in d y y 7and arrange in correc IF form d d e y (8e Acos Bsin)d M Find IF, muliply boh sides by IF and show inenion o inegrae y e ( AsinBcos ) A [] (iii) 7 A M Use condiion on a GS 0 B M Use condiion on a GS e (6cossin ) A cao y e (6sin cos) A cao [] [] 0

69 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iv) y, so k M F FT wrong consan erm(s) and wrong form of epressions for and y y 6sin cos (6 cos sin ) M The eponenial erm needs o have been cancelled 0sin 8cos an.8 A Or equivalen e.g. an( 0.9) which has infiniely many roos E This can be awarded for a correc jusificaion following a wrong value for an (i.e. can ge A0 E) []

70 GCE Mahemaics (MEI) Advanced GCE Uni 78: Differenial Equaions Mark Scheme for January 0 Oford Cambridge and RSA Eaminaions

71 78 Mark Scheme January 0 Quesion Answer Marks Guidance (i) 60 M 6 0 E Allow if implici in facorisaion of cubic ( )( )( ) 0 M Aemp roos (any mehod) (),, A CF Ae Be Ce F PI y asin bcos B y acos bsin y asin bcos y acos bsin ( acos bsin ) ( asin bcos ) M Differeniae and subsiue ( acos bsin ) 6( asin bcos ) sin 6a8b0 M Compare coefficiens and solve a, b 0 8a6b A GS y cos sin Ae Be Ce F 0 [0] (ii) bounded so A 0 F 0, y B C F Use condiion 0 y sin cos Be Ce 0 M Differeniae 0, y 0 0 B C M Use condiion 9, C B A y cos sin e e F [6] (iii) y cos sin F 0 ampliude 0 0 M A B [] Skech showing oscillaions wih heir ampliude. More han one oscillaion, ignore origin

72 78 Mark Scheme January 0 Quesion Answer Marks Guidance (iv) bounded so BC 0 B 7 0, y A A 0 0 M dy M Or A So no such soluion A www [] (i) dv NL: m 9.8m mkv d dv 9.8 kv d E EITHER dv d 9.8 kv M Separae and inegrae A LHS ln 9.8 kv c k A RHS (including consan on one side) 9.8 kv Ae k M Rearrange, dealing properly wih consan 0, v0 9.8 A M Use condiion 9.8 v k k A cao OR Inegraing facor e k M ve k k 9.8e d A Muliply boh sides by IF and recognise derivaive on LHS k 9.8 k ve e A k M Inegrae boh sides A Mus include consan 9.8 0, v0 A k M Use condiion 9.8 v k k A cao OR Auiliary equaion k 0 M CF v Ae k A 9.8 PI vb, v 0b k M 6

73 78 Mark Scheme January 0 Quesion Answer Marks Guidance 9.8 GS v Ae k k A 9.8 0, v0 A k M Use condiion 9.8 v k k A cao [7] (ii) 9.8 k. 7 B [] (iii) dv m 9.8m 0.mv d M v dv d M 9 v dvc A RHS (including consan on one side) d vc 7 v 7 v M Inegrae ln 7 v ln 7 v c A LHS 7 v ln c 7 v 7 v / Rearrange ino a form wihou ln, dealing B e M 7 v properly wih consan 0, v0 B M Use condiion / 7 ve 7 v M Rearrange o ge v in erms of / / e e v 7 7 / / e e A oe, 7 0 v 7 E as 0 [0] 7

74 78 Mark Scheme January 0 Quesion Answer Marks Guidance (iv) / v g 0.9v E v v hv M Use algorihm A v (0.) A v (0.) a leas d.p A v (0.) =.9 o s.f. [] (v) / v 0 g 0.9v v7.00 ( sf) E Or (a) [] I ep an M ep ln sec or ep ln cos A cos A dy cos ysin sin cos d ycos sin cos M Muliply and recognise derivaive ycos sin cos M Aemp inegral M Use ideniy, subsiuion or inspecion on RHS sin d cos c sin k ) A oe (bu mus include consan) 0, y c M Use condiion cos y or cos sin y or cos cos y cos A oe [9] 8

75 78 Mark Scheme January 0 Quesion Answer Marks Guidance (b) p'( ) f( )p( ) g( ) M Mus be p ( ) oe c'( ) f( )c( ) 0 M Mus be c ( ) oe dy f( ) y p'( ) Ac'( ) f( ) p( ) Ac( ) M Subsiue in DE p'( ) f( )p( ) Ac'( ) f( )c( ) M Separae p and c erms g( ) A0 g( ) E Complee argumen [] (c) (i) d y y e e B so LHS of DE e e M e e E [] (c) (ii) dy y dy y M ln y ln c A LHS A RHS including consan y A A cao OR Inegraing facor = B d y 0 M y A A y A A cao [] 9

76 78 Mark Scheme January 0 Quesion Answer Marks Guidance (c) (iii) y e A Here A combines he arbirary consans of (b) B and (c) (ii) ino a single arbirary consan. e A M Use condiion y e e A [] y M (i) y M Differeniae M Subsiue A oe AE 0 M A j CF e Acos Bsin M Correc form F FT wrong roos PI a b B b, 0b( ab) M Differeniae and subsiue ba M a, b b A Equae coefficiens and solve GS e Acos Bsin F [] 0

77 78 Mark Scheme January 0 Quesion Answer Marks Guidance y M (ii) e Acos Bsin e Asin Bcos M F y e Asin Bcos A [] (iii), 0 A A M y 0, 0 0 B B M (iv) e cos sin e sin cos y A Boh 6 e sin cos [] Mus be using produc rule Mus be GS from (i) y M Adding and aemping he limi 6 e 0 6 sin cos 0 an y E FT for finie limi y M Esablish equaion and indicae mehod Correcly invesigae he eisence of a which occurs (infiniely ofen) E soluion, bu eplici soluion for no required. []

78 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) 0 M Auiliary equaion, A Correc roos / CF A e B e F FT roos PI a co s b sin B a sin b co s, a co s b sin ( a cos b sin ) ( a sin b cos ) ( a cos b sin ) cos M Differeniae and subsiue a b M Compare and solve a, b 0 0 b a 0 A Correc values / GS sin co s A e B e F FT CF wih arbirary consans + PI 0 [8] (ii) 0, 0 0 A B M Use condiion 0 / c o s sin e e M Differeniae GS from (i) A B 0 0 0, 0 0 A B M Use condiion / sin co s e e A Correc epression Graph 0 B A leas one and a half oscillaions; scales no required Approimaely consan ampliude and B consan frequency over a leas wo oscillaions; scales no required [6] (iii) sin 0 co s M Use 0 in eiher or soi A FT incorrec a and b from (i) c o s 0 sin 0 A FT incorrec a and b from (i) (iv) 0 0 M 0 or sae roos are real and disinc. Overdamped A Mus give a reason [] [] Consider auiliary equaion; eiher discriminan or naure of roos found in (i). Or sae he CF again. Noe ha only large required, bu accep any evidence of oscillaions FT values of a,b,a,b and boh values of negaive

79 78 Mark Scheme June 0 Quesion Answer Marks Guidance (v) / C e D e B FT from (i) (i) 0 0 C e D e M Use condiion. Accep applied a 0 C e D e C / 0 e, D e 0 C e D e 0 M Use condiion. Accep applied a 0 A cao 0 ( 0 ) / e e A EITHER: d v d k 00 d v g k v M NL d 0 v A k 00 [] 6 /. 0 e.6 0 e FT heir C and D if correc used sc Award for clearly defined. dv d M Separae and inegrae 0 v 00 k k 00 ln 0 v c A LHS / e e if new A RHS / 0 0 Rearrange, dealing properly wih M 00 consan M Use condiion 0 k v A e k 0, v 0 A 0 v 000 / 0 0 e k E k OR: d v d IF: d v g k v M NL d k 00 0 v A 00 e k k 0 0 k 0 0 ve e. 0 d M [8] B Muliply hrough by IF and recognise LHS 6

80 78 Mark Scheme June 0 Quesion Answer Marks Guidance 000 e k 00 A A Inegrae RHS k 0, v 0 A v 000 M Use condiion k 000 / 0 0 e k E k OR: d v d M Rearrange d v g k v M Use NL d k 00 0 v A k Auiliary equaion: 0 M k 00 CF: v Ae PI: v 000 k 00 M Find PI k 00 GS: v Ae 000 A k 0, v 0 A v 000 M Use condiion k 000 / 0 0 e k E k (ii) k/ 0 0 d e ( ) M Aemp o inegrae boh erms k k v c 0, M Use condiion c k k k/ 0 0 e k k A cao [8] A [8] [] 7

81 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iii) / k., e M.0 ( d p ) i.e. consisen E v 000. / 0 0 e 9. o dp B cao. (iv) d v 0 0 v g 0. v M NL (v) (a) (i) d v v d d v E v d v d v M Separae variables: aemp o inegrae A LHS ln v c A RHS v B M Rearrange, dealing properly wih consan M Use correc condiion e, v 9. B 9.8 v e A oe [] [] Graph B Increasing and condiion shown Saring from (, 9.) FT B Asympoe a (awr) [8] 0 M Auiliary equaion or IF : d y e 0 d CF A e A y a co s b sin B y a sin b co s a sin b cos ( a cos b sin ) sin M Differeniae and subsiue a b a b a 0, b M A Compare and solve cao co s sin e F CF wih one arbirary consan + PI y A [7] 8

82 78 Mark Scheme June 0 Quesion Answer Marks Guidance (a) (ii) A M Use condiion 9 co s sin e A cao y (b) (i) I Graph B Sars from (0, ) wih negaive gradien B Oscillaions wih consan ampliude and frequency for large. Scales no required e p d e B e d y e y e d d d e y e y e d e M M Inegrae e c A All correc y e c e F Divide hrough by IF [] (b) (ii) 0, y c c M Use condiion y e e A cao [] (c) I e M d d e y e a n [] M A 0 0 A LHS wih limis RHS 0 y () e e B LHS e y e a n d y () A 0.69 or beer [6] (i) z A e B Condiions z e B [] Their funcion mus have eponenial decay erm 9

83 78 Mark Scheme June 0 Quesion Answer Marks Guidance (ii) e (iii) y y y e M Subsiue y e M Rearrange y e M Differeniae e e e M Subsiue e A cao 0 M Auiliary equaion j A Correc roos CF e B co s C sin M Correc form of CF for heir roos F FT heir roos PI a e B Correc form of PI a e, a e a e a e a e e a M A valid mehod o find a GS e e B co s C sin A cao y e M [] Subsiue and ino epression for y e e co s sin e sin co s M Differeniae using produc rule B C B C e e ( ) co s ( ) sin A y B C B C [] (iv), 0 B M Use condiion y 0, 0 B C 0 M Use condiion cao aef, simplified o a maimum of erms FT if eponenial erm on RHS of d.e. 9, C B e e co s sin A e e co s sin A y [] 0

84 78 Mark Scheme June 0 Quesion Answer Marks Guidance (v) y e e c o s sin e e c o s sin e e sin e sin B Be convinced For 0, 0 e so is graph M Consider graphs oe will mee he graph of s in infiniely ofen E Complee argumen []

85 78 Mark Scheme June 0. Annoaions and abbreviaions Annoaion in scoris Meaning Blank Page his annoaion mus be used on all blank pages wihin an answer bookle (srucured or unsrucured) and on each page of an addiional objec where here is no candidae response. and BOD Benefi of doub FT Follow hrough ISW Ignore subsequen working M0, M Mehod mark awarded 0, A0, A Accuracy mark awarded 0, B0, B Independen mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighing Oher abbreviaions in mark Meaning scheme E Mark for eplaining U Mark for correc unis G Mark for a correc feaure on a graph M dep* Mehod mark dependen on a previous mark, indicaed by * cao Correc answer only oe Or equivalen ro Rounded or runcaed soi Seen or implied www Wihou wrong working

86 78 Mark Scheme June 0. Subjec-specific Marking Insrucions for GCE Mahemaics (MEI) Mechanics srand a Annoaions should be used whenever appropriae during your marking. The A, M and B annoaions mus be used on your sandardisaion scrips for responses ha are no awarded eiher 0 or full marks. I is vial ha you annoae sandardisaion scrips fully o show how he marks have been awarded. For subsequen marking you mus make i clear how you have arrived a he mark you have awarded. b An elemen of professional judgemen is required in he marking of any wrien paper. Remember ha he mark scheme is designed o assis in marking incorrec soluions. Correc soluions leading o correc answers are awarded full marks bu work mus no be judged on he answer alone, and answers ha are given in he quesion, especially, mus be validly obained; key seps in he working mus always be looked a and anyhing unfamiliar mus be invesigaed horoughly. Correc bu unfamiliar or unepeced mehods are ofen signalled by a correc resul following an apparenly incorrec mehod. Such work mus be carefully assessed. When a candidae adops a mehod which does no correspond o he mark scheme, award marks according o he spiri of he basic scheme; if you are in any doub whasoever (especially if several marks or candidaes are involved) you should conac your Team Leader. c The following ypes of marks are available. M A suiable mehod has been seleced and applied in a manner which shows ha he mehod is essenially undersood. Mehod marks are no usually los for numerical errors, algebraic slips or errors in unis. However, i is no usually sufficien for a candidae jus o indicae an inenion of using some mehod or jus o quoe a formula; he formula or idea mus be applied o he specific problem in hand, eg by subsiuing he relevan quaniies ino he formula. In some cases he naure of he errors allowed for he award of an M mark may be specified. A Accuracy mark, awarded for a correc answer or inermediae sep correcly obained. Accuracy marks canno be given unless he associaed Mehod mark is earned (or implied). Therefore M0 A canno ever be awarded. B Mark for a correc resul or saemen independen of Mehod marks.

87 78 Mark Scheme June 0 E A given resul is o be esablished or a resul has o be eplained. This usually requires more working or eplanaion han he esablishmen of an unknown resul. Unless oherwise indicaed, marks once gained canno subsequenly be los, eg wrong working following a correc form of answer is ignored. Someimes his is reinforced in he mark scheme by he abbreviaion isw. However, his would no apply o a case where a candidae passes hrough he correc answer as par of a wrong argumen. d e When a par of a quesion has wo or more mehod seps, he M marks are in principle independen unless he scheme specifically says oherwise; and similarly where here are several B marks allocaed. (The noaion dep * is used o indicae ha a paricular mark is dependen on an earlier, aserisked, mark in he scheme.) Of course, in pracice i may happen ha when a candidae has once gone wrong in a par of a quesion, he work from here on is worhless so ha no more marks can sensibly be given. On he oher hand, when wo or more seps are successfully run ogeher by he candidae, he earlier marks are implied and full credi mus be given. The abbreviaion f implies ha he A or B mark indicaed is allowed for work correcly following on from previously incorrec resuls. Oherwise, A and B marks are given for correc work only differences in noaion are of course permied. A (accuracy) marks are no given for answers obained from incorrec working. When A or B marks are awarded for work a an inermediae sage of a soluion, here may be various alernaives ha are equally accepable. In such cases, eacly wha is accepable will be deailed in he mark scheme raionale. If his is no he case please consul your Team Leader. Someimes he answer o one par of a quesion is used in a laer par of he same quesion. In his case, A marks will ofen be follow hrough. In such cases you mus ensure ha you refer back o he answer of he previous par quesion even if his is no shown wihin he image zone. You may find i easier o mark follow hrough quesions candidae-by-candidae raher han quesion-by-quesion. f Unless unis are specifically requesed, here is no penaly for wrong or missing unis as long as he answer is numerically correc and epressed eiher in SI or in he unis of he quesion. (e.g. lenghs will be assumed o be in meres unless in a paricular quesion all he lenghs are in km, when his would be assumed o be he unspecified uni.) We are usually quie fleible abou he accuracy o which he final answer is epressed and we do no penalise overspecificaion. When a value is given in he paper Only accep an answer correc o a leas as many significan figures as he given value. This rule should be applied o each case. When a value is no given in he paper

88 78 Mark Scheme June 0 Accep any answer ha agrees wih he correc value o s.f. f should be used so ha only one mark is los for each disinc error made in he accuracy o which working is done or an answer given. Refer cases o your Team Leader where he same ype of error (e.g. errors due o premaure approimaion leading o error) has been made in differen quesions or pars of quesions. There are some misakes ha migh be repeaed hroughou a paper. If a candidae makes such a misake, (eg uses a calculaor in wrong angle mode) hen you will need o check he candidae s scrip for repeiions of he misake and consul your Team Leader abou wha penaly should be given. There is no penaly for using a wrong value for g. E marks will be los ecep when resuls agree o he accuracy required in he quesion. g Rules for replaced work If a candidae aemps a quesion more han once, and indicaes which aemp he/she wishes o be marked, hen eaminers should do as he candidae requess. If here are wo or more aemps a a quesion which have no been crossed ou, eaminers should mark wha appears o be he las (complee) aemp and ignore he ohers. NB Follow hese mahs-specific insrucions raher han hose in he assessor handbook. h For a genuine misreading (of numbers or symbols) which is such ha he objec and he difficuly of he quesion remain unalered, mark according o he scheme bu following hrough from he candidae s daa. A penaly is hen applied; mark is generally appropriae, hough his may differ for some unis. This is achieved by wihholding one A mark in he quesion. Marks designaed as cao may be awarded as long as here are no oher errors. E marks are los unless, by chance, he given resuls are esablished by equivalen working. Fresh sars will no affec an earlier decision abou a misread. Noe ha a miscopy of he candidae s own working is no a misread bu an accuracy error. i If a graphical calculaor is used, some answers may be obained wih lile or no working visible. Allow full marks for correc

89 78 Mark Scheme June 0 answers (provided, of course, ha here is nohing in he wording of he quesion specifying ha analyical mehods are required). Where an answer is wrong bu here is some evidence of mehod, allow appropriae mehod marks. Wrong answers wih no supporing mehod score zero. If in doub, consul your Team Leader.

90 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) Auiliary equaion m m 0 M m i A CF e Acos Bsin F Their roos PI Psin Qcos B cao M Differeniae wice ( Ps Qc) (Pc Qs) ( Ps Qc) 0c M Subsiue P Q P 0 Q P Q 0 M Compare coefficiens and solve 6sin cos A Implied by P = 6 and Q = cwo GS: e A cos B sin 6sin cos F Their PI + heir CF wih arbirary consans. Mus be on RHS [9] (ii) 0, 0 A M Use given condiion e Asin Bcos e Acos Bsin cos 6sin M Differeniae, produc rule 0, 0 B M Use given condiion e cos sin 6sin cos A cao [] (iii) 6sin cos B FT Use large Ampliude = 6 = 6.7 B FT [] (iv) Auiliary equaion m m m 0 : B Some working mus be shown m = 0 is a roo m i B FT roos from (i) [] 6

91 78 Mark Scheme June 0 Quesion Answer Marks Guidance (v) Follow CF from (i), or new CF if differen roos in (iv), plus a non-zero = e Pcos Qsin C B consan. 0, 0 0 P C M Use condiion, dependen on CF including he non-zero consan 0, 0 0 Q P M Use condiion and use produc rule, 0 P Q P M Use condiion and use produc rule Q, P, C e cos sin A cao [] (vi) Sars a origin wih posiive gradien B Ignore negaive Tends o for large values of B FT heir value of C if correc form of soluion in (v) [] (i) ln 0. P A or (ii) 0. P B e M Separae and inegrae (need ln on LHS) 0, P 00 A ln00 or B 00 M Use condiion 0. P 00e A cao No suiable as ends o infiniy as increases Alernaives for firs M: 0. Auiliary eqn gives m 0. : P B e Inegraing facor e : Pe B : P B e B Allow if consan of inegraion no found 'Increases for ever/always' ges B; 'eponenial growh' ges B 'Suiable' ges B0; 'Suiable for small ' ges B [] 0. CF: Ae B PI: P 0. k e B Correc form P k e = 0.k e 8e M Differeniae and subsiue M Compare coefficiens and solve k A GS: P = Ae + e F Their PI + heir CF wih one arbirary consan [6] 7

92 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iii) 0, P 00 A 76 M Use condiion P = 76e + e A cao Unsuiable, no bounded (or equivalen) F [] (iv) M Separae variables (v) P 00 P 00 P 00 P ln P P dp 6 0 d P ln 6 0 ( B ) M A 'Increases for ever/always' ges B; 'eponenial growh' ges B 'Suiable for small ' ges B' 'Beer/more suiable han previous one' ges F0; Use parial fracions wih correc denominaors M Inegrae LHS (dependen on previous M) A P B ln M Use condiion (no dependen on parial fracions) 0, P P [ ln ] M Rearrange (dependen on parial fracions) 6 00 P P 00 e 6 T, P 00 or 6 00e P 6 A cao aef e [8] 600 T ln ln E AG (vi) Tends o a limi B Dependen on correc form of soluion in (iv) limi of 00 B cao (Noe ha P 00 ges BB) [] [] 8

93 78 Mark Scheme June 0 Quesion Answer Marks Guidance (a) (i) e y y B Divide by + IF e e M Spliing ino A e ln( ) B Seen ln( ) e e e E Show answer d ye ( ) e M Muliply boh sides by given IF A RHS mus be simplified ye ( ) e C M Inegrae boh sides, including + C y e e C M A [] (a) (ii) 0, y C M y e e A A [] Make y he subjec (need o divide all erms by coeff of y on LHS) cao aef cao 9

94 78 Mark Scheme June 0 Quesion Answer Marks Guidance (b) (i) Circle, cenre O B Radius B [] (b) (ii) One correc isocline B All correc isoclines B Raios of radii mus be correc Correc direcion indicaors B A leas four on each of he hree circles [] (b) (iii) Aemp a a soluion curve B Correc soluion curve B [] (b) (iv) One correc use of algorihm M.0 A.09 A.09 or beer.7 A []. or beer. Do no penalise lack of accuracy in inermediae values if final answer correc 0

95 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) y e M Differeniae wih respec o y e M Rearrange e y e M Subsiue for y e 8e e M Subsiue for y e A Auiliary equaion 0 M, A CF Ae B e F PI a e B Correc form for heir RHS ae ; a e : a = M Differeniae and equae a A Ae Be e F Their PI + heir CF wih wo arbirary consans [] (ii) y e M Rearrange firs given equaion (all erms mus be presen) y ( Ae Be e ) e 6 M Subsiue for M Subsiue for y Ae Be e A cwo in (i) and (ii) []

96 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iii) A B M Use given condiion A B 0 A B 0 A 8B 0 M Use given condiion A ; B 9 M Solve e e e 9 A cao y e e e A cao [] (iv) e e e e e e > 9 e e 9 M Use condiion and simplify e M Take logs ln A cao []

97 GCE Mahemaics (MEI) Uni 78: Differenial Equaions Advanced GCE Mark Scheme for June 0 Oford Cambridge and RSA Eaminaions

98 OCR (Oford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualificaions o mee he needs of candidaes of all ages and abiliies. OCR qualificaions include AS/A Levels, Diplomas, GCSEs, Cambridge Naionals, Cambridge Technicals, Funcional Skills, Key Skills, Enry Level qualificaions, NVQs and vocaional qualificaions in areas such as IT, business, languages, eaching/raining, adminisraion and secrearial skills. I is also responsible for developing new specificaions o mee naional requiremens and he needs of sudens and eachers. OCR is a no-for-profi organisaion; any surplus made is invesed back ino he esablishmen o help owards he developmen of qualificaions and suppor, which keep pace wih he changing needs of oday s sociey. This mark scheme is published as an aid o eachers and sudens, o indicae he requiremens of he eaminaion. I shows he basis on which marks were awarded by eaminers. I does no indicae he deails of he discussions which ook place a an eaminers meeing before marking commenced. All eaminers are insruced ha alernaive correc answers and unepeced approaches in candidaes scrips mus be given marks ha fairly reflec he relevan knowledge and skills demonsraed. Mark schemes should be read in conjuncion wih he published quesion papers and he repor on he eaminaion. OCR will no ener ino any discussion or correspondence in connecion wih his mark scheme. OCR 0

99 78 Mark Scheme June 0 Annoaions and abbreviaions Annoaion in scoris Meaning and BOD Benefi of doub FT Follow hrough ISW Ignore subsequen working M0, M Mehod mark awarded 0, A0, A Accuracy mark awarded 0, B0, B Independen mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighing Oher abbreviaions in Meaning mark scheme E Mark for eplaining U Mark for correc unis G Mark for a correc feaure on a graph M dep* Mehod mark dependen on a previous mark, indicaed by * cao Correc answer only oe Or equivalen ro Rounded or runcaed soi Seen or implied www Wihou wrong working

100 78 Mark Scheme June 0 Subjec-specific Marking Insrucions for GCE Mahemaics (MEI) Mechanics srand a Annoaions should be used whenever appropriae during your marking. The A, M and B annoaions mus be used on your sandardisaion scrips for responses ha are no awarded eiher 0 or full marks. I is vial ha you annoae sandardisaion scrips fully o show how he marks have been awarded. For subsequen marking you mus make i clear how you have arrived a he mark you have awarded. b An elemen of professional judgemen is required in he marking of any wrien paper. Remember ha he mark scheme is designed o assis in marking incorrec soluions. Correc soluions leading o correc answers are awarded full marks bu work mus no be judged on he answer alone, and answers ha are given in he quesion, especially, mus be validly obained; key seps in he working mus always be looked a and anyhing unfamiliar mus be invesigaed horoughly. Correc bu unfamiliar or unepeced mehods are ofen signalled by a correc resul following an apparenly incorrec mehod. Such work mus be carefully assessed. When a candidae adops a mehod which does no correspond o he mark scheme, award marks according o he spiri of he basic scheme; if you are in any doub whasoever (especially if several marks or candidaes are involved) you should conac your Team Leader. c The following ypes of marks are available. M A suiable mehod has been seleced and applied in a manner which shows ha he mehod is essenially undersood. Mehod marks are no usually los for numerical errors, algebraic slips or errors in unis. However, i is no usually sufficien for a candidae jus o indicae an inenion of using some mehod or jus o quoe a formula; he formula or idea mus be applied o he specific problem in hand, eg by subsiuing he relevan quaniies ino he formula. In some cases he naure of he errors allowed for he award of an M mark may be specified. A Accuracy mark, awarded for a correc answer or inermediae sep correcly obained. Accuracy marks canno be given unless he associaed Mehod mark is earned (or implied). Therefore M0 A canno ever be awarded. B Mark for a correc resul or saemen independen of Mehod marks. E A given resul is o be esablished or a resul has o be eplained. This usually requires more working or eplanaion han he esablishmen of an unknown resul.

101 78 Mark Scheme June 0 Unless oherwise indicaed, marks once gained canno subsequenly be los, eg wrong working following a correc form of answer is ignored. Someimes his is reinforced in he mark scheme by he abbreviaion isw. However, his would no apply o a case where a candidae passes hrough he correc answer as par of a wrong argumen. d e When a par of a quesion has wo or more mehod seps, he M marks are in principle independen unless he scheme specifically says oherwise; and similarly where here are several B marks allocaed. (The noaion dep * is used o indicae ha a paricular mark is dependen on an earlier, aserisked, mark in he scheme.) Of course, in pracice i may happen ha when a candidae has once gone wrong in a par of a quesion, he work from here on is worhless so ha no more marks can sensibly be given. On he oher hand, when wo or more seps are successfully run ogeher by he candidae, he earlier marks are implied and full credi mus be given. The abbreviaion f implies ha he A or B mark indicaed is allowed for work correcly following on from previously incorrec resuls. Oherwise, A and B marks are given for correc work only differences in noaion are of course permied. A (accuracy) marks are no given for answers obained from incorrec working. When A or B marks are awarded for work a an inermediae sage of a soluion, here may be various alernaives ha are equally accepable. In such cases, eacly wha is accepable will be deailed in he mark scheme raionale. If his is no he case please consul your Team Leader. Someimes he answer o one par of a quesion is used in a laer par of he same quesion. In his case, A marks will ofen be follow hrough. In such cases you mus ensure ha you refer back o he answer of he previous par quesion even if his is no shown wihin he image zone. You may find i easier o mark follow hrough quesions candidae-by-candidae raher han quesion-by-quesion. f Unless unis are specifically requesed, here is no penaly for wrong or missing unis as long as he answer is numerically correc and epressed eiher in SI or in he unis of he quesion. (e.g. lenghs will be assumed o be in meres unless in a paricular quesion all he lenghs are in km, when his would be assumed o be he unspecified uni.) We are usually quie fleible abou he accuracy o which he final answer is epressed and we do no penalise overspecificaion. When a value is given in he paper Only accep an answer correc o a leas as many significan figures as he given value. This rule should be applied o each case. When a value is no given in he paper Accep any answer ha agrees wih he correc value o s.f. f should be used so ha only one mark is los for each disinc error made in he accuracy o which working is done or an answer given. Refer cases o your Team Leader where he same ype of error (e.g. errors due o premaure approimaion leading o error) has been made in differen quesions or pars of quesions.

102 78 Mark Scheme June 0 There are some misakes ha migh be repeaed hroughou a paper. If a candidae makes such a misake, (eg uses a calculaor in wrong angle mode) hen you will need o check he candidae s scrip for repeiions of he misake and consul your Team Leader abou wha penaly should be given. There is no penaly for using a wrong value for g. E marks will be los ecep when resuls agree o he accuracy required in he quesion. g Rules for replaced work If a candidae aemps a quesion more han once, and indicaes which aemp he/she wishes o be marked, hen eaminers should do as he candidae requess. If here are wo or more aemps a a quesion which have no been crossed ou, eaminers should mark wha appears o be he las (complee) aemp and ignore he ohers. NB Follow hese mahs-specific insrucions raher han hose in he assessor handbook. h For a genuine misreading (of numbers or symbols) which is such ha he objec and he difficuly of he quesion remain unalered, mark according o he scheme bu following hrough from he candidae s daa. A penaly is hen applied; mark is generally appropriae, hough his may differ for some unis. This is achieved by wihholding one A mark in he quesion. Marks designaed as cao may be awarded as long as here are no oher errors. E marks are los unless, by chance, he given resuls are esablished by equivalen working. Fresh sars will no affec an earlier decision abou a misread. Noe ha a miscopy of he candidae s own working is no a misread bu an accuracy error. i j If a graphical calculaor is used, some answers may be obained wih lile or no working visible. Allow full marks for correc answers (provided, of course, ha here is nohing in he wording of he quesion specifying ha analyical mehods are required). Where an answer is wrong bu here is some evidence of mehod, allow appropriae mehod marks. Wrong answers wih no supporing mehod score zero. If in doub, consul your Team Leader. If in any case he scheme operaes wih considerable unfairness consul your Team Leader. 6

103 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) Auiliary equaion: m 8m 0 M m i A e A cos B sin F From heir roos (ii) (iii) CF: A 0 M Use condiion e B cos e. Bsin M Differeniae, produc rule 0. B : B e sin M A Use condiion (mus use 0.) F [8] 0 d e cos sin 0 M Equae heir o zero an : 0. M Obain epression for an and aemp o solve A Dependen on using correc answer o (i) 0.0 A cao [] Oscillaions wih decreasing ampliude B Accep ampliude 0 ; accep 0. accep "oscillaions wih small ampliude" Do no accep "oscillaions" [] 7

104 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iv) CF: e C cos D sin (v) F PI: Psin Q cos B Correc form P cos Q sin P sin Q cos P 0Q P Q 0P Q 0 : P 0, Q 8 M M A Differeniae wice and subsiue Compare coefficiens and solve e C cos D sin cos Their CF wih arbirary consans + heir PI 8 F 0, y 0 : C M Use condiion 8 e C sin D cos e C cos Dsin sin 8 0., 0: 0. D C ( D ) e cos sin cos 8 8 Oscillaions of approimaely consan ampliude 8 M M A [0] B [] Differeniae using produc rule Use condiion (mus use 0.) cao FT heir ampliude 8

105 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) n y y B Divide hrough by n d IF = e B ln (= e n ) = n B d n n n y M Muliply boh sides by heir IF n n n M Inegrae boh sides y A n n A Mus include arbirary consan M Divide boh sides, including an arbirary consan, by heir IF n y A n n A cao [8] (ii) y 0,, n : ( A ) M Use condiion o find a value for A y A Curve in s and h quadrans hrough (,0) B Mus use correc form of soluion. Skech going hrough (, 0) wih posiive gradien a (, 0) Correc behaviour as 0 and B Ignore curve for < 0 [] (iii) d y M Find and muliply by IF A y ln B M Inegrae boh sides A Mus include arbirary consan y ln B A cao [] 9

106 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iv) B M Use condiion, y 0 y ln 0 M Differeniae, equae o zero ln : One soluion E [] (v) y values: M NB he DE is used in he form d y y or A Agree o s.f A Agree o s.f. y value: 0.78 A 0.79 (or beer) [] (i) dv 0v 00 v M* Use NL wih accn in erms of v and d vv Mdep v * Separae variables ln A ln v A v e v e E When = 0, v.9() B [9] Mdep * Inegrae boh sides A lhs A rhs, including +A Mdep * Use condiion Mdep * rearrange cao 0

107 78 Mark Scheme June 0 Quesion Answer Marks Guidance (ii) dv 0 00 v M* Use NL wih accn in erms of v and d dv d Mdep Separae variables v * (iii) v ln B v Mdep * Inegrae boh sides. The inegral may be quoed from MF or PF used. v v Correc epression A B = 0 Mdep * Use condiion v ln v A aef When v =.9, =.689 M Use answer from (i) [7] dv 0 00 v d M Use NL wih accn in erms of v and 0ln(0 v) C M Separae and inegrae C 0ln 0 M Use condiion 0 0ln 0 v A 0 v 0e M Make v subjec A Correc epression Terminal velociy = 0ms B [7]

108 78 Mark Scheme June 0 Quesion Answer Marks Guidance dv OR: 0 00 v d M Use NL wih accn in erms of v and [ IF: 0. d e : ve e ] d ve 0e A M Muliply hrough by IF and inegrae A A 0 M Use condiion 0 v 0e M Make v subjec A Correc epression Terminal velociy = 0ms B [7] dv OR: 0 00 v d M Use NL wih accn in erms of v and 0. Auiliary eqn 0m 0 : CF: v Ae M PI: v B: B 0 M 0. GS: v0 Ae A A 0 M Use condiion 0 v 0e A Correc epression Terminal velociy = 0ms B [7] (iv) When v =.9, 0ln..0 s B []

109 78 Mark Scheme June 0 Quesion Answer Marks Guidance (i) d d d y M Differeniae d d d 60e M Subsiue for dy d d d 60e A AG Rearrange d d (ii) Auiliary equaion m m 0 M m, A CF: Ae Be Pe PI: B Correc form for heir CF Pe, Pe M Differeniae and subsiue P 0 A Solve Ae Be 0e F PI + CF wih arb cons [0] y d M Rearrange Subsiue for and d d M 7 y Ae Be e A cao. As final answer [] F

110 78 Mark Scheme June 0 Quesion Answer Marks Guidance (iii) 0, 0 0 A B 0 M Use condiion 7 M Use condiion y 0, 0 0 A B A 0, B 0 0e 0e 0e A cao 7 A cao y e e e [] (iv) T T When 0 0e 0e 0 0 M Muliply by e T T T e e 0 M Aemp o solve as a quadraic e (or -) A T ln (=.0) A cao M Subsiue T in epression for y y.6 A cao [6] (v) Unsuiable, X is negaive B []

111 OCR (Oford Cambridge and RSA Eaminaions) Hills Road Cambridge CB EU OCR Cusomer Conac Cenre Educaion and Learning Telephone: Facsimile: general.qualificaions@ocr.org.uk For saff raining purposes and as par of our qualiy assurance programme your call may be recorded or moniored Oford Cambridge and RSA Eaminaions is a Company Limied by Guaranee Regisered in England Regisered Office; Hills Road, Cambridge, CB EU Regisered Company Number: 866 OCR is an eemp Chariy OCR (Oford Cambridge and RSA Eaminaions) Head office Telephone: 0 Facsimile: 0 OCR 0

112 GCE Mahemaics (MEI) Uni 78: Differenial Equaions Advanced GCE Mark Scheme for June 06 Oford Cambridge and RSA Eaminaions

113 OCR (Oford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualificaions o mee he needs of candidaes of all ages and abiliies. OCR qualificaions include AS/A Levels, Diplomas, GCSEs, Cambridge Naionals, Cambridge Technicals, Funcional Skills, Key Skills, Enry Level qualificaions, NVQs and vocaional qualificaions in areas such as IT, business, languages, eaching/raining, adminisraion and secrearial skills. I is also responsible for developing new specificaions o mee naional requiremens and he needs of sudens and eachers. OCR is a no-for-profi organisaion; any surplus made is invesed back ino he esablishmen o help owards he developmen of qualificaions and suppor, which keep pace wih he changing needs of oday s sociey. This mark scheme is published as an aid o eachers and sudens, o indicae he requiremens of he eaminaion. I shows he basis on which marks were awarded by eaminers. I does no indicae he deails of he discussions which ook place a an eaminers meeing before marking commenced. All eaminers are insruced ha alernaive correc answers and unepeced approaches in candidaes scrips mus be given marks ha fairly reflec he relevan knowledge and skills demonsraed. Mark schemes should be read in conjuncion wih he published quesion papers and he repor on he eaminaion. OCR will no ener ino any discussion or correspondence in connecion wih his mark scheme. OCR 06

114 78 Mark Scheme June 06 Annoaions and abbreviaions Annoaion in scoris and Meaning Benefi of doub Follow hrough Ignore subsequen working,,, Mehod mark awarded 0, Accuracy mark awarded 0, Independen mark awarded 0, Special case Omission sign Highlighing Misread Oher abbreviaions in Meaning mark scheme E Mark for eplaining U Mark for correc unis G Mark for a correc feaure on a graph M dep* Mehod mark dependen on a previous mark, indicaed by * cao Correc answer only oe Or equivalen ro Rounded or runcaed soi Seen or implied www Wihou wrong working

115 78 Mark Scheme June 06. Subjec-specific Marking Insrucions for GCE Mahemaics (MEI) Mechanics srand a Annoaions should be used whenever appropriae during your marking. The A, M and B annoaions mus be used on your sandardisaion scrips for responses ha are no awarded eiher 0 or full marks. I is vial ha you annoae sandardisaion scrips fully o show how he marks have been awarded. For subsequen marking you mus make i clear how you have arrived a he mark you have awarded. b An elemen of professional judgemen is required in he marking of any wrien paper. Remember ha he mark scheme is designed o assis in marking incorrec soluions. Correc soluions leading o correc answers are awarded full marks bu work mus no be judged on he answer alone, and answers ha are given in he quesion, especially, mus be validly obained; key seps in he working mus always be looked a and anyhing unfamiliar mus be invesigaed horoughly. Correc bu unfamiliar or unepeced mehods are ofen signalled by a correc resul following an apparenly incorrec mehod. Such work mus be carefully assessed. When a candidae adops a mehod which does no correspond o he mark scheme, award marks according o he spiri of he basic scheme; if you are in any doub whasoever (especially if several marks or candidaes are involved) you should conac your Team Leader. c The following ypes of marks are available. M A suiable mehod has been seleced and applied in a manner which shows ha he mehod is essenially undersood. Mehod marks are no usually los for numerical errors, algebraic slips or errors in unis. However, i is no usually sufficien for a candidae jus o indicae an inenion of using some mehod or jus o quoe a formula; he formula or idea mus be applied o he specific problem in hand, eg by subsiuing he relevan quaniies ino he formula. In some cases he naure of he errors allowed for he award of an M mark may be specified. A Accuracy mark, awarded for a correc answer or inermediae sep correcly obained. Accuracy marks canno be given unless he associaed Mehod mark is earned (or implied). Therefore M0 A canno ever be awarded. B Mark for a correc resul or saemen independen of Mehod marks.

116 78 Mark Scheme June 06 E A given resul is o be esablished or a resul has o be eplained. This usually requires more working or eplanaion han he esablishmen of an unknown resul. Unless oherwise indicaed, marks once gained canno subsequenly be los, eg wrong working following a correc form of answer is ignored. Someimes his is reinforced in he mark scheme by he abbreviaion isw. However, his would no apply o a case where a candidae passes hrough he correc answer as par of a wrong argumen. d e When a par of a quesion has wo or more mehod seps, he M marks are in principle independen unless he scheme specifically says oherwise; and similarly where here are several B marks allocaed. (The noaion dep * is used o indicae ha a paricular mark is dependen on an earlier, aserisked, mark in he scheme.) Of course, in pracice i may happen ha when a candidae has once gone wrong in a par of a quesion, he work from here on is worhless so ha no more marks can sensibly be given. On he oher hand, when wo or more seps are successfully run ogeher by he candidae, he earlier marks are implied and full credi mus be given. The abbreviaion f implies ha he A or B mark indicaed is allowed for work correcly following on from previously incorrec resuls. Oherwise, A and B marks are given for correc work only differences in noaion are of course permied. A (accuracy) marks are no given for answers obained from incorrec working. When A or B marks are awarded for work a an inermediae sage of a soluion, here may be various alernaives ha are equally accepable. In such cases, eacly wha is accepable will be deailed in he mark scheme raionale. If his is no he case please consul your Team Leader. Someimes he answer o one par of a quesion is used in a laer par of he same quesion. In his case, A marks will ofen be follow hrough. In such cases you mus ensure ha you refer back o he answer of he previous par quesion even if his is no shown wihin he image zone. You may find i easier o mark follow hrough quesions candidae-by-candidae raher han quesion-by-quesion. f Unless unis are specifically requesed, here is no penaly for wrong or missing unis as long as he answer is numerically correc and epressed eiher in SI or in he unis of he quesion. (e.g. lenghs will be assumed o be in meres unless in a paricular quesion all he lenghs are in km, when his would be assumed o be he unspecified uni.) We are usually quie fleible abou he accuracy o which he final answer is epressed and we do no penalise overspecificaion. When a value is given in he paper Only accep an answer correc o a leas as many significan figures as he given value. This rule should be applied o each case.

117 78 Mark Scheme June 06 When a value is no given in he paper Accep any answer ha agrees wih he correc value o s.f. f should be used so ha only one mark is los for each disinc error made in he accuracy o which working is done or an answer given. Refer cases o your Team Leader where he same ype of error (e.g. errors due o premaure approimaion leading o error) has been made in differen quesions or pars of quesions. There are some misakes ha migh be repeaed hroughou a paper. If a candidae makes such a misake, (eg uses a calculaor in wrong angle mode) hen you will need o check he candidae s scrip for repeiions of he misake and consul your Team Leader abou wha penaly should be given. There is no penaly for using a wrong value for g. E marks will be los ecep when resuls agree o he accuracy required in he quesion. g Rules for replaced work If a candidae aemps a quesion more han once, and indicaes which aemp he/she wishes o be marked, hen eaminers should do as he candidae requess. If here are wo or more aemps a a quesion which have no been crossed ou, eaminers should mark wha appears o be he las (complee) aemp and ignore he ohers. NB Follow hese mahs-specific insrucions raher han hose in he assessor handbook. h For a genuine misreading (of numbers or symbols) which is such ha he objec and he difficuly of he quesion remain unalered, mark according o he scheme bu following hrough from he candidae s daa. A penaly is hen applied; mark is generally appropriae, hough his may differ for some unis. This is achieved by wihholding one A mark in he quesion. Marks designaed as cao may be awarded as long as here are no oher errors. E marks are los unless, by chance, he given resuls are esablished by equivalen working. Fresh sars will no affec an earlier decision abou a misread. Noe ha a miscopy of he candidae s own working is no a misread bu an accuracy error. 6

118 78 Mark Scheme June 06 i j If a graphical calculaor is used, some answers may be obained wih lile or no working visible. Allow full marks for correc answers (provided, of course, ha here is nohing in he wording of he quesion specifying ha analyical mehods are required). Where an answer is wrong bu here is some evidence of mehod, allow appropriae mehod marks. Wrong answers wih no supporing mehod score zero. If in doub, consul your Team Leader. If in any case he scheme operaes wih considerable unfairness consul your Team Leader. 7

119 78 Mark Scheme June 06. (i) AE: m 8m 0 M A m, F CF: Ae Be PI: ke B k M A Differeniae and subsiue o find k Ae Be e F PI + CF wih arbirary consans [7] (ii) (iii) 0, 6 6 A B 0, d 7 AB A, B Minimum when e 8e 9 0 M M M e e e A [] Use condiion Differeniae and use condiion Solve 9 e e e 0 M Differeniae and equae o 0 M Muliply hrough by e e 9 0 M Solve ln9 = ln A cao o.e (.97 ) [] e 8

120 78 Mark Scheme June 06 (iv) CF: Ce De FT from (i) PI: 0. Qe B 0. Q( 0. )e : 0. Q( 0. e ) M Differeniae using produc rule and subsiue Q M Compare coefficiens and solve Q A 0. e + Ce De A cao 0. Ce De e 0. M Differeniae using produc rule = 0, = 6: CD 6 M Use condiion correcly 0, : d C 0 M Use condiion correcly (independen of use of produc rule) and solve C, D 0. e + e e A cao [9]. (a)(i) Divide hrough by B Boh sides IF: e d M Aemp o find inegraing facor ln = e M Inegrae and simplify log erm = A d y cos M Muliply boh sides by IF and epress LHS as differenial o.e. RHS = sin cos c M Inegrae by pars A Including + c y sin cos c F Divide hrough by heir inegraing facor [8] 9

121 78 Mark Scheme June 06 (ii) (b)(i) y = 0 when 0 M Use condiion c : y sin cos A cao [] dy y sec B Separae he variables M Aemp o inegrae = cos d y A LHS = sin c A Inegral of RHS including + c y when M Use condiion M Make y he subjec y sin A cao Final answer [7] (ii) Ma value when sin M Correc saemen for heir soluion o (b)(i) Or any oher valid mehod e.g. differeniaion, second derivaive no required Ma value = /(-) = AG E Correcly shown; /(-) mus be seen [] SC for saing BOTH ma value = and min value = /7 (c) dy ( )cos d y B May be implied by correc values y y hy M Use algorihm A y(.0) = Agreemen o sf A y (.0) = 0.79 Agreemen o sf A 0.0 or beer [] 0

122 78 Mark Scheme June 06. (i) dv 90v d g v M Use NL, erms, allow sign errors, allow any form for accn, including a dv v d v E [] (ii) vv d v M* Separae variables ln v A A Inegrae o obain LHS A RHS including consan on one side Use v = 0, = 0: A ln Mdep* Use condiion e v Mdep* Rearrange E [6] (iii) Increasing graph hrough (0, 0) B Asympoe v 0 B Allow 9. [] (iv) m MA Accep.s.f. [] (v) dv d g v M Use NL, erms, allow sign errors dv d g 0.8 v A dv d v M Separae variables ln v B 0.8 A Inegrae o obain LHS A RHS, including consan on one side

123 78 Mark Scheme June 06 Use = 0, v = 8: B ln M Use condiion..7e 0.8 v M Rearrange A cao [8] Alernaive mehod : dv d g v dv d g 0.8 v M A Use NL, erms, allow sign errors Inegraing facor: e 0.8 d ve ge d ve.e A 0.8 v. Ae Use = 0, v = 8: A.7..7e 0.8 v Alernaive mehod : dv d g v dv d g 0.8 v CF: A e 0.8 PI: v =. 0.8 GS: v Ae. Use = 0, v = 8: A.7..7e 0.8 v B M A M M A [8] M A MA M A M A [8] Muliply boh sides by IF and epress LHS as differenial o.e. Rearrange Use condiion cao Use NL, erms, allow sign errors Use condiion

124 78 Mark Scheme June 06 (vi) e D 0.8 M Inegrae A cao.7 Use = 0, = 0: D 0.8 M Use condiion When =, disance = 0 m A cao [] Quesion Answer Marks Guidance. (i) y sin M Differeniae y sin sin M Subsiue for y y cos M Subsiue for y 9sin cos A oe AE m 0 M m j A CF Acos B sin F PI Psin Qcos B Pcos Qsin Psin Qcos M Differeniae wice and subsiue P9 Q M Equae coefficiens and solve P, Q M Acos B sin sin cos A cao [] (ii) Asin Bcos cos sin M Differeniae Subsiue and in y cos M y ( A B)cos ( A B)sin sin cos A []

125 78 Mark Scheme June 06 Quesion Answer Marks Guidance (iii) 0, y 0: A B 0 M Use condiion dy 0, : A ( A B) d M Use condiion Solve o give A0, B sin sin cos y cos sin sin cos A A [] (iv) M Equae heir epressions for and y cos sin A cao sin sin 0 : sin sin M Use double angle formula and aemp o solve quadraic equaion or 6 6 A One correc value A cao []

126 OCR (Oford Cambridge and RSA Eaminaions) Hills Road Cambridge CB EU OCR Cusomer Conac Cenre Educaion and Learning Telephone: Facsimile: general.qualificaions@ocr.org.uk For saff raining purposes and as par of our qualiy assurance programme your call may be recorded or moniored Oford Cambridge and RSA Eaminaions is a Company Limied by Guaranee Regisered in England Regisered Office; Hills Road, Cambridge, CB EU Regisered Company Number: 866 OCR is an eemp Chariy OCR (Oford Cambridge and RSA Eaminaions) Head office Telephone: 0 Facsimile: 0 OCR 06