2610 MEI Differential Equations January 2005 Mark Scheme packs/MS_MEI_DE.pdf · 2018-04-04 · 2610 MEI Differential Equations January 2005 Mark Scheme 1(i) I =exp 2 d()∫ xx M1

2610 MEI Differential Equations January 2005 Mark Scheme

1(i) ( )exp 2 dI x x= ∫ M1

2ex= A1

2 2 2 2( 2)de 2 e ed

x x x xy x yx

− −+ = M1 multiply

( )2 4 4d ed

x xy ex

−= F1 follow their integrating factor

2 4 4e e dx xy x−= ∫ M1 integrate

4 414 e x A−= + A1

( )2 4 414e ex xy A− −= + F1 divide by their I (must divide constant also)

7 (ii) ( )1 1 1

4 40 e A A−= + ⇒ = − M1 condition on y

( )2 4 414 e e 1x xy − −= − A1 cao

4 41 4 4 0 e 1xx x −> ⇒ − > ⇒ > M1 attempt inequality for y and

214 e 0x− > so y > 0 E1 fully justified

B1 through (1,0) and y > 0 for x > 1

B1 asymptotic to y = 0

6 (iii)

maximum d 0dyx

⇒ = M1

2( 2)2 e xxy − −⇒ = M1 must use DE

142 4 1x y y= ⇒ = ⇒ = E1

( )4 41 14 4e e A−= + M1 substitute into GS for y

2( 2)140 e xA y − −⇒ = ⇒ = A1 cao

B1 general shape consistent with their solution

B1 maximum labelled at 14(2, )

7

1

¼ 2


2(i) for x < 0, aux.eq. 2 9 0 3 jα α+ = ⇒ = ± M1 imaginary root or recognise SHM equation

cos3 sin 3y A x B x= + A1

for x > 0, 2

2d 16 0d

y yx

− = B1 may be implied

2 16 0α − = M1 4α = ± A1 4 4e ex xy C D−= + F1 accept A,B again here but not in (ii)

6 (ii) (1) 0A y= F1 0C D y+ = F1

(2) d0, 3 sin 3 3 cos3dyx A x B xx

< = − + M1 differentiate

4 4d0, 4 e 4 ed

x xyx C Dx

−> = − + M1 differentiate

3 4 4B C D= − + A1 (3) only 4e x is unbounded so 0D = B1 hence 0C y= B1 4

03B y= − B1

8 (iii) B1 curve for x < 0 B1 curve for x > 0 B1

continuous gradient at x = 0 (must have reasonable attempt at each of x < 0 and x > 0)

3 (iv) 2

2d 4 0 cos 2 sin 2d

y y y E x F xx

+ = ⇒ = + B1

bounded so (3) provides no equation M1 hence insufficient (3 equations, 4 unknowns) A1 3


3(i)

rate of flow = area × speed d 0.0004 2dV gxt

⇒ = − M1 accept either in words or symbols for M1 but need both for E1

M1 use of chain rule

d d 0.0004 2d dV x gxx t

⇒ = − E1 complete argument

3 (ii) 5 5

3 3d d 0.0004 2d dV x gxx t= ⇒ = − B1

125

3 d 0.0004 2 dx x g t− = −∫ ∫ M1 separate

1210

3 0.0004 2x g t A= − + M1 integrate

1030, 2 2t x A= = ⇒ = M1 condition on x

( )22 1 0.00012x g t= − A1 cao

0 2660x t= ⇒ ≈ A1 6 (iii) ( )2 d1 2 0.0004 2

dxx x gxt

+ − = − M1 substitute for ddVx

31 12 2 2( 2 )d 0.0004 2 dx x x x g t− + − = −∫ ∫ M1 separate

M1 integrate

3 512 2 24 2

3 52 0.0004 2x x x g t B+ − = − + A1

46150, 2 2 4.337t x B= = ⇒ = ≈ M1 condition on x

0 2450x t= ⇒ ≈ A1 6 (iv) (0) 0.0004( 4 0) /1 0.00250x g= − + = − E1 must show working

(0.1) 2 0.00250 0.1x = − × M1 use of algorithm =1.99975 E1 must show working (0.1) 0.00251x = − B1 (0.2) 1.99950x = B1 5


4(i) M1 differentiate

2

2d d d8 3

d ddx x y

t tt= −

A1 d8 3( 2 7 )

dx x yt

= − − + M1 substitute for ddyt

d 21 d8 6 8d 3 dx xx xt t

⎛ ⎞= + − −⎜ ⎟⎝ ⎠

M1 substitute for y

2

2d d15 50 0

ddx x x

tt− + = E1

5 (ii) 2 15 50 0α α− + = M1 auxiliary equation 5 or 10α = A1 5 10e et tx A B= + F1 CF for their roots ( )1

3 8y x x= − M1 rearrange equation (1)

( )( )5 10 5 1013 8 e 8 e 5 e 10 et t t tA B A B= + − + M1 substitute for x and x

5 1023e et tA B= − A1 cao

6 (iii) e 8 e 3 e et t t ta a b− − − −− = − + M1 substitute in (3) e 2 e 7 e et t t tb a b− − − −− = − + + M1 substitute in (4) 9 3 1, 2 8 1a b a b− + = − = M1 compare coefficients and solve 1

6a = − A1

16b = − A1

5 (iv) substituting f(t), g(t) into LHS gives zero B1 stated (or substitution) substituting solutions in (iii) gives e t− B1 as equations linear, substituting sums gives sum 0 e et t− −+ = E1 (or verified by substitution) General solutions because two arbitrary constants as expected for two first order equations. E1 4

2610 Mark Scheme June 2005

1(i) 2

2d d4 6

d dddx x y

t tt= − + M1 differentiate

d4 6( 3 2 26dx x yt

= − + − + + ) M1 substitute for ddyt

d 12 d4 18 4 28 1d 6 dx xx xt t

⎛ ⎞= − + − + + − +⎜ ⎟⎝ ⎠

56 M1 substitute for y

2

2d d2 10 100

ddx x x

tt+ + = E1

4 (ii) 01022 =++ αα M1 auxiliary equation 1 3 jα = − ± A1 CF e ( cos3 sin 3 )tx A t B t−= + F1 CF for their solutions

PI 10

10100

==x B1

GS 10 e ( cos3 sin 3 )tx A t B t−= + + F1 their CF + their PI

d e ( cos3 sin 3 3 sin 3 3 cos3 )d

tx A t B t A t B tt

−= − − − + M1 differentiate

16 ( 4 28y x x= + − ) M1 rearrange and substitute

(122 e ( )cos3 ( )sin 3t )y A B t B A t−= + + + − A1 cao

8 (iii) 10010 −=⇒=+ AA M1 use t = 0 with their x = 0 60)(2 2

1 =⇒=++ BBA M1 use t = 0 with their y = 0 (or their 28x = )

10 e (6sin 3 10cos3 )tx t t−= + −

2 e (8sin 3 2cos3 )ty t−= + − t A1 both (cao)

3 (iv) B1 initial condition and asymptote for one

graph

B1 both generally correct (must start at origin)

B1 long-term values

long-term values can be found by setting 0== yx in DE’s M1 and solving the resulting equations A1 5


2(i) B1 r starts at 56 and decreases, tending

to 4

B1 I starts at 0, gradient is positive but decreases

2 (ii) M1

d ( 4)dr k rt= − − A1

d d4

r k tr

= −−∫ ∫ M1 separate and integrate

1|4|ln cktr +−=− A1 all correct 4 e ktr A −= + M1 rearranging 5256,0 =⇒== Art M1 use initial condition ( )4 1 13 ktr e−= + E1

7 (iii)

213d 4 e ktI r t t ck

−⎛ ⎞= = − +⎜ ⎟⎝ ⎠∫ M1 integrate r

k

cIt 520,0 2 =⇒== M1 use initial condition

( )524 1 e ktI tk

−= + − A1 cao

( )620523000 4 620 1 e kk

−= × + − M1 condition on their I

62010 1 e kk −⇒ = − E1 must follow correct I unless k very small, 620e 0k− ≈ 1.0110 ≈⇒≈⇒ kk E1 independent of other marks 6 (iv) 1.001.0 −=⇒=+ αα so CF 0.1e tr B −= B1

for given PI d 2 sin 2 2 cos 2dr b t ct= − +

4

t

ttcbtbca 2cos2.04.02cos)21.0(2sin)21.0(1.0 +=++−+ M1 differentiate and substitute 0.1 0.4a = 021.0 =− bc M1 compare at least two coefficients and

solve

2.021.0 =+ cb 2 44, ,

401 401a b c= = =

0 A1

conditions , so 995.51=⇒ B

( )0.1 251.995e 4 cos 2 20sin 2401

tr t−= + + + t A1

5


3(a)(i) d cos 2sin

d 1y x yx x

−=

+ M1 may be implied

0.16 0.625629 M1 use of algorithm 0.18 0.143637 0.591150 A1 y(0.18) 0.20 0.155460 M1 use of algorithm A1 y(0.20) 5 (a)(ii) dy/dx decreases B1 B1 sketch showing curve and step by

step solution

For each step, gradient used is greater than dy/dx over interval, hence overestimates y. E1 convincing argument

3 (b)(i) d 2 cos

d 1 1y xyx x x+ =

+ + M1 rearrange

M1 attempt integrating factor

22exp d ( 1)1

I xx

⎛ ⎞= =⎜ ⎟+⎝ ⎠∫ x + A1

2 d( 1) 2( 1) ( 1)cosdyx x y xx

+ + + = + x M1 multiply

2( 1) ( 1)cos d ( 1)sin sin dx y x x x x x x+ = + = + −∫ x∫ M1 attempt integration by parts

Axxx +++= cossin)1( A1 10,0 −=⇒== Ayx M1

2)1(1cossin)1(

+−++

=x

xxxy E1

1517.02.0 =⇒= yx B1 9 (b)(ii)

Since sin y < y, replacing sin y by y in d cos 2sind 1y x yx x

−=

+ M1 consider effect on dy/dx

gives an underestimate for dy/dx. A1 Hence (while y > 0), the approx. DE will underestimate y. E1 3


4(i) 022 23 =−−+ ααα M1 auxiliary equation 0)23)(1( 2 =++− ααα M1 factorise or demonstrate 1 is a root

2,1,1 −−=α A1 CF 2e et ty A B C− −= + + et

F1 CF for their roots (must have 3 constants)

PI 3e ty a −= B1 correct form

3 33 e , 9 e , 27 et ty a y a y a− −= − = = − 3t− M1 differentiate and substitute 4231827 =−++− aaaa M1 compare coefficients

21−=a A1

2 312e e e et t ty A B C− − −= + + − t F1 their CF + their PI

9 (ii) decays 0=⇒ C B1

21

23

23,0 −+=−⇒−== BAyt M1 condition on y

2 332e 2 e et ty A B− −= − − + t− M1 differentiate

23

23

23 2,0 +−−=⇒== BAyt M1 condition

1,2 =−= BA so 2 3122e e et ty − − −= − + − t E1

5 (iii) let so e tu −= )42(2 2

213

212 +−−=−+−= uuuuuuy

e tu −= ≠ 0 and 03)1(42 22 >+−=+− uuu M1 consider quadratic (any valid method)

hence for all t 0≠y E1 if discriminant used, value or working must be shown must indicate u non-zero

( )2 3 23 3 4 42 2 32e 2e et t ty u− − −= − + = − + 3u u

( )( )23 822 3 9 0u u= − + > M1 consider quadratic (any valid method)

hence no turning points E1 B1 starts at 2

3− and asymptote y = 0 B1 Shape (increasing)

6

1(i) 2 6 8 0λ λ+ + = M1 2 or 4λ = − − A1 CF 2 4e et tI A B− −= + F1 PI e tI a −= B1 e 6 e 8 e 6et t t ta a a− − − −− + = M1 differentiate and substitute 3 6a = dM1 compare 2a = A1 2 42e e et t tI A B− − −= + + F1 CF + PI 8 (ii) 2 6 9 0λ λ+ + = M1 3λ = − (repeated) A1 CF 3( ) e tI C Dt −= + F1 PI e tI b −= B1 e 6 e 8 e 6et t t tb b b− − − −− + = M1 substitute and compare 3

2b = A1 33

2 e ( ) et tI C Dt− −= + + F1 CF + PI 3

21.5 0C C= + ⇒ = M1 condition on I 3 33

2 e 3( )e et t tI C Dt D− − −= − − + + M1 differentiate 3 3

2 20 3C D D= − − + ⇒ = M1 condition on I ( )33

2 e et tI t− −= + A1 cao as , 0t I→∞ → F1 recognise e 0t− → 12 (iii)

2 6 0 3 9k kλ λ λ+ + = ⇒ = − ± − M1

0 9 9 3k k< < ⇒ − < ⇒ two negative roots hence 1 2e e 0t tI A Bλ λ− −= + → E1 9 3k jλ β> ⇒ = − ±

M1 complex roots with negative real part (or CF)

3e ( cos sin ) 0t A t B tβ β− + → E1 4

2(i)

21 d 1

1 3 dy

y x x=

− M1 separate

2

1 1d d1 3

y xy x

=−∫ ∫ M1 integrate

A1 ± LHS

13

1ln 1 3y cx

− − = − + A1 ± RHS

31 3 e xy A− = M1 rearrange

30, 1 ey x A −= = ⇒ = M1 condition ( )31

3 1 exp( 3)xy = − − A1 2 0.259x y= ⇒ ≈ A1 from correct solution 8 (ii)

2d 3 1 cosdy y xx x x+ = M1 divide

3exp( d )xI x= ∫ M1 attempt integrating factor 3x= A1 ( )3d cos

dx y x x

x= F1 follow their I

3 cos d sin sin dx y x x x x x x x= = −∫ ∫ M1 integrate (by parts) sin cosx x x B= + + A1 RHS (or multiple) constant not

required here

2 3sin (cos )y x x x x B− −= + + M1 divide to get y 1, 0 cos1 sin1x y B= = ⇒ = − − M1 use condition 2 3sin (cos cos1 sin1)y x x x x− −= + − − A1 M1 substitute 2x =

2 0.00258x y= ⇒ ≈ A1 cao

11 (iii)

( )( )2 2cos 3 0.1y x x y y x−′ = − + B1 seen or implied by correct numerical value

x y y 1.8 0.034411 –0.12767 M1 use algorithm 1.9 0.021644 –0.12380 A1 y(1.9) 2.0 0.00926(

3…) A1 y(2.0)

Using smaller h would give greater accuracy B1

5

3(i) 2d 0.001

dvF ma mv mg mvx

= ⇒ = − M1 N2L (accept just ma for M1)

2d 0.001dvv g vx

⇒ = − E1

down positive so weight positive and resistance negative as it opposes motion B1

3

(ii) M1 separate 2

0.002 d 0.002d0.001

v v xg v−

= −−∫ ∫

M1 integrate 2ln 0.001 0.002g v x c− = − + A1 LHS (or multiple) ( )2 0.0021000 e xv g A −= − M1 rearrange 0, 0x v A g= = ⇒ = M1 use condition ( )0.0021000 1 e xv g −= − A1 cao

50 30.54x v= ⇒ = E1 must follow correct work 7 (iii) d 2

dvmv mg mvx= − M1

d 2dvv g vx= − A1

d d

2v v x

g v=

−∫ ∫ M1 separate

A1 x c+ 1

2 1 d2

g v x cg v

⎛ ⎞− + = +⎜ ⎟−⎝ ⎠

∫ M1 attempt to integrate LHS

1 12 4 ln 2v g g v x c− − − = + A1

50, 30.54 74.91...x v c= = ⇒ = − M1 use condition (correct value of v at least) 5 76.36v x= ⇒ = M1 so 26.4 m deep A1 9 (iv) terminal velocity when acceleration zero M1 4.9v⇒ = F1 follow their DE B1 increasing from (0,0) B1 decreasing to asymptote at 4.9 (or follow

their value)

B1 cusp/max at (50, 30.54) (both coordinates shown)

5

4.9

4(i) 2 cosx x y t= − + M1 differentiate 2(4 3 cos ) cosx x y t t= − − + + M1 substitute 8 6 cosx x y t= − + − 1

2 ( sin )y x t x= + − M1 y in terms of ,x x 8 3( sin ) cosx x x x t x t= − + + − − M1 substitute 2 5 3sin cosx x x t t+ + = − E1 5 (ii) 2 2 5 0λ λ+ + = M1 auxiliary equation M1 solve to get complex roots

1 2 jλ = − ± A1

CF ( )e cos 2 sin 2tx A t B t−= + F1

CF for their roots (for complex roots must be in exp/trig form, not complex exponentials)

PI sin cosx a t b t= + B1 cos sin , sin cosx a t b t x a t b t= − = − − M1 differentiate twice and substitute 2 5 3a b a− − + = M1 compare 2 5 1b a b− + + = − M1 solve 1 1

2 2,a b= = − A1 ( ) ( )1

2e cos 2 sin 2 sin costx A t B t t t−= + + − F1 CF + PI 10 (iii) ( )1

2 siny x t x= + − M1 y in terms of ,x x ( ) (

( )12

e cos 2 sin 2 e 2 sin 2 2 cos 2

cos sin

t tx A t B t A t B

t t

− −= − + + − +

+ +

M1 differentiate x

M1 substitute for ,x x A1 CF part

( ) ( )12e ( ) cos 2 ( )sin 2 sin costy A B t A B t t t−= − + + + −

A1 PI part

5 (iv) ( )1

2~ sin cosx t t− F1 ( )1

2~ sin cosy t t− F1 hence for large ,t x y≈ B1 must follow correctly from their

solutions

but unless 0, or so A B A B A A B B x y= = − ≠ + ≠ ≠

B1 must follow correctly from their solutions

4


1(i) 0λ = B1 cos 5 sin 5x A t B t= + M1 cos 5t or sin 5t or cos sinω ω+A t B t seen

or GS for their λ

A1 3 (ii) 2(2 ) 4 5 0λ − ⋅ < M1 Use of discriminant A1 Correct inequality 0 5λ< < A1 Accept lower limit omitted or 5− 3 (iii) 2 2 5 0α α+ + = M1 Auxiliary equation 1 2 jα = − ± A1 ( )e cos 2 sin 2tx C t D t−= + F1 CF for their roots 3 (iv) 0x C= M1 Condition on x ( ) ( )e cos 2 sin 2 e 2 sin 2 2 cos 2t tx C t D t C t D t− −= − + + − +& M1 Differentiate (product rule) 0 2C D= − + M1 Condition on x& 1

02D x= ( )1

0 2e cos 2 sin 2tx x t t−= + A1 cao 4 (v) 1

2cos 2 sin 2 0t t+ = M1 tan 2 2t = − M1 1.017t = A1 cao 3 (vi) 2 6 5α α+ + M1 Auxiliary equation 1, 5α = − − A1 5e et tx E F− −= + F1 CF for their roots 0x E F= + M1 Condition on x 5e 5 et tx E F− −= − −& 0 5E F= − − M1 Condition on x& 5 1

0 04 4,E x F x= = − ( )51

04 5e et tx x − −= − A1 cao ( )41

04 e 5 et tx x − −= − M1 Attempt complete method 4

00 5 e , 0, e 0 0− −> ⇒ > > > ⇒ >t tt x x i.e. never zero E1 Fully justified (only 0≠ required) 8


2(i) 2 0 2λ λ+ = ⇒ = − M1 CF 2e tx A −= A1 PI x at b= + B1 2( ) 1a at b t+ + = + M1 Differentiate and substitute 2 1, 2 1a a b= + = M1 Compare 1 1

2 4,a b= = A1 21 1

2 4 e tx t A −= + + F1 CF + PI 1

40, 1 1t x A= = ⇒ = + M1 Condition on x 231 1

2 4 4 e tx t −= + + F1 Follow a non-trivial GS Alternatively: M1 ( ) 2exp 2 d e tI t= =∫

A1 Integrating factor

( )2 2 2de 2e e 1d

t t tx x tt

+ = + B1 Multiply DE by their I

( )2 2e e 1 dt tx t t= +∫ M1 Attempt integral ( )2 21 1

2 2e 1 e dt tt t= + − ∫ M1 Integration by parts ( )2 2 21 1

2 4e e 1 et t tx t A= + − + A1 21 1

2 4 e tx t A −= + + F1 Divide by their I (must also divide constant) 1

40, 1 1t x A= = ⇒ = + M1 Condition on x 231 1

2 4 4 e tx t −= + + F1 Follow a non-trivial GS 9 (ii) 2 d 1

dy

y x x= M1 Separate

2 1d dy xy x

=∫ ∫ M1 Integrate

2ln lny x c= + y B x= M1 Make y subject, dealing properly with constant ( )0 , 1, 4 4t x y y x= = = ⇒ = M1 Condition 231 1

2 4 44 e ty t −= + + F1 y = 4√(their x in terms of t)

5 (iii) d 2 6

dz zx x

+ = M1 Divide DE by x

( )2exp dxI x= ∫ M1 Attempt integrating factor 2x= A1 Simplified ( )2 2d 6

dx z x

x= F1 Follow their integrating factor

2 32x z x C= + A1 22z x Cx−= + F1 Divide by their I (must also divide constant) ( )0 , 1, 3 1t x z C= = = ⇒ = M1 Condition on z 22z x x−= + A1 cao (in terms of x) 1 0.852t x= ⇒ = 3.69y = B1 Any 2 values (at least 3sf) 3.08z = B1 All 3 correct (and 3sf) 10


3(i) d 1 f( )d

v xx v

= so ( ) dunless f( ) 0 , 0dvx vx

= → ⇒ → ±∞ M1

Consider dd

vx

or dd

xv

when v = 0, but not if

d 0d

vx

=

i.e. gradient parallel to v-axis (vertical) E1 Must conclude about direction

2 2d 1 14000 0d 5000 5000

= ⇒ = − =vx vx

M1 Consider dd

vx

when x = 4000

so if 0≠v then gradient parallel to x-axis (horizontal) E1 Must conclude about direction M1 Add to tangent field A1 Several vertical direction indicators on x-axis 6 (ii) M1 Attempt one curve A1 M1 Attempt second curve

A1

0 0.05V = ⇒ probe reaches B B1 Must be consistent with their curve 0 0.025V = ⇒ probe returns to A B1 Must be consistent with their curve N.B. Cannot score these if curve not drawn 6 (iii) ( )2 2d (9000 ) (1000 ) dv v x x x− −= − − +∫ ∫ M1 Separate M1 Integrate B1 LHS

212

1 19000 1000

v cx x

= + +− +

A1 RHS

21 1 102 9000 1000V c= + + M1 Condition

2 2 10 450

2 29000 1000

v Vx x

= + + −− +

A1

6 (iv) minimum when x = 4000 B1 Clearly stated M1 Substitute their x into v or 2v

2 22 2 1min 05000 5000 450v V= + + −

F1 Their 2v or v when x = 4000 need 2

min 0v > M1 For 2min 0v >

2 2 1 4min 0 450 50000 if v V> > − M1 Attempt inequality for 2

0V 0 0.0377V > A1 cao 6


4(i) 2x x y= −&& & & M1 Differentiate first equation 2 (5 4 18)x x y= − − +& M1 Substitute for &y 2 3y x x= + − & M1 y in terms of , &x x 2 5 4(2 3 ) 18x x x x x= − + + − −&& & & M1 Substitute for y 2 3 6x x x+ − = −&& & E1 LHS E1 RHS 6 (ii) 2 2 3 0λ λ+ − = M1 Auxiliary equation 1 or 3λ = − A1 CF 3e et tx A B−= + F1 CF for their roots PI x a= B1 Constant PI 3 6 2a a− = − ⇒ = B1 PI correct 32 e et tx A B−= + + F1 Their CF + PI 2 3y x x= + − & M1 y in terms of , &x x 3 34 2 e 2 e 3 ( 3 e e )t t t tA B A B− −= + + + − − + M1 Differentiate x and substitute

37 5 e et ty A B−= + + A1 Constants must correspond with those in x

9 (iii) 4 2 A B= + + M1 Condition on x 17 7 5A B= + + M1 Condition on y 2, 0A B= = M1 Solve 32 2e tx −= + F1 Follow their GS 37 10e ty −= + F1 Follow their GS

B1 Sketch of x starts at 4 and decreases

B1 Asymptote x = 2

B1 Sketch of y starts at 17 and decreases

B1 Asymptote y =7

9

7

4758 Mark Scheme Jan 2007

40

1(i) 2 2 0λ λ− − = M1 Auxiliary equation 1 or 2λ = − A1 CF 2e et ty A B−= + F1 CF for their roots PI 2e ty a −= B1 2 22 e , 4 et ty a y a− −= − = M1 Differentiate twice ( )2 2 2 24 e 2 e 2 e et t t ta a a− − − −− − − = M1 Substitute 4 1a = M1 Compare and solve 1

4a = A1 2 21

4e e et t ty A B− −= + + F1 Their CF with 2 constants + their PI 9(ii) 1

40 A B= + + M1 Use initial condition 2 2e 0,e 0,et t tt − −→ ∞ ⇒ → → → ∞ so 0 0y B→ ⇒ = M1 Use asymptotic condition ( )21

4 e et ty − −= − A1 cao M1 Valid method to establish 0 is only root

20 e e e 1 0t t ty t− −= ⇔ = ⇔ = ⇔ = E1 Complete argument

B1 Curve satisfies both conditions

B1 0y ≠ for 0t > and consistent with their solution

7(iii) CF 2e et ty C D−= + F1 Correct or same as in (i) PI e ty bt −= B1 e e , 2 e et t t ty b bt y b bt− − − −= − = − + ( )2 e e e e 2 e et t t t t tb bt b bt b− − − − − −− + − − − = M1 Differentiate (product) and substitute 1

32 1b b b⇒ − − = ⇒ = − A1 cao GS 2 1

3e e et t ty C D t− −= + − F1 Their CF + their non-zero PI 0, 0 0y t C D= = ⇒ + = M1 Use condition 0 0y D→ ⇒ = M1 Use condition 1

3 e ty t −= − A1 cao 8


41

2(i)

( )d 1ln sin cos cotd sin

x x xx x

= = E1 Differentiate (chain rule)

1(ii) 1 d 2cot 2

dy x

y x= − M1 Rearrange

1 d 2cot 2 dy x xy

= −∫ ∫ M1 Integrate

ln ln sin 2y x c= − + A1 One side correct (ignore constant) A1 All correct, including constant cosec 2y A x= M1 Rearrange, dealing properly with constant A1 6(iii) d 2 cot 2

dy y x kx

+ =

( )exp 2cot 2 dI x x= ∫ M1 Attempt integrating factor ( )exp ln sin 2x= M1 Integrate sin 2x= A1 Simplified form of IF d sin 2 2 cos 2 sin 2

dy x y x k xx

+ = M1 Multiply by their IF

sin 2 sin 2 dy x k x x= ∫ M1 Integrate both sides 1

2 cos 2k x A= − + A1 cao 1

2cosec 2 cot 2y A x k x= − E1 7(iv) 1

4 , 0 0x y Aπ= = ⇒ = M1 Use condition 1

2 cot 2y k x= − A1 B1 Increasing and through ( )1

4 ,0π B1 Asymptote 0x =

4(v) B1 Both double angle formulae correct (or small

angle approximations or series expansion)

M1 Use expressions in general solution

( )21122

1 2sincos 2sin 2 2sin cos

A k xA k xy

x x x

− −−= =

A1 M1 Identify value of A

121

2

sincosk x

A k yx

= ⇒ = E1 Correct solution, fully justified which tends to zero as 0x → B1 Must be from correct solution 6


42

3(i) d

dvm mg Rt

= − B1 N2L equation (accept ma, allow sign errors)

1

ddv g k vt

= − E1 Must follow from correct N2L

1

1 d dv tg k v

=−∫ ∫ M1 Separate and integrate

1 1

1

1 ln g k v t ck

− − = + A1 LHS

11 e k tg k v A −− = M1 Rearrange (dealing properly with constant)

Alternatively M1 Attempt integrating factor

A1 ( )1 1d e ed

k t k tv gt

=

M1 Integrate Alternatively M1 Auxiliary equation A1 CF 1e k tA − M1 Constant PI ( )1/g k 0, 0t v A g= = ⇒ = M1 Use condition ( )1

11 e k tgv

k−= − E1

7(ii) M1 Integrate v 1

1 1

1d e k tgx v t t Bk k

−⎛ ⎞= = + +⎜ ⎟

⎝ ⎠∫

A1 cao (including constant)

1

10, 0t x Bk

= = ⇒ = − M1 Use condition

1

1 1 1

1 1e k tgx tk k k

−⎛ ⎞= + −⎜ ⎟

⎝ ⎠ A1 cao

4(iii) 2

2ddvmv mg mk vx

= − B1 N2L with 22mk v (accept d or

dvma mt

)

2

2

d 1d

v vxg k v

=−

E1 Must follow from correct N2L

2

2d dv v x

g k v=

−∫ ∫ M1 Integrate

22 2

2

1 ln2

g k v x ck

− − = + A1 LHS

2222 e k xg k v C −− = M1 Rearrange (dealing properly with constant)

0, 0x v C g= = ⇒ = M1 Use condition

( )22

21 e k xgv

k−= − A1 cao

7(iv) t v v B1 First line 0 0 9.8 M1 Use algorithm 0.1 0.98 8.6115

7 A1 0.98

0.2 1.84116

M1 Use algorithm

A1 1.84116 (accept 3sf or better) 5


43

(v) 32

32

3 30 when 4 1.2254

gg k v v k− = = ⇒ = = E1 Deduce or verify value (must relate to resultant force or acceleration being zero)

1


44

4(i) subtracting 5 5 0x⇒ − + = M1 Solve simultaneously 1x = A1 7y = A1 3(ii) 3x x y= − − M1 Differentiate 3 (2 5)x x y= − − − + M1 Substitute for y M1 y in terms of ,x x

( )3 2 3 10 5x x x x= − − + − − + − M1 Substitute

4 5 5x x x+ + = E1 5(iii) 2 4 5 0λ λ+ + = M1 Auxiliary equation M1 Solve to get complex roots 2 jλ = − ± A1 CF ( )2e cos sintx A t B t−= + F1 CF for their roots PI 5

5 1x = = B1 GS ( )2e cos sin 1tx A t B t−= + + F1 Their CF with 2 constants + their PI 3 10y x x= − − + M1 y in terms of ,x x M1 Differentiate their x

( ) ( )( )

2 2

2

e sin cos 2e cos sin

3e cos sin 3 10

t t

t

A t B t A t B t

A t B t

− −

−

= − − + + +

− + − +

M1 Substitute ( ) ( )( )2e cos sin 7t A B t A B t−= − − + − + A1 cao 10(iv) 0, 0 1 0t x A= = ⇒ + = M1 Use condition on x 0, 0 7 0t y A B= = ⇒ − + + = M1 Use condition on y 1, 8A B= − = ( )2e 8sin cos 1tx t t−= − + ( )2e 7cos 9sin 7ty t t−= − + + A1 Both correct 3(v) B1 Through origin B1 Positive gradient at t = 0 B1 Asymptote x = 1, or their non-zero constant

PI (accept oscillatory or non-oscillatory)

NB Oscillates about 1x = , but not apparent at this scale due to small amplitude

3


0 1(i) 2 4 29λ λ+ + = M1 Auxiliary equation M1 Solve for complex roots

2 5 jλ = − ± A1 CF ( )2e cos5 sin 5ty A t B−= + t F1 CF for their roots (if complex, must

be exp/trig form)

PI cos siny a t b t= + B1 Correct form for PI sin cos , cos siny a t b t y a t b= − + = − − t M1 Differentiate twice

( )( )

cos sin 4 sin cos

29 cos sin 3cos

a t b t a t b t

a t b t t

− − + − +

+ + = M1 Substitute

4 28 34 28b aa b

+ =− + = 0

M1 Compare coefficients (both sin and cos)

0.105a = M1 Solve for two coefficients 0.015b = A1 Both ( )2e cos5 sin 5 0.105cos 0.015sinty A t B t t−= + + + t F1 GS = PI + CF (with two arbitrary

constants)

110, 0 0 0.105t y A= = ⇒ = + M1 Use condition on y (ii)

0.105A⇒ = − F1 ( )

( )

2

2

2e cos5 sin 5

e 5 sin 5 5 cos5 0.105sin 0.015cos

t

t

y A t B t

A t B t t

−

−

= − +

+ − + − + t M1 Differentiate (product rule)

0, 0 0 2 5 0.015t y A B= = ⇒ = − + + M1 Use condition on y 0.045B⇒ = −

( )2e 0.105cos5 0.045sin 5 0.105cos 0.015sinty t t t−= − + + + t A1 cao For large t, 0.105cos 0.015siny t≈ + t M1 Ignore decaying terms M1 Calculate amplitude from solution of

this form amplitude 2 20.105 0.015 0.106≈ + ≈

A1 cao 8 (iii) (10 ) 0.105y π ≈ B1 Their a from PI, provided GS of

correct form

(10 ) 0.015y π ≈ B1 Their b from PI, provided GS of correct form

2 (iv) F1 Correct or follows previous CF

Must not use same arbitrary constants as before

( )2e cos5 sin 5ty C t D−= + t

oscillations B1 with decaying amplitude (or tends to zero) B1 Must indicate that y approaches

zero, not that for 0y ≈ 10t π>

3

46

4758 Mark Scheme June 2007 2(i) 1d 2 1

dny y x

x x x−− = + M1 Rearrange

2exp dI xx

⎛ ⎞= −⎜ ⎟⎝ ⎠∫ M1 Attempt IF

( )exp 2ln x= − M1 Integrate to get lnk x

2x−= A1 Simplified form of IF ( )2 3d

dn 3yx x x

x− −= + − M1 Multiply both sides by IF and recognise

derivative

M1 Integrate 2 2 21 12 2

nnyx x x− − −

−= − + + A A1 RHS including constant

21 12 2

nny x−= − + + Ax F1 Their integral (with constant) divided by IF

8 (ii) From solution, 1

20x y→ ⇒ → − B1 Limit consistent with their solution From DE, 0 0 2x y= ⇒ − = 1 M1 Use DE with 0x =

12y⇒ = − E1 Correctly deduced

3 (iii) 1 1 1 1

2 2 2 2, 1 ny x −= − = ⇒ − = − + + A M1 Use condition 1

2nA −⇒ = −

( )21 12 2

nny x−= − + − x F1 Consistent with their GS and given

condition

2121,n y x x= = − − +

B1 Shape for x > 0 consistent with their solution (provided not y = constant)

B1 Through ( )121, − or (0, their value from (ii))

–½ (1, –½)

4 (iv) M1 Use result from (i) or attempt to solve from

scratch ( )2 3dd

1yx x xx

− −= + − F1 Follow work in (i)

2 212 lnyx x x B− −= − + + M1 Integrate

A1 RHS (accept repeated error in first term from (i))

2 212 lny x x B= − + + x M1 Divide by IF, including constant (here or

later)

12(1)y B= − + M1 Use condition at 1x =

12(2) 4ln 2 4y B= − + + M1 Use condition at 2x =

43(1) (2) 3 4ln 2 ln 2y y B B= ⇒ = − ⇒ = − M1 Equate and solve

( )21 42 3ln ln 2y x x= − + − A1 cao

9

47

4758 Mark Scheme June 2007 3(i) ( )

12 d 1 0.1cos 25 dy y k t− = − +∫ ∫ t M1 Separate

M1 Integrate A1 LHS ( )

122 0.004sin 25y k t t= − + + c

A1 RHS (condone no constant) M1 Use condition (must have constant) 0, 1 2t y c= = ⇒ =

F1 M1 Rearrange, dealing properly with constant ( )( )21

21 0.004sin 25y k t t= − + A1 cao 8 (ii) ( ) ( )

121, 0.5 2 0.5 1 0.004sin 25 2t y k= = ⇒ = − + + M1 Substitute

0.586k⇒ ≈ E1 Calculate k (must be from correct solution) M1 Substitute ( )( )21

22 1 0.586 2 0.004sin 50 0.172t y= ⇒ = − × + ≈ A1 cao 4 (iii) solution curve on insert M1 Reasonable attempt at curve A1 From (0,1) and decreasing A1 Curve broadly in line with tangent field tank empty after 3.0 minutes F1 Answer must be consistent with their curve 4 (iv) M1 ( )(0.1) 1 0.1 0.6446x = + −

A1 –0.6446 = 0.93554 E1 Must be clearly shown M1 ( )(0.2) 0.93554 0.1 0.51985x = + −

A1 –0.51985 = 0.88356 A1 awrt 0.884 6 (v) 0.01 0.1 0.1cos 25 0y y y< ⇒ < ⇒ + <t for

some t M1 Consider size of y and sign of

0.1cos 25y t+

d 0dyt

⇒ > for some values of t E1 Complete argument

2

48

4758 Mark Scheme June 2007 4(i) 25 4 2e tx x y −= − + − M1 Differentiate ( )2 25 4 9 7 3e 2et tx x y − −= − + − + + − M1 Substitute for y M1 y in terms of ,x x

( )2 22845 36 5 e 10et tx x x x − −= − − + + − +

M1 Substitute for y 22 3e tx x x −− + = E1 5 (ii) 2 2 1λ λ− + = 0 M1 Auxiliary equation 1λ = (repeated) A1 CF ( )etx A Bt= + F1 CF for their roots PI 2e tx a −= B1 Correct form for PI 2 22 e , 4 et tx a x a− −= − = M1 Differentiate twice ( )2 2 24 e 2 2e e 3et t ta a− − −− − + = 2t− M1 Substitute and compare 1

3a = A1 GS GS = PI + CF (with two arbitrary

constants) ( )21

3 e et tx A Bt−= + + F1

8 (iii) ( )21

4 5 e ty x x −= + − M1 y in terms of ,x x M1 Differentiate x F1 x follows their x (but must use

product rule) ( ) ( )( )2 251 2

4 3 3e e e e 5 e et t t t t tB A Bt A Bt− −= − + + + + + + − 2−

( )1

4 e 6 6ty A B= + + Bt A1 cao 4 (iv) 1

3 0A+ = M1 Condition on x

( )14 6 0A B+ = M1 Condition on y

13 , 2A B= − =

( )21 13 3e 2

3 e

t t

t

x t

y t

−= + −

=

e A1 Both solutions correct

0 1,t x y= ⇒ = = 3 B1 Both values correct B1 x through origin and consistent with

their solution for large t (but not linear)

B1 y through origin and consistent with their solution for large t (but not linear)

B1 Gradient of both curves at origin consistent with their values of ,x y

7

49

4758 Mark Scheme January 2008

27

4758 Differential Equations

1(i)

2 2 1 0α α+ + = M1 Auxiliary equation

1α = − (repeated) A1 CF ( )e ty A Bt −= + F1 CF for their roots PI y a= B1 Constant PI in DE 2y⇒ = B1 PI correct ( )2 e ty A Bt −= + + F1 Their PI + CF (with two

arbitrary constants)

0, 0 0 2 2t y A A= = ⇒ = + ⇒ = − M1 Condition on y ( )e ty B A Bt −= − − M1 Differentiate (product rule) 0, 0 0 2t y B A B= = ⇒ = − ⇒ = − M1 Condition on y ( )2 2 2 e ty t −= − + A1 10(ii) Both terms in CF hence will give zero if substituted in

LHS E1

PI 2 e ty bt −= B1 ( ) ( )2 22 e , 2 4 et ty bt bt y b bt bt− −= − = − + in DE ( )( )2 2 22 4 2 2 e et tb bt bt bt bt bt − −⇒ − + + − + = M1 Differentiate twice and

substitute

12b⇒ = A1 PI correct

( )212 e ty C Dt t −= + + F1 Their PI + CF (with two

arbitrary constants)

0, 0 0t y C= = ⇒ = M1 Condition on y ( )21

2 e ty D t C Dt t −= + − − − 0, 0 0 0t y D C D= = ⇒ = − ⇒ = M1 Condition on y 21

2 e ty t −= A1 8(iii) 21

20 0 and e 0 0tt t y−> ⇒ > > ⇒ > E1 ( )21

2 e ty t t −= − so 2120 0 0 or 2y t t t= ⇔ − = ⇔ = M1 Solve 0=y

Maximum at 22, 2et y −= = A1 Maximum value of y B1 Starts at origin B1 Maximum at their value of y B1 0y >

6


28

2(i)

d 3 3d 1 1v v gt t t

+ = −+ +

M1 Rearrange

M1 Attempt integrating factor A1 Correct

( ) ( ) ( )33ln 131exp d e 1t

tI t t++= = = +∫

A1 Simplified

( ) ( ) ( ) ( )3 2 3 2d1 3 1 1 3 1dvt t v g t tt

+ + + = + − + F1 Multiply DE by their I

( )( ) ( ) ( )3 3 2d 1 1 3 1

dt v g t t

t+ = + − +

( ) ( ) ( )( )3 3 21 1 3 1 dt v g t t x+ = + − +∫ M1 Integrate

( ) ( )4 314 1 1g t t A= + − + + A1 RHS

( ) ( ) 314 1 1 1v g t A t −= + − + + F1 Divide by their I (must also divide constant)

140, 0 0 1t v g A= = ⇒ = − + M1 Use condition

( ) ( )( ) 31 14 41 1 1 1v g t g t −= + − + − + E1 Convincingly shown

10(ii)

( ) ( )d1 5 1dvt v t gt

+ + = + M1 Rearrange

d 5d 1v v gt t

+ =+

M1 Attempt integrating factor ( ) ( ) ( )55ln 15

1exp d e 1ttI t t+

+= = = +∫ A1 Simplified

( ) ( ) ( )5 4 5d1 5 1 1

dvt t v g tt

+ + + = + F1 Multiply DE by their I

( )( ) ( )5 5d 1 1

dt v g t

t+ = +

( ) ( )5 51 1 dt v g t x+ = +∫ M1 Integrate ( )61

6 1g t B= + + A1 RHS ( ) ( ) 51

6 1 1v g t B t −= + + + F1 Divide by their I (must also divide constant) 1

60, 0 0t v g B= = ⇒ = + M1 Use condition ( )( )51

6 1 1v g t t −= + − + F1 Follow a non-trivial GS

9(iii)

First model: ( )( ) 41 14 4

d 3 1 1dv g g tt

−= − − + M1 Find acceleration

As ( ) 4, 1 0t t −→ ∞ + → B1 Identify term(s) → 0 in their solution for either model

Hence acceleration tends to 14 g A1

Second model ( )( )61

6d 1 5 1dv g tt

−= + + M1 Find acceleration

Hence acceleration tends to 16 g A1

5


29

3(i) 0.5e tP A= M1 Any valid method 0, 2000 2000t P A= = ⇒ = M1 Use condition 0.52000e tP = A1 3(ii) CF 0.5e tP A= F1 Correct or follows (i) PI cos 2 sin 2P a t b t= + B1 2 sin 2 2 cos 2P a t b t= − + M1 Differentiate ( )2 sin 2 2 cos 2 0.5 cos 2 sin 2 170sin 2a t b t a t b t t− + = + +

M1 Substitute

2 0.5 170a b− = + M1 Compare coefficients 2 0.5b a= M1 Solve solving 80, 20a b⇒ = − = − A1 GS 0.5e 80cos 2 20sin 2tP A t t= − − F1 Their PI + CF (with one arbitrary

constant)

8(iii) 20802000,0 =⇒== APt M1 Use condition 0.52080e 80cos 2 20sin 2tP t t= − − F1 Follow a non-trivial GS 2(iv) t P P M1 Use of algorithm 0 2000 1000 A1 2100 0.1 2100 1082.58 A1 1082.5… 0.2 2208 A1 2208 4(v) (A) Limiting value 0P⇒ = M1 Set 0P = 1

2

1 012000

PP⎛ ⎞

⇒ − =⎜ ⎟⎝ ⎠

M1 Solve

(as limit non-zero) limiting value = 12000 A1 3 (B) Growth rate max when

( )12

f 112000

PP P⎛ ⎞

= −⎜ ⎟⎝ ⎠

max M1 Recognise expression to maximise

( )

1 12 21f 1 1

12000 2 12000 12000P PP P

−⎛ ⎞ ⎛ ⎞

′ = − − −⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ M1 Reasonable attempt at derivative

( ) 1f 0 1 0

12000 2 12000PP P

⎛ ⎞′ = ⇔ − − =⎜ ⎟ ×⎝ ⎠

M1 Set derivative to zero

8000P⇔ = A1 4


30

4(i) 3x x y= − + M1 Differentiate first equation 3 ( 5 15)x x y= − + − + + M1 Substitute for y 3 9y x x= − + M1 y in terms of ,x x ( )3 5 3 9 15x x x x x= − − + − + + M1 Substitute for y

2 2 6x x x+ + = E1 5(ii) 2 2 2 0λ λ+ + = M1 Auxiliary equation 1 jλ = − ± A1 CF ( )e cos sintx A t B t−= + M1 CF for complex roots

F1 CF for their roots PI x a= B1 Constant PI 2 6 3a a= ⇒ = B1 PI correct GS ( )3 e cos sintx A t B t−= + + F1 Their CF + PI (with two arbitrary

constants)

7(iii) 3 9y x x= − + M1 y in terms of ,x x ( )

( ) ( )9 3e cos sin 9

e cos sin e sin cos

t

t t

A t B t

A t B t A t B t

−

− −

= + + −

− + + − + M1 Differentiate x and substitute

( ) ( )( )e 2 cos 2 sinty A B t B A t−= + + − A1 Constants must correspond with those in x

3(iv) 0 3 3A A= + ⇒ = − M1 Condition on x 0 2 6A B B= + ⇒ = M1 Condition on y ( )3 3e 2sin costx t t−= + − F1 Follow their GS

15e sinty t−= F1 Follow their GS

4(v) B1 Sketch of x starts at origin B1 Asymptote x = 3

B1 Sketch of y starts at origin B1 Decaying oscillations (may

decay rapidly)

B1 Asymptote y = 0

5


38

4758 Differential Equations 1 (i)

( )2 2 8 0.25 2x g x g kv= − + − M1 N2L equation with all forces using given expressions for tension and resistance

Weight positive as down, tension negative as up. B1

Resistance negative as opposes motion. B1 4 0x kx x⇒ + + = E1 Must follow correct N2L equation 4 (ii) cos 2 sin 2x A t B t= + B1 0, 0.1 0.1t x A= = ⇒ = M1 Find the coefficient of cos 2 sin 2 2 cos 2x A t B t= − + so 0, 0 0t x B= = ⇒ = M1 Find the coefficient of sin 0.1cos 2x t= A1 cao 4 (iii) 2 2 4 0α α+ + = M1 Auxiliary equation 1 3 jα = − ± A1 M1 CF for complex roots ( )e cos 3 sin 3tx C t D t−= + F1 CF for their roots 0, 0.1 0.1t x C= = ⇒ = M1 Condition on x ( )

( )e cos 3 sin 3

e 3 sin 3 3 cos 3

t

t

x C t D t

C t D t

−

−

= − +

+ − + M1 Differentiate (product rule)

0 3C D= − + M1 Condition on x 0.1

3D =

( )13

0.1e cos 3 sin 3tx t t−= + A1 cao

B1 Curve through (0,0.1) with zero gradient B1 Oscillating B1 Asymptote x = 0

11 (iv) 2 4 1 4 0k − ⋅ ⋅ > M1 Use of discriminant A1 Correct inequality (As k is positive) 4k > A1 Accept 4k < − in addition (but not k > –4) B1 Curve through (0,0.1) B1 Decays without oscillating (at most one

intercept with positive t axis)

5


39

2 (i)

2e tx A −= M1 Any valid method

0, 8 8t x A= = ⇒ = M1 Condition on x 28e tx −= A1 3 (ii) 216e ty y −+ = M1 Substitute for x 1 0 1α α+ = ⇒ = − M1 Auxiliary equation CF e ty B −= A1 PI 2e ty a −= B1 2 2 22 e e 16et t ta a− − −− + = M1 Differentiate and substitute 16a = − A1 cao GS 216e et ty B− −= − + F1 Their PI + CF (with one arbitrary

constant)

0, 0 16t y B= = ⇒ = M1 Condition on y ( )216 e et ty − −= − F1 Follow a non-trivial GS Alternative mark scheme for first 7 marks: M1 Substitute for x I = e t M1 Attempt integrating factor A1 IF correct d(y e t )/dt = 16e –t B1 M1 Integrate y e t = –16e –t + B A1 cao y = –16e –2t + Be –t F1 Divide by their I (must divide constant) 9 (iii) ( )16e 1 et ty − −= − M1 Or equivalent (NB e –t > e –2t needs

justifying)

16e 0t− > and 0 e 1tt −> ⇒ < hence y > 0 E1 Complete argument B1 Starts at origin B1 General shape consistent with their

solution and y > 0

B1 Tends to zero 5 (iv) ( ) ( ) ( ) ( )d 2 2 0

dx y z x x y y

t+ + = − + − + = M1 Consider sum of DE’s

x y z c⇒ + + = E1 Hence initial conditions 8x y z⇒ + + = E1 8z x y= − − M1 Substitute for x and y and find z ( ) ( )228 1 2e e 8 1 et t tz − − −= − + = − E1 Convincingly shown (x, y must be correct)

5 (v) ( )2

0.99 8 8 1 e t−× = − B1 Correct equation (any form)

0.690638 or 5.29581t = − 99% is Z after 5.30 hours B1 Accept value in [5.29, 5.3] 2


40

3 (i) 1ky y

t+ = M1 Divide by t (condone LHS only)

( ) ( )exp d exp ln kktI t k t t= = =∫ M1 Attempt integrating factor

A1 Integrating factor 1k k kt y kt y t−+ = F1 Multiply DE by their I ( )d

dk kyt t

t= M1 LHS

dk kyt t t= ∫ M1 Integrate At k

k += ++

11

1 A1 cao (including constant) 1

1k

ky t At−+= + F1 Divide by their I (must divide constant)

1 11 11, 0 0 k kt y A A+ += = ⇒ = + ⇒ = − M1 Use condition

( )11

kky t t−

+= − F1 Follow a non-trivial GS

10 (ii) ( )21

3y t t−= − B1 Shape consistent with their solution for t > 1 B1 Passes through (1, 0) B1 Behaviour for large t

3 (iii) 1 1 dyt t t− −= ∫ M1 Follow their (i) ln t B= + A1 cao ( )lny t t B= + F1 Divide by their I (must divide constant) 1, 0 0 lnt y B y t t= = ⇒ = ⇒ = A1 cao 4 (iv) 1d 1 sin

dy t yt

−= + M1 Rearrange DE (may be implied)

t y dy/dt 1 0 1 M1 Use algorithm 1.1 0.1 1.0908 A1 y(1.1) 1.2 0.2091 A1 y(1.2) 4 (v) 0.2138 as smaller step size B1 Must give reason Decreasing step length has increased

estimate. Assuming this estimate is more accurate, decreasing step length further will increase estimate further, so true value likely to be greater.

M1 Identify effect of decreasing step length

Hence underestimates. A1 Convincing argument Alternative mark scheme for last 2 marks: dy/dt seems to be increasing, hence Euler’s

method M1 Identify derivative increasing

will underestimate true value + sketch (or explanation). A1 Convincing argument

3

1


41

4 (i)

4 6 9cosx x y t= − − M1 Differentiate first equation

4 6(3 5 7sin ) 9cosx x y t t= − − − − M1 Substitute for y ( )1

6 4 9siny x x t= − − M1 y in terms of ,x x

4 18 5(4 9sin ) 42sin 9cosx x x x x t t t= − + − − + − M1 Substitute for y 2 3sin 9cosx x x t t+ − = − − E1 LHS E1 RHS 6 (ii) 2 2 0α α+ − = M1 Auxiliary equation 1 or 2α = − A1 CF 2e et tx A B −= + F1 CF for their roots PI cos sinx a t b t= + B1 PI of this form ( ) ( ) ( )2 3 9ac bs as bc ac bs s c− − + − + − + = − − M1 Differentiate twice and substitute

2 9a b a− + − = − M1 Compare coefficients (2 equations)

2 3b a b− − − = − M1 Solve (2 equations) 3, 0a b⇒ = = A1 23cos e et tx t A B −= + + F1 Their PI + CF (with two arbitrary

constants)

9 (iii) ( )1

6 4 9siny x x t= − − M1 y in terms of ,x x

( )2 216 12cos 4 e 4 e 3sin e 2 e 9sint t t tt A B t A B t− −= + + + − + − M1 Differentiate x and substitute

2122cos sin e et ty t t A B −= − + + A1 Constants must correspond with

those in x

3 (iv) x bounded 0A⇒ = M1 Identify coefficient of exponentially

growing term must be zero

y⇒ bounded E1 Complete argument 2 (v) 0, 0 0 2 2t y B B= = ⇒ = + ⇒ = − M1 Condition on y 23cos 2e tx t −= − , 22cos sin 2e ty t t −= − − F1 Follow their (non-trivial) general

solutions

3cosx t= A1 cao 2cos siny t t= − A1 cao 4


29


1(i) 022 23 =−−+ ααα B1 02)2()2(2)2( 23 =−−−−+− E1 Or factorise 0)1)(2( 2 =−+ αα M1 Solve 1 ,2 ±−=α A1 xxx CeBeAey ++= −−2 M1 Attempt CF F1 CF for their three roots 6

(ii) PI 12

2−=

−=y M1 Constant PI

A1 Correct PI GS xxx CeBeAey +++−= −−21 F1 GS = PI + CF 3 (iii) ∞→∞→ xex as M1 Consider ∞→xas 0limit finite so =⇒ C F1 Must be shown, not just stated BAx ++=⇒== -100y ,0 M1 Use condition BAyx 2

141100,2ln ++−=⇒== M1 Use condition

Solving gives A = -2, B = 3 M1 132 2 −+−= −− xx eey E1 Convincingly shown 6 (iv) )1)(12( −−−= −− xx eey 1or 0 2

1=⇔= − xey M1 Solve 0or 2ln=⇔ x E1 Convincingly show no other roots

)34(34 2 −=−= −−−− xxxx eeeedxdy

0 as 0 43 ≠=⇔= −− xx ee

dxdy M1 Solve

34ln=⇔ x E1 Show only one root

Stationary point at ),(ln 81

34 A1

5 (v) B1 Through (0, 0) B1 Through (ln 2, 0)

B1 Stationary point at their answer to (iv)

B1 ∞→−→ xy as 1

4

y (ln(4/3),1/8) ln2 x —1


30

2(i) xxxydxdy costan =+ M1 Rearrange

∫= xdxI tanexp M1 Attempt IF xseclnexp= A1 Correct IF xsec= A1 Simplified

xxydxd

=)sec( M1 Multiply and recognise derivative

Axxy += 221sec M1 Integrate

A1 RHS

xAxy cos)( 221 += F1 Divide by their IF (must divide

constant)

11,0 =⇒== Ayx M1 Use condition xxy cos)1( 2

21 += F1 Follow their non-trivial GS

10(ii) B1 Shape correct for ππ 2

121 <<− x

B1 Through (0,1)

2

(iii) x

xyxxxycos

sinsincos −=′ M1 Rearrange

0)0( =′y B1 1)1.0( =y B1 090351.0)1.0( −=′y B1 990965.0090351.01.01)2.0( =−×+=y M1 Use of algorithm for second step A1 3sf or better 6

(iv) xI sec= M1 Same IF as in (i) or attempt from scratch

xxxydxd tan)sec( = A1

[ ] ∫===

2.0

0

2.00 tansec xdxxxy x

x M1 Integrate

A1 Accept no limits 002688.00sec1)2.0sec()2.0( ≈×−y M1 Substitute limits (both sides) 982701.0)2.0( ≈⇒ y A1 Awrt 0.983 6

y 1

—π/2 π/2 x


31

3(i) 2416060 vg

dxdvv −= M1 N2L

A1 Correct N2L equation

2401

240 2 =− dx

dvvg

v E1 Convincingly shown

∫∫ =−

dxdvvg

v2401

240 2 M1 Integrate

cxvg +=−−2401240ln

21 2 A1 2240ln vg − seen

A1 RHS

1202240x

Aevg =− M1 Rearrange, dealing properly with constant

gAvx 2400,0 =⇒== M1 Use condition

)1(240 1202x

egv−

−= A1 Cao 9

(ii) 71.13)1(24010 12010

≈−=⇒=−

egvx E1 Convincingly shown 1

(iii) gvgdtdv 90606060 −−= M1 N2L

gvdtdvvg

dtdv

21or

21

−=+−−= A1 Correct DE

Solving DE (three alternative methods):

∫ ∫ −=+

dtgv

dv

21

M1 Separate

ktgv +−=+ 21ln M1 Integrate

A1 LHS

tAegv −=+ 21 M1 Rearrange, dealing properly

with constant

or 101 −=⇒=+ αα M1 Solve auxiliary equation CF tAe− M1 CF for their root

PI g21− M1 Attempt to find constant

PI

gAev t21−= − A1 All correct

or

teI = M1 Attempt integrating factor

tt gevedtd

21)( −= M1 Multiply

Ageve tt +−=21 M1 Integrate

gAev t

21

−= − A1 All correct

61.1871.130,71.132

1 =⇒−=⇒== AgAtv M1 Use condition 9.461.18 −= −tev E1 Complete argument 8


32

(iv) At greatest depth, v = 0 M1 Set velocity to zero and attempt to solve

3345.161.189.4

=⇒=⇒ − te t A1

Depth ∫ −= −3345.1

0)9.461.18( dte t M1 Integrate

[ ] 3345.109.461.18 te t −−= − A1 Ignore limits

M1 Use limits (or evaluate constant and substitute for t)

= 7.17 m A1 All correct 6

4(i) 41

022073

==

⇔⎭⎬⎫

=+−=+−−

yx

yxyx

B1 B1

2(ii) yxx −−= 3 M1 Differentiate )22(3 +−−−= yxx M1 Substitute for y xxy −+−= 73 M1 y in terms of xx, 27323 −−+−−−= xxxxx M1 Substitute for y 554 =++⇒ xxx E1 Complete argument 5(iii) 0542 =++ αα M1 Auxiliary equation i±−=⇒ 2α A1 M1 CF for complex roots CF )sincos(2 tBtAe t +− F1 CF for their roots

PI 155

==x B1

GS )sincos(1 2 tBtAex t ++= − F1 GS = PI + CF with two arbitrary constants

6(iv) xxy −+−= 73 M1 y in terms of xx,

)cossin()sincos(2 22 tBtAetBtAex tt +−++−= −− M1 Differentiate their x (product rule)

)cos)(sin)((4 2 tBAtBAey t +−−+= − A1 Constants must correspond 3(v) 1 + A = 4 M1 Use condition on x 4 – A – B = 0 M1 Use condition on y A = 3, B = 1 )sincos3(1 2 ttex t ++= − )cos4sin2(4 2 ttey t −+= − A1 Both solutions 3(vi) B1 (0, 4) B1 1→ B1 (0, 0) B1 4→

As the solutions approach the asymptotes, the gradients approach zero. B1 Must refer to gradients

5

4 1


44


1(i) 2 25 0α + = M1 Auxiliary equation 5jα = ± A1 CF cos5 sin5y A t B t= + F1 CF for their roots PI cos5 sin5y at t bt t= + B1

cos5 5 sin5 sin5 5 cos5y a t at t b t bt t= − + + 10 sin5 25 cos5 10 cos5 25 sin5y a t at t b t bt t= − − + − M1 Differentiate twice In DE 10 cos5 10 sin5 20cos5b t a t t⇒ − = M1 Substitute and compare coefficients 2, 0b a⇒ = = A1 PI 2 sin5y t t= GS 2 sin5 cos5 sin5y t t A t B t= + + F1 8 (ii) 0, 1 1t y A= = ⇒ = B1 From correct GS 2sin5 10 cos5 5 sin5 5 cos5y t t t A t B t= + − + M1 Differentiate

0, 0 0t y B= = ⇒ = M1 Use condition on y 2 sin5 cos5y t t t= + A1 4 (iii) Curve through (0,1) B1 Curve with zero gradient at (0,1) B1 Oscillations B1 Oscillations with increasing amplitude B1 4 (iv) 2sin5 , 10cos5 , 50sin5y t y t y t= = = −

2 25 50sin5 20cos5 50sin5y y y t t t+ + = − + + M1 Substitute into DE 20cos5t= E1

2 2 25 0α α+ + = M1 Auxiliary equation

1 24 jα = − ± A1

CF ( cos 24 sin 24 )te C t D t− + F1 CF for their complex roots

GS 2sin5y t= + ( cos 24 sin 24 )te C t D t− + F1 Their PI + their CF with two arbitrary

constants 6�

(v) Oscillations of amplitude 2 B1 or bounded oscillations; or both oscillate Compared to unbounded oscillations in first model B1 � or equivalent;

2 or one bounded, one unbounded


45

2(i) 2d 3 sindy xyx x x

+ = M1 Rearrange

=I exp3x∫ dx M1 Attempting integrating factor

= exp(3lnx) A1

= 3x A1 Correct and simplified

( )3dd

x yx

= sinx x M1 Multiply and recognise

derivative

3 sin cos cosx y x xdx x x xdx= = − +∫ ∫ M1 Integrate

A1 cos sinx x A= − + + A1 All correct

3( cos sin ) /y x x x A x= − + + F1 Must include constant 9�

(ii) 2 3 31 11 /2 6

y x x x x A x⎛ ⎞⎛ ⎞≈ − − + − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

M1 Substitute given approximations

F1

= 313

Ax

+ M1 Use finite limit to deduce A�

� A = 0 A1

3(sin cos ) /y x x x x= = B1 Correct particular solution

0

1lim3x

y→

= B1 Correct limit

6�� (iii) 0 sin cos 0y x x x= ⇒ − = M1 Equate to zero and attempt to get tanx tanx x⇒ = E1 Convincingly shown 2�

(iv) 3d 3 1 1 , multiply by d 6y y x I xx x x

+ = − = M1 Rearrange and multiply by IF

B1 Same IF as in (i) or correct IF

3 2 4d 1( )d 6

x y x xx

= − A1 Recognise derivative and RHS

correct

3 3 51 13 30

x y x x B= − + M1 Integrate

23

1 13 30

By xx

= − + A1 cao

Finite limit ⇒ B = 0 M1 Use condition to find constant


46

0

1lim3x

y→

= E1 Show correct limit (or same limit

as (ii) 7� 3(a)(i) 2 4 0 2α α+ = ⇒ = − M1 Find root of auxiliary equation

CF 2e tA − A1 PI cos2 sin2I a t b t= + B1 2 sin2 2 cos2I a t b t= − + M1 Differentiate 4 sin2 4 cos2 4 cos2 4 sin2 3cos2a t b t a t b t t− + + + = M1 Substitute

34 4 0,4 4 38

a b b a a b− + = + = ⇒ = = M1 Compare coefficients and solve

PI 3(cos2 sin2 )8

I t t= + A1

GS 2 3e (cos2 sin2 )8

tI A t t−= + + F1 8 Their PI + their CF with one

arbitrary constant

(ii) 3 30, 0 08 8

t I A A= = ⇒ = + ⇒ = − M1 Use condition

23(cos2 sin2 e )8

tI t t −= + − A1 2 cao

(iii) For large t, 3(cos2 sin2 )8

I t t≈ + M1 Consider behaviour for large t�

� � � � � � � � � (may be implied)

Amplitude = 2 23 31 1 28 8

+ = A1

Curve with oscillations with constant amplitude B1 Their amplitude clearly indicated B1 4

(b)(i) (A) t = 0, y = 0 0d 2 2(0) edyt

⇒ = − + M1 Substitute into DE

Gradient =3 A1

(B) At stationary point, d 90,d 8y yt

= = M1 Substitute into DE

9 10 2 2( ) e e8 4

t t− −⇒ = − + ⇒ = M1 Solve for t�

t⇒ = ln4 A1

(C) d 0,e 0d

tyt

−→ → M1 Substitute into DE

Giving 0 2 2 0, so 1y y= − + → A1 7�

(ii) Curve through origin with positive gradient B1 With maximum at (ln4, 9/8) B1 Follow their ln4 With 1 as y x→ → ∞ B1 3 Follow their (C)

4(i) 37 6 6e tx x y −= + − M1 Differentiate

= 37 6( 12 10 5sin ) 6e tx x y t −+ − − + − M1 Substitute for y

31( 7 2e )6

ty x x −= − − M1 y in terms of , ,x x t


47

3 37 72 10( 7 2e ) 30sin 6et tx x x x x t− −= − − − − + − M1 Substitute for y

33 2 14e 30sintx x x t−+ + = + E1 Complete argument 5�

(ii) 3e 9cos 3sintx a t t−= − +

3

3

3 e 9sin 3cos

9 e 9cos 3sin

t

t

x a t t

x a t t

−

−

= − + +

= + − M1 Differentiate twice

In DE gives M1 Substitute

39 e 9cos 3sinta t t− + −

+3( 33 e 9sin 3costa t t−− + + )

+2( 3e 9cos 3sinta t t− − + )

32 e 30sinta t−= + So PI with 2a =14 E1 Correct form shown 7a⇒ = A1

AE 2 3 2 0α α+ + = M1 Auxiliary equation 1, 2α = − − A1

CF 2e et tA B− −+ F1 CF for their roots

GS 2 3e e 7e 9 3sint t tx A B cost t− − −= + + − + F1 Their PI + their CF with two arbitrary 8 constants

(iii) 31 ( 7 2e )6

tx x x −= − − M1 in terms of , ,y x x t

2 3e 2 e 21e 9sin 3cost t tx A B t t− − −= − − − + + M1 Differentiate GS for x F1 Follow their GS

2 34 3e e 12e 11cos 2sin3 2

t t ty A B t t− − −= − − − + − A1 4 cao

(iv) 3sin 9cosx t t≈ − B1 Follow their x 11cos 2siny t t≈ − B1 Follow their y 11cos 2sin 3sin 9cosx y t t t t= ⇒ − ≈ − M1 Equate 20cos 5sin tan 4t t t⇒ ≈ ⇒ ≈ A1 4 Complete argument

(v) Amplitude of 2 23 9 3 10x ≈ + = M1 Attempt both amplitudes

Amplitude of 2 211 2 5 5y ≈ + = A1 One correct

Ratio is 5 26

A1 3 cao (accept reciprocal)

2

2 (b) (i)

x

y

–1

–1 0

1

2

1

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© OCR 2010 4758/01 Ins Jan10


29

4758 Differential Equations 1(i) 2 6 9 0α + α + = M1 Auxiliary equation 3α = − (repeated) A1

( )-3e ty A Bt= + F1 CF for their roots

PI sin cosy a t b t= + B1 cos siny a t b t= − sin cosy a t b t= − −

( )sin cos 6 cos sina t b t a t b t− − + −

( )9 sin cos 0.5 sina t b t t+ + = M1 Differentiate twice and substitute

8 6 0.5a b− = M1 Compare coefficients 8 6 0b a+ = M1 Solve Solving gives 0.04, 0.03a b= = − A1

GS ( )3e 0.04 sin 0.03 costy A Bt t t−= + + − F1 PI + CF with two arbitrary constants

9(ii) 0, 0 0.03t y A= = = M1 Use condition

( )3e 3 3 0.04 cos 0.03 sinty B A Bt t t−= − − + + M1 Differentiate

F1 Follows their GS 0, 0 0 3 0.04t y B A= = = − + M1 Use condition

( )( )30.01 e 3 5 4 sin 3 costy t t t−= + + − A1 Cao

5(iii) For large t , the particle oscillates B1 Oscillates With amplitude constant ( 0.05)≈ B1 Amplitude approximately constant 2(iv) 6020 et ππ −= very small M1 0.03y ≈ − A1 0.04y ≈ A1 3

(v) ( )3e ty C Dt−= + M1 CF of correct type or same type as in (i) A1 Must use new arbitrary constants B1√ 0.03y ≈ − at 20t π= B1 Gradient at 20π consistent with (iv) B1 Shape consistent

5

20π -0.03


30

2(a)(i) exp tan dI x x= − M1 Attempt IF

( )exp ln sec x= − A1 Correct IF

( ) 1sec cosx x−= = A1 Simplified

dcos sin sindyx y x xx

− = M1 Multiply by IF

( )d cos sind

y x xx

= M1 Recognise derivative

M1 Integrate cos cosy x x A= − + A1 RHS (including constant) ( sec 1y A x= − ) A1 LHS 8

( )d 1+ tan

d=y y x

x

ln(1+y)=lnsecx + A

M1A1 M1A1 M1A1 M1A1

Rearrange equation Separate variables RHS LHS

8

(ii) 0, 0 0 1x y A= = = − M1 Use condition sec 1y x= − A1 B1 Shape and through origin B1 Behaviour at ±½π

4 (b)(i) M1 Attempt one curve A1 Reasonable attempt at one curve M1 Attempt second curve A1 Reasonable attempt at both curves 4

(ii) ( )2' 1 tany y x= + M1 Rearrange

0, 1 ' 0x y y= = =

( )0.1 1 0.1 0 1y = + × = M1 Use of algorithm

0.1, 1 ' 0.201x y y= = = A1

( )0.2 1 0.1 0.201y = + × M1 Use of algorithm for second step

1.0201= E1 5

(iii) tan2π undefined so cannot go past

2π M1

So approximation cannot continue to1.62π> A1

2 (iv) Reduce step length B1 1

– π/2 π/2 x

y

1


31

3(i) e ktx A −= M1Any valid method (or no method shown)

A1 1 10,t x v A v= = = M1 Use condition

1ektx v −= A1

1e dktx v t−= M1 Integrate

1 e ktv Bk

−= − + A1

10, 0 vt x Bk

= = = M1 Use condition

( )1 1 e ktvxk

−= − E1

8

(ii) d d

/y k t

y g k= −

+

M1 Separate and integrate

In gy kt Ck

+ = − + A1 LHS

A1 RHS

e ktgy Dk

−+ = M1Rearrange, dealing properly with constant

2 20, gt y v D vk

= = = + M1 Use condition

2ktg gy v e

k k− = + −

A1

2 e dktg gy v tk k

− = + −

M1 Integrate

21 e ktg gv t Ek k k

− = − + − +

A1

210, 0 0 gt y v Ek k = = = − + +

M1 Use condition

( )( )22

1 1 e kt gy kv g tk k

−= + − − E1

10

(iii) 1

1 e−− =kt kxv

M1

1

1 ln 1

= − −

kxtk v

A1

M1 Substitute

2

21 1

ln 1 += + −

kv g g kxy xkv k v

E1 Convincingly shown

4 (iv) 8 4.686x y= = M1 Hence will not clear wall A1 2


32

4(i) 4 3 23y x x= − + − M1 y or 4y in terms of ,x x 4 3y x x= − − M1 Differentiate

( ) ( )1 13 2 3 23 74 4

x x x x x− − = + − + − − M1 Substitute for y

3 8 3 23 28x x x x x− − = − + − − M1 Substitute for y 2 5 5x x x + + = E1 5 (ii) 2 2 5 0α α+ + = M1 Auxiliary equation 1 2iα = − ± A1 M1 CF for complex roots

CF ( )e cos 2 sin 2− +t A t B t F1 CF for their roots

PI 5 15

x = = B1 Constant PI

B1 Correct PI

GS ( )1 e cos 2 sin 2tx A t B t−= + + F1 PI + CF with two arbitrary constants

7

(iii) ( )1 3 234

y x x= − + − M1

( )1 3 3e cos2 sin 2 234

t A t B t−= − − + +

( )e cos2 sin 2t A t B t−+ +

( )e 2 sin 2 2 cos2t A t B t−− − +

M1 F1

Differentiate and substitute Expression for x follows their GS

( ) ( )( )15 e cos 2 sin 22

ty A B t B A t−= − + + − A1

4 (iv) 0, 8 1 8 7t x A A= = + = = M1 Use condition

( )10, 0 5 0 32

t y A B B= = − + = = M1 Use condition

( )1 e 7cos2 3sin 2tx t t−= + + A1

( )5 e 5cos2 2sin 2ty t t−= − − A1

4 (v) For large t, e-t tends to 0 M1 y→5 B1 x→1 B1 y x > E1 Complete argument 4

Oxford Cambridge and RSA Examinations

GCE

Mathematics (MEI) Advanced GCE 4758

Differential Equations

Mark Scheme for June 2010


1

1(i) 2α 4α 8 0 M1 Auxiliary equation α 2 2j A1 CF 2e ( cos2 sin 2 ) x A x B x M1 CF for complex roots F1 CF for their roots PI 2 y ax bx c B1 2 , 2 y ax b y a

2 22 4(2 ) 8( ) 32a a x b a x bx c x M1 Differentiate twice and substitute

8 32a M1 Compare coefficients 8 8 0 a b 2 4 8 0 a b c M1 Solve 4, 4, 1 a b c A1

GS 2 24 4 1 e ( cos2 sin 2 ) xy x x A x B x F1 PI + CF with two arbitrary constants 10

(ii) 0, 0 1 x y A M1 Use condition

28 4 e ( 2 sin 2 2 cos2xy x A x B x

2 cos2 2 sin 2 )A x B x M1 Differentiate (product rule)

0, 0 0 4 (2 2 ) 1 x y B A B M1 Use condition 2 24 4 1 e (sin 2 cos2 ) xy x x x x A1 Cao 4 (iii) x y oscillates B1 Oscillates With (exponentially) growing amplitude B1 Amplitude growing 2 (iv) 2(2 1)y x or 24 4 1 x x B1 1 (v) B1 Minimum point at origin

B1 Oscillates for x<0 with growing amplitude

B1 Approximately parabolic for x>0

3

(vi) At stationary point d 0d

yx

M1 Set first derivative (only) to zero in DE

So 2

22

d 32 8d

y x y

x A1

2

2

d0 0d

yy

x M1 Deduce sign of second

derivative

minimum E1 Complete argument 4


2

2(a)(i) exp 2d IF t M1 Attempt IF 2e t A1

2 2de 2e 1d

t ty yt

M1* Multiply by IF

2d (e ) 1d

t yx

A1

2e t y t A *M1A1 Integrate both sides 6 2[ e ( )]ty t A

Alternative method: CF 2e ty E PI 2e ty Ft In DE: 2 2 2e ( 2 ) 2 e et t tF Ft Ft F = 1

2e ( ) ty t E

B1 B1

M1

M1A1 F1

(ii) 2d 2 e ( )d

tz z t At

2e tI B1 Correct or follows (i)

2d (e )d

t z t At

M1 Multiply by IF and integrate

2 212e t z t At B A1

2 212e ( )tz t At B

0, 1 1t z B M1 Use condition 2 21

22e ( ) e ( )t tz t At B t A M1 Differentiate (product rule) 0, 0 0 2 2t z B A A M1 Use condition 2 21

2e ( 2 1)tz t t A1 7

Alternative method: PI 2 2( )e tx Pt Qt P = A and Q = 0.5

2 212e ( )tz t At B

Then as above

B1 M1A1

Correct form of PI Complete method

(b)(i) 2 0 2α α CF 2e tx C B1 CF correct PI sin cosx a t b t B1 Correct form of PI cos sinx a t b t In DE: cos sin 2 sin 2 cos sina t b t a t b t t M1 Differentiate and substitute 2 0, 2 1a b b a M1 Compare and solve 2 1

5 5,a b A1

GS 215 (2sin cos ) e tx t t C F1 Their PI + CF 6

(ii) 0, 0 0x t x (from DE) M1 Or differentiate 1

50 C M1 Use condition

215 (2sin cos e tx t t A1 3

(iii) For large t, 1 15 5(2sin cos ) 5 sinx t t (t-) M1 Complete method

So x varies between 15 5 and 1

5 5 A1 Accept 15 5x 2


3

3(i) 12 d dy y k t M1 Separate and integrate

122y kt B A1 LHS

A1 RHS 0, 1 2t y B M1 Use condition 2, 0.81 1.8 2 2t y k M1 Use condition 0.1k A1 1

2 1 0.05y t 2(1 0.05 )y t A1 Valid for 1 0.05 0t , i.e. 20t B1√ √ on arithmetical error in k B1 Shape B1 Intercepts 10

(ii) 32 d 0.4dπy y t M1 Separate and integrate

522

5 0.4πy t C A1 LHS A1 RHS 2

50, 1t y C π M1 Use condition 0.81 1.287y t A1 5

(iii) 2

0.4(2 )

yy

π y y

M1 Rearrange (implied by

correct values)

t y y hy M1 Use algorithm 0 1 –0.12732 –0.01273 A1 (0.1)y (awrt 0.987) 0.1 0.987268 –0.12653 –0.01265 M1 Use algorithm 0.2 0.974614 A1 (0.2)y (0.974 to 0.975) 5

(iv) If V volume, v velocity, A horizontal cross-sectional area,

then 1ddV k vt M1 Rate of change of volume

2v k y

d dVd dyAt t M1

Relate rates of change of y and volume

1 2ddyA k k yt

M1 Eliminate volume and/or velocity

ddy k yt

E1 Complete argument 4

t 20

1

y


4

4(i) 25 2 9e ty x x M1 y or 5y in terms of ,x x 25 2 18e ty x x M1 Differentiate M1 Substitute for y

215 (2 18e )tx x

2 245 (2 9e ) 3et tx x x M1 Substitute for y

22 3 3e tx x x E1 5 (ii) 2 2 3 0α α M1 Auxiliary equation 1, 3α A1 CF 3e et tA B F1 CF for their roots PI 2e tx a B1 PI of correct form 2 22 e , 4 et tx a x a M1 Differentiate and substitute

2 24 4 3 3 t ta a a e e M1 Compare coefficients and solve

1a A1

GS 3 2e e et t tx A B F1 PI + CF with two arbitrary constants 8

(iii) 215 (2 9e )ty x x M1

M1 Differentiate and substitute

3 2 215 (2 e 2 e 2e 9e t t t tA B

3 2( e 3 e 2e ))t t tA B F1 Expression for x follows their GS

3 215 e e e t t ty A B A1 4

(iv) 0, 0 0 1t x A B M1 Use condition 1

50, 2 2 1t y A B M1 Use condition 0, 1A B 3 2e et tx A1 3 2e et ty A1 4 (v) As , 0, 0t x y B1

(0) 2 0y A M1 Consider coefficient(s) of et and mention of y < 2

,x y as t E1 Complete argument 3

GCE


Mathematics (MEI)

Unit 4758: Differential Equations

Advanced GCE

Mark Scheme for January 2011


Question Answer Marks Guidance 1 (a) (i) 2 2 5 0 M1 Auxiliary equation 1 2j A1 CF )t e ( cos 2 sin 2t A t B M1 CF for complex roots F1 CF for their roots PI etx a B1 e , et tx a x a M1 Differentiate twice and substitute e 2 e 5 e 4t t t ta a a e M1 Compare coefficients and solve 1

2a A1 GS 1

2 e e ( cos 2 sin 2 )t tx A t B t F1 PI + CF with two arbitrary constants [9] 1 (a) (ii) 1

20, 0 0t x A M1 Use condition 1

2 e -e ( cos2 sin 2 )

e ( 2 sin 2 2 cos 2 )

t t

t

x A t B t

A t B t

M1 Differentiate (product rule)

120, 0 0 2t x A B M1 Use condition

1 12 2e - e (cos 2 sin 2 )t tx t t A1 cao

[4] 1 (b) (i) AE 3 24 6 0 B1 3 21 4 1 1 6 0 E1 2( 1)( 5 6) 0 M1 Factorise (or solve by other means) 1, 2 or 3 A1 GS 2 3e e ex x xy A B C F1 GS = CF with three arbitrary constants [5] 1 (b) (ii) 0 0y A B1 (0) 1 1y B C M1 Use condition (0) 4 2 3 4y B C M1 Use condition 1, 2B C 3 22e ex xy A1 cao [4]

4


Question Answer Marks Guidance 1 (b) (iii) 30 e (2 e )x xy 0 e 2x M1 ln 2x A1 [2] 2 (a) (i) exp 2 dI x x M1 Attempt IF

2

ex A1 Correct IF 2 2de 2 e sin

dx xy x y x

x

M1 Multiply by IF

2dd ( e ) sinxx y x M1 Recognise derivative

2

e sinxy x x d M1 Integrate

cos x A A1 RHS (including constant) 0, 0 1 1x y A M1 Use condition 2A A1 2

e (2 cosxy x ) F1 Divide by their IF, including constant

[9] 2 (a) (ii) 2

e 0,cos 1x x M1

2 cos 0x y 0 E1 2 2d 2 e (2 cos ) e (sin )

dx xy x x x

x

Or use DE

d0 0dyxx

E1

0x y B1

B1 B1

Through (0,1) Shape consistent with results shown

[6]

5


Question Answer Marks Guidance 2 (b) (i) d 1 2

dy xyx

x y y 0 1 1 B1 First row 0.1 1.1 0.78 M1 Use algorithm 0.2 1.178 A1 1.1 A1 1.178 [4] 2 (b) (ii) 2

exI F1 2 2d ( e ) e

dx xy

x

M1

2 20.2 0.2

00e e d

xx x

xy x

A1

20.2(0.2)e 1 0.2027y M1

(0.2) 1.15(55)y A1 [5] 3 (i) 0k k M1 Root of auxiliary equation CF e kxA A1 PI xcos3 sin 3y a x b B1 d 3 sin3 3 cos3

dy a x b xx

M1 Differentiate

3 sin 3 3 cos3( cos3 sin3 ) cos3a x b x

k a x b x x

M1 Substitute and compare

3 03 1

a kbb ka

A1

2 2

3,9 9

ka bk k

A1

2

1e ( cos3 3sin 3 )9

kxy A k x xk

F1 PI + CF with one arbitrary constant

[8]

6


Question Answer Marks Guidance 3 (ii) 0, 1 0x y y (from DE)

OR differentiate y M1A1

20

9kAk

M1

Use condition

2

1 ( cos3 3sin3 e )k 9

kxy k x xk

A1

[4] 3 (iii) CF e kxB F1 PI e kxy cx B1 e ekx kxy c kcx M1 Differentiate e (1 ) e 2ekx kx kxc kx kcx M1 Substitute and compare 2c A1 e 2 ekx kxy B x F1 PI + CF with one arbitrary constant [6] 3 (iv) d (previous DE) with 2

dk

x

M1 Recognise relationship

2 2e 2 ex xy B x C F1 0, 0 0x y B C B1 Condition 2 2 22 e 2e 4 ex x xy B x M1 Differentiate 0, 1 1 2 2x y B M1 Use condition 1 1

2 2,B C NEED ALTERNATIVE SOLUTION 2 21 1

2 2e 2 ex xy x A1

OR for first 2 marks

2 2m m 0 ; CF 2

2xy Be C

and PI xy pxe giving 2p GS 2 2e 2 ex xy B x C

[6]

M1 A1

Complete method

7


8

Question Answer Marks Guidance 4 (i) 10y x x M1 10y x x M1 10 0.2 3 0.3x x x x x M1 Eliminate y 0.4 0.05 0x x x M1

E1 Eliminate y

[5] 4 (ii) 2 0.4 0.05 0 M1 Auxiliary equation 0.2 0.1j A1 0.2e ( cos0.1 sin 0.1 )tx A t B t M1

F1 CF for complex roots CF for their roots

[4] 4 (iii) 0.2

0.2

0.2e ( cos0.1 sin 0.1 )0.1e ( sin 0.1 cos0.1 )

t

t

x A t B tA t B

t

M1 A1

Differentiate (product rule)

0.2

10e (( )cos0.1 ( )sin 0.1t

y x xA B t B A t)

M1 A1

Substitute to find y

[4] 4 (iv) 0x A B1 0y A B M1 Use condition 0.2

0 0 0e ( cos0.1 ( )sin 0.1t )x x t y x t A1 0.2

0 0 0e ( cos0.1 ( 2 )sin 0.1ty y t y x t ) A1 [4] 4 (v) 0

0 0

0 when tan 0.1 1.252

yy ty x

M1 F1

So (for least positive t), 22.5t A1 Or compare values of tan 0.1t 0

0 0

10 when tan 0.19

xx ty x

M1 F1

So (for least positive t), 30.3t A1 Or compare values of tan 0.1t Hence rabbits die out first A1 Complete argument [7]

GCE


Mathematics (MEI)


Advanced GCE



1(i) 2 4 3 0 M1 Auxiliary equation 1 or -3 A1 CF 3e et tA B F1 CF for their roots PI cos 2 sin 2y a t b t B1

2 sin 2 2 cos 2y a t b t M1 Differentiate twice and substitute

4 cos 2 4 sin 2y a t b t 4 cos 2 4 sin 2 8 sin 2 8 cos 2 3 cos 2 3 sin 3 13cos 2a t b t a t b t a t b t t M1 Compare coefficients

8 1b a 3

8 0b a A1

1 8, 5 5

a b A1

GS 315 (8sin 2 cos 2 ) e et ty t t A B F1 PI + CF with two arbitrary

constants

9 (ii) 1

50, y 0 0t A B M1 Use condition

3sin 2 ) e 3 et tt A 15 (16cos 2 2y t B M1 Differentiate

F1 16

50, 0 0 3t y A B M1 Use condition

13 3, B=10 2

A A1

313 315 10 2(8sin 2 cos 2 ) e et ty t t A1 Cao

6 (iii) If , differentiating (*) gives new DE z y c M1 Recognise derivative and has 3 arbitrary constants so must be GS A1 or Integrating gives (*) with on RHS k M1 PI will be previous PI 1

3 k , CF as before, so GS y c A1 SC1 for showing that correct y from (i) + c satisfies new DE 2 (iv) 31

5 (8sin 2 cos 2 ) e et tz t t D E c

150, 2 2t z D E c M1 Use condition

315 (16cos 2 2sin 2 ) e 3 et tz t t D E F1 Derivative

1650, 0 0 3t z D E M1 Use condition

315 ( 32sin 2 4cos 2 ) e 9 et tz t t D E F1

Second derivative: condone, for this mark only, +c appearing

450, 13 13 9t z D E M1 Use condition

13 3, , 210 2

D E c B1

313 315 10 2(8sin 2 cos 2 ) e e 2t tz t t A1 Cao

7

1


2(a)(i) 2exp( d )xI x M1 Attempt integrating factor

exp( 2ln )x A1 2x A1

3

22 3d 2dyx x y xx

M1 Multiply both sides by IF

3

22d ( )d

x y xx

M1

122 2x y x A M1 Integrate both sides

A1 3

2 22y x Ax F1 Must divide constant 8 (ii) 0 2 A M1 3

22y 2 2x x A1 2 (iii) 0, 0x y F1

1

2d 94 3 0 (as 0)d 16y x x x xx M1

d0, 0dyxx

F1

B1 Behaviour at origin B1 Through and shape for 1x B1 Stationary point at 6 (b)(i) Circle centre origin B1 Radius 1 B1 2 (ii) B1 One isocline correct B1 All three isoclines correct

B1 Reasonably complete and accurate direction indicators

3 (iii) B1 Solution curve 1 (iv) B1 Solution curve B1 Zero gradient at origin 2

1

2


3(a)(i) N2L: 22ma k x M1

2d2 2d

vv kx x M1 Acceleration

2ddvv kx x E1

3 (ii) 2 dvdv k x x M1 Separate and integrate

2 2 21 12 2v k x A A1 LHS

A1 RHS 2 21

2, 0x a v A k a M1 Use condition 2 2 2 2( )v k a x A1

So for 2 2d0, dxv k at

x E1

6

(iii) 2 2

1 d dx k ta x


arcsin xa B k t A1 LHS

A1 RHS 1

2, 0x a t B M1 Use condition

12sin( ) cosx a kt a kt A1 Either form

5 (b)(i) d 9sin d M1 Separate and integrate

212 9cos C A1 LHS

A1 RHS 91

3 2, 0 C M1 Use condition So 2 9(2cos 1) A1

d 3 2cos 1dt (decreasing) E1

6 (ii) 1

3 0 M1

So estimate 1 13 30 A1

The algorithm will keep giving 13 B1

but is not constant so not useful B1 4

3


4

4(i) 3 31

2 2 2y x x t M1

3 312 2y x x 2 M1

3 31 3 312 2 2 2 2 22 ( ) 2x x x x x t t M1 Eliminate

M1 Eliminate 2 5 1x x x t E1 5 (ii) 2 2 1 0 M1 Auxiliary equation 1 (repeated) A1 Root CF: ( )e tA Bt F1 CF for their root(s) (with two constants) PI: x at b B1 , 0x a x In DE: 0 2 5 1a at b t M1 Differentiate and substitute 5a 2 1a b M1 Compare and solve 5, 9a b A1 GS: 9 5 ( )e tx t A Bt F1 GS = PI + CF with two arbitrary constants 8 (iii) 3 31

2 2 2y x x t M1

M1 Differentiate (product rule)

12 [ 5 e ( )e ]t tB A Bt

3 32 2[9 5 ( )e ]tt A Bt t M1 Substitute

1 t29 11 ( )et A B Bt A1

4 (iv) 0, 9 0t x A M1 Use condition 1

20, 0 0 11 22t y B B M1 Use condition 9 5 22 e tx t t A1 9 11 (11 22 )e ty t t A1 4 (v) e 0t M1 9 5x t F1 9 11y t F1 3


Question Answer Marks Guidance 1 (i) 2 6 9 0 M1 3 (repeated) A1 3e xy A Bx F1 2y ax bx c B1 2 , 2y ax b y a 2 22 6(2 ) 9( )a ax b ax bx c x M1 Differentiate and substitute 9 1a 12 9 0a b M1 Compare at least two coefficients 2 6 9 0a b c M1 Solve for at least two unknowns

1 4 2, ,9 27 27

a b c A1

2 31 4 2 ( ) e

9 27 27xy x x A Bx F1 Non-zero PI + CF with two arbitrary constants

[9] 1 (ii)

20, 0

27x y A M1 Use condition

3 3d 2 4 e 3( )e

d 9 27x xy x B A Bx

x M1

F1 Differentiate using product rule FT only c from (i)

40, 0 0 327

x y B A M1 Use condition

227

B

2 31 4 2 2 (1 )e

9 27 27 27xy x x x A1 cao

[5]

1 (iii) Both appear in CF B1 Or any clear statement or reason [1] 1 (iv) 2 3e xy x B1 Allow for 3 3 2 3e ex xy x x

3 2 3d 2 e 3 ed

x xy x xx

M1 Differentiate twice using product rule

5


Question Answer Marks Guidance

2

3 3 2 32

d 2 e 12 e 9 ed

x x xy x xx

A1

in DE:

3 2 2 2 3e 2 12 9 12 18 9 ex xx x x x x M1 Substitute into DE

12 1

2 M1 Compare coefficients

A1 Correct PI correct working only 2 31

2 e xy A Bx x F1 Non-zero PI + CF with two arbitrary constants

[7] 1 (v) (A) 3(e 0)x , quadratic has +ve coefficient of 2x FT for any three term quadratic so can have for all x for suitable 0y A and B B1 Reasonably complete explanation or show for a specific example 1 (v) (B) would need quadratic < 0 for all x, but for large x it will FT for any three term quadratic be positive B1 Reasonably complete explanation For both marks to be awarded in (v), ust be stated 3e 0x m [2] 2 (i)

2ddvmv mg mkvx M1 N2L 3 terms, allow sign errors and any form for accn, including a

2d

dvv g kvx E1

2 d dv v x

g kv

M1* Separating variables. Integrating factor attempt gets zero

2

11 ln

2g kv x c

k A1 LHS

A1 RHS (including constant on one side) 2 2e kxg kv A M1dep* Rearrange, dealing properly with constant 0, 0x v A g M1dep* Use condition

2 21 e kxgvk

E1

[8] 2 (ii)

2553025

g gkk

B1 Or 0.00324: cao

[1]

6


Question Answer Marks Guidance 2 (iii)

d 0.1dvm mg mg vt M1 N2L 3 terms, allow sign errors and any form for accn, including a

d (1 0.1 )dv g vt A1

EITHER

1 d d1 0.1

v g tv

M1 Separating variables

210 ln 1 0.1v gt c A1 LHS A1 RHS (including constant on one side) 0.11 0.1 e gtv B M1 Rearrange 0, 54 4.4t v B M1 Use condition 0.110 44e gtv A1 [8] OR Integrating factor 0.1e gt B1 FT their equation 0.1 0.1e 10egt gtv A M1 Multiply through by IF and integrate both sides A1 FT 0.110 e gtv A M1 Divide through by IF 0, 54 44t v A M1 Use condition 0.110 44e gtv A1 [8] OR

d 0.1dv gv gt

Auxiliary equation CF 0.1 0m g 0.1e gtA M1 A1

PI v =10 B1 GS 0.110 e gtv A M1 0, 54 44t v A M1 Use condition 0.110 44e gtv A1 [8]

7


Question Answer Marks Guidance 2 (iv)

1 12 10ln 3.15 s

0.1 44t

g

B1 FT FT for v obtained by a correct method

[1] 2 (v)

0.13

440d 10 e gtx v t t cg

M1 F1 FT attempt to integrate their answer to part (iii)

3

4400, 0t x cg

M1 A1 FT

FT for x obtained by a correct method

3.15 74.4t x M1 A1 cao [6] 3 (i)

2d 2 sindy y x x x x M1 Divide both sides by x (NOTE that a MR (eg sinx missing) can earn 7/8)

2exp dI xx

M1 Allow

2lne x A1 2x A1

2d sin

dx y x

x

2 sin dx y x x

M1 B1

Multiply by their IF and attempt to integrate both sides LHS must be their IF x y

2 cosx y x A A1 + A is needed 2 cosy x x A F1 Divide every term by the multiplier of y (including a constant) [8] 3 (ii) 20 cos A M1 Use condition 1A 2 (cos 1)y x x A1 cao

8



B1 Oscillations corresponding to their expression B1 Changing amplitude corresponding to their expression B1 0y with maxima on x-axis; cusps get B0; cao Max B1 if MR has made graph simpler [5]

3 (iii)

21 d 2

dyx xy M1 Attempt to separate variables Attempt at IF, M0 for this part

2

1 2d dy xxy

M1 Integrate both sides

1 2 ln x By

A1 LHS

A1 RHS (including constant on one side)

1

2lny

x B

A1 Any correct form of answer

[5] 3 (iv)

3 2d sin 2

dy x x yx x

B1 May be implied by correct values

M1 Use algorithm can get this even if B0 A1 y(3.15) = 0.000157… Agreement to 3sf A1 –0.08342… Agreement to 3sf Award for -0.000834

x y y’ hy’ 3.14 0 0.015703 0.00015703 3.15 0.00015703 –0.083421 –0.00083421 3.16 –0.000677

A1 –0.000676…Agreement to 3sf. ONLY award if this is the FINAL answer

[5] 3 (v) Reduce step length (and carry out more steps) B1 No contradictions [1]

9


Question Answer Marks Guidance 4 (i) 2 6y x x M1 2y x x M1 2 2 4 12 7x x x x x M1 Substitute for y M1 Substitute for y 4 5x x x 5 A1 cao AE 2 4 5 0 M1 2 j A1 cao CF )2e ( cos sint A t B t M1 CF for complex roots B1 CF for their roots

PI 5 15

x B1 B1

Appropriate form of PI Correct PI

GS )21 e ( cos sintx A t B t F1 Non-zero PI + CF with 2 arbitrary constants

[12]

4 (ii) 2 2d 2e ( cos sin ) e ( sin cos )

dt tx A t B t A t B t

t M1 Differentiate using product rule

2 6y x x

24 e ( sin cos )t A t B t

M1 A1

Substitute cao

[3]

Alternative method: 2d1 e cos sin 2 7

dty

A t B t yt

2d

2 8 e cos sind

tyy A t B t

t M1 Substitute for x in

d2 7

d

yx y

t and arrange in correct IF form

2 2e (8e cos sint ty A t B t t )d M1 Find IF, multiply both sides by IF and show intention to integrate 24 e ( sin cost )y A t B t A1 [3] 4 (iii) 7 1 A M1 Use condition on a GS 0 4 B M1 Use condition on a GS 21 e (6cos 4sin )tx t t A1 cao 24 e (6sin 4cost )y t t A1 cao [4]

10


11

Question Answer Marks Guidance 4 (iv)

4yx , so 4k M1

F1 FT wrong constant term(s) and wrong form of expressions for x and y 4 6sin 4cos 4(6cos 4sin )y x t t t t M1 The exponential term needs to have been cancelled 10sin 28cost t tan 2.8t A1 Or equivalent e.g. tan( 0.59) 4t which has infinitely many roots E1 This can be awarded for a correct justification following a wrong value for

tant (i.e. can get A0 E1) [5]


GCE

Mathematics (MEI) Advanced GCE


Mark Scheme for January 2013


5

Question Answer Marks Guidance 1 (i) 3 22 5 6 0 M1 3 22 2 2 5 2 6 0 E1 Allow if implicit in factorisation of cubic ( 2)( 1)( 3) 0 M1 Attempt roots (any method) (2), 1, 3 A1 CF 2 3e e ex x xA B C F1 PI sin cosy a x b x B1 cos siny a x b x sin cosy a x b x cos siny a x b x ( cos sin ) 2( sin cos )a x b x a x b x M1 Differentiate and substitute 5( cos sin ) 6( sin cos ) sina x b x a x b x x

3225 50

6 8 0,

8 6 1a b

a ba b

M1 Compare coefficients and solve

A1 GS 2 33 2

50 25cos sin e e ex x xy x x A B C F1 [10]

1 (ii) bounded so 0A F1 3

500, 1 1x y B C F1 Use condition 33 2

50 25sin cos e 3 ex xy x x B C M1 Differentiate 2

250, 0 0 3x y B C M1 Use condition 29 51

20 100,B C A1 33 29 512

50 25 20 100cos sin e ex xy x x F1 [6]

1 (iii) 3 250 25cos siny x x F1

amplitude 2 21 150 103 4 M1

A1

B1

Sketch showing oscillations with their amplitude. More than one oscillation, ignore origin

[4]


6

Question Answer Marks Guidance 1 (iv) bounded so 0B C B1 3 47

50 500, 1 1x y A A M1 472

25 50d0 2 0dyxx

M1 Or 1

25A

So no such solution A1 www [4]

2 (i) N2L: d 9.8

dvm m mkvt

d 9.8dv kvt

E1

EITHER 1 d d

9.8v t

kv


1 ln 9.8 kv t ck

A1 A1

LHS RHS (including constant on one side)

9.8 e ktkv A M1 Rearrange, dealing properly with constant 0, 0 9.8t v A M1 Use condition 9.8 1 e ktv

k A1 cao

OR Integrating factor ekt M1 e 9.8e dkt ktv t A1 Multiply both sides by IF and recognise

derivative on LHS

9.8e ekt ktv Ak

M1 Integrate both sides

A1 Must include constant 9.80, 0t v A

k M1 Use condition

9.8 1 e ktvk

A1 cao

OR Auxiliary equation 0k M1 CF e ktv A A1

PI 9.8, 0v b v bk

M1


7


GS 9.8e ktv Ak

A1

9.80, 0t v Ak

M1 Use condition

9.8 1 e ktvk

A1 cao

[7] 2 (ii) 9.8 1.4

7k B1

[1] 2 (iii) 2d 9.8 0.2

dvm m mvt M1

2

1 d d9.8 0.2

v tv

M1

22

5 d49

v t cv

A1 RHS (including constant on one side)

2

5 1 1 d14 7 7

v t cv v

M1 Integrate

5214 ln 7 ln 7v v t c A1 LHS

14

257ln7

v t cv

14 /57 e7

tv Bv

M1 Rearrange into a form without ln, dealing

properly with constant

0, 0 1t v B M1 Use condition 14 /57 e 7tv v M1 Rearrange to get v in terms of t 14 /5 14 /5

14 /5 14 /5e 1 1 e7 7e 1 1 e

t t

t tv

A1 oe

as 1 01 0, 7 7t v E1

[10]


8

Question Answer Marks Guidance 2 (iv) 3/20.529v g v E1 t v v hv M1 Use algorithm 0 0 9.8 0.98 A1 (0.1)v 0.1 0.98 9.2868 0.92868 A1 (0.1)v at least 3d.p. 0.2 1.9087 A1 (0.2)v =1.91 to 3s.f. [5]

2 (v) 3/20 0.529 7.00v g v v (3 sf) E1 Or 3

29.8 0.529 7 0 [1]

3 (a) exp tan dI x x M1 exp ln sec x or exp ln cos x A1 cos x A1 dcos sin sin cos

dyx y x x xx

d cos sin cosd

y x x xx

M1 Multiply and recognise derivative

cos sin cos dy x x x x M1 Attempt integral 1

2 sin 2 dx x M1 Use identity, substitution or inspection on RHS 1

14 cos2x c (or 212 sin x k ) A1 oe (but must include constant)

1140, 1 1x y c M1 Use condition

5 cos 24cos

xyx

or

2sin 22cos

xyx

or 23 cos

2cosxy

x

A1 oe

[9]


9

Question Answer Marks Guidance 3 (b) p '( ) f ( )p( ) g( )x x x x M1 Must be ( )p x oe c '( ) f ( )c( ) 0x x x M1 Must be ( )c x oe d f ( ) p '( ) c '( ) f ( ) p( ) c( )

dy x y x A x x x A xx M1 Substitute in DE

p '( ) f ( )p( ) c '( ) f ( )c( )x x x A x x x M1 Separate p and c terms g( ) 0 g( )x A x E1 Complete argument [5]

3 (c) (i) 2 2de 2 ed

x xyy xx

B1

so LHS of DE

2 222 e ex xxx

M1

2 2 21 12e 2ex x xxx x

E1

[3] 3 (c) (ii) d 2 1 2d d

dy y y xx x y x M1

2ln 2lny x c A1 LHS A1 RHS including constant 2y Ax A1 cao OR Integrating factor = 2x B1 2d 0

dyx

x M1

2yx A A1 2y Ax A1 cao [4]


10

Question Answer Marks Guidance 3 (c) (iii) 2 2exy Ax B1 Here A combines the arbitrary constants of (b)

and (c) (ii) into a single arbitrary constant.

11 e A M1 Use condition

2 2e 1 exy x A1 [3]

4 (i) 2 13 2y x x t M1

2 13 2 1y x x M1 Differentiate

32 1 1 2 13 2 2 2 3 21 2x x x x x t t M1 Substitute

2 2 5 2 5x x x t A1 oe AE 22 2 5 0 M1 31

2 2 j A1 CF 1

2 3 32 2e cos sint A t B t M1 Correct form

F1 FT wrong roots PI x a bt B1 , 0 2 5( ) 2 5x b x b a bt t M1 Differentiate and substitute

45

2 5 2, 1

5 5b a

a bb

M1

A1 Equate coefficients and solve

GS 12 3 34

5 2 2e cos sintx t A t B t F1

[13]


11

Question Answer Marks Guidance 4 (ii) 2 1

3 2y x x t M1

12

12

3 312 2 2

3 3 3 32 2 2 2

1 e cos sin

e sin cos

t

t

x A t B t

A t B t

M1

F1 Must be using product rule Must be GS from (i)

12 3 32

5 2 2e sin costy t A t B t A1

[4] 4 (iii) 4 1

5 51, 0 1x t A A M1 2 2

5 50, 0 0y t B B M1 1

2 3 34 1 25 5 2 5 2e cos sintx t t t

12 3 32 1 2

5 5 2 5 2e sin costy t t t A1 Both

[3] 4 (iv) 1

26 3 3 315 5 2 5 2e sin costx y t t M1 Adding and attempting the limit

12 6

5e 0tt x y E1 FT for finite limit 6 3 3 3 31 1

5 5 2 5 2 2 3sin cos 0 tanx y t t t M1 Establish equation and indicate method which occurs (infinitely often) E1 Correctly investigate the existence of a

solution, but explicit solution for t not required.

[4]


5


1 (i) 22 3 1 0 M1 Auxiliary equation

1

21, A1 Correct roots

CF / 2e e

t tx A B

F1 FT roots

PI co s s inx a t b t B1

s in co sx a t b t , co s s inx a t b t

2( cos sin ) 3( sin cos ) ( cos sin ) cosa t b t a t b t a t b t t M1 Differentiate and substitute

31

1 0 1 0

3 1,

3 0

a ba b

b a

M1 Compare and solve

A1 Correct values

GS / 21

1 03 s in co s e e

t tx t t A B

F1 FT CF with 2 arbitrary constants + PI

[8]

1 (ii) 1

1 00, 0 0x t A B M1 Use condition

/ 21 1

1 0 23 c o s s in e e

t tx t t A B

M1 Differentiate GS from (i)

3 1

1 0 20 , 0 0x t A B M1 Use condition

/ 21 1 2

1 0 2 53 s in co s e e

t tx t t

A1 Correct expression

Graph B1 At least one and a half oscillations;

scales not required

Note that only large t

required, but accept any

evidence of oscillations

B1

Approximately constant amplitude and

constant frequency over at least two

oscillations; scales not required

FT values of a,b,A,B and

both values of negative

[6]

1 (iii) 1 1

1 0 1 03 sin 1 0 co s 1 0x M1 Use 1 0t in either x or x soi

A1 FT incorrect a and b from (i)

31

1 0 1 03 c o s 1 0 s in 1 0x A1 FT incorrect a and b from (i)

[3]

1 (iv) 2

3 4 2 1 1 M1

Consider auxiliary equation; either

discriminant or nature of roots found in

(i). Or state the CF again.

0 or state roots are real and distinct. Overdamped A1 Must give a reason

[2]


6


1 (v) / 2e e

t tx C D

B1 FT from (i)

1 0 51

1 0e eC D

M1 Use condition. Accept applied at 0t

/ 21

2e e

t tx C D

1 0 53 1

1 0 2e eC D

M1 Use condition. Accept applied at 0t

1 0 51 2

2 5e , eC D

A1 cao

1 0 (1 0 ) / 21 2

2 5e e

t tx

A1

13 6 / 22 .2 10 e 2 .65 10 e

t t

FT their C and D if correct t used

sc Award for / 21 2e e

2 5

t tx

if new t

clearly defined.

[5]

2 (i) EITHER:

d5 0 0 1 0 0 0 0 5 0 0

d

vg k v

t M1 N2L

5 0 0

d1 0

d

kv

vt A1

5 0 0

1d d

1 0k

v tv


5 0 0

5 0 0ln 1 0

k

kv t c A1 LHS

A1 RHS

/ 5 0 0

5 0 01 0 e

k tkv A

M1

Rearrange, dealing properly with

constant

0 , 0 1 0t v A M1 Use condition

/ 5 0 05 0 0 0

1 ek t

kv

E1

[8]

OR:

d5 0 0 1 0 0 0 0 5 0 0

d

vg k v

t M1 N2L

5 0 0

d1 0

d

kv

vt A1

IF: 5 0 0

ek t

B1

5 0 0 5 0 0e e .1 0 d

k t k tv t M1

Multiply through by IF and recognise

LHS


7


5 0 05 0 0 0e

k tA

k A1 Integrate RHS

5 0 0 00 , 0t v A

k M1 Use condition

M1 Rearrange

/ 5 0 05 0 0 0

1 ek t

kv

E1

[8]

OR:

d5 0 0 1 0 0 0 0 5 0 0

d

vg k v

t M1 Use N2L

5 0 0

d1 0

d

kv

vt A1

Auxiliary equation: 0

5 0 0

k M1

CF: 5 0 0e

k t

v A

A1

PI:

5 0 0 0v

k M1 Find PI

GS: 5 0 0

5 0 0 0e

k t

v Ak

A1

5 0 0 00 , 0t v A

k M1 Use condition

/ 5 0 05 0 0 0

1 ek t

kv

E1

[8]

2 (ii) / 5 0 05 0 0 0 5 0 0

2d e ( )

k t

k kx v t t c

M1 Attempt to integrate both terms

5 0 0 0 5 0 0

20 , 0 0

k kt x c M1 Use condition

/ 5 0 05 0 0 0 5 0 0e 1

k t

k kx t

A1 cao

[3]


8


2 (iii) 1 2 .5 / 5 0 05 0 0 0 5 0 0

2 .5 2 .52 .5 , 5 5 e 1k t x

M1

1 2 4 .0 (1 d p ) i.e. consistent E1

1 2 .5 / 5 0 05 0 0 0

2 .51 e 4 9 .4v

to 1dp B1 cao

[3]

2 (iv) 2d5 0 0 1 0 0 0 0 5 0 0 0 .4

d

vv g v

x M1 N2L

2d1 0 0 .0 0 0 8

d

vv v

x E1

[2]

2 (v) 2

d d

1 0 0 .0 0 0 8

vv x

v

M1 Separate variables: attempt to integrate

21

30 .0 0 1 6ln 1 0 0 .0 0 0 8 v x c

A1 LHS

A1 RHS

2 0 .0016

10 0 .0008 ex

v B

M1 Rearrange, dealing properly with

constant

1 2 4 , 4 9 .4 9 .8x v B M1 Use correct condition

0 .0 0 1 61 2 5 0 1 0 9 .8 e

xv

A1 oe

Graph B1 Increasing and condition shown Starting from (124, 49.4) FT

B1 Asymptote at (awrt) 112

[8]

3 (a) (i) 2 0 2 M1 Auxiliary equation or IF :

2d

e 0

d

x

y

x

CF 2e

xA

A1

co s 2 sin 2y a x b x B1

2 sin 2 2 co s 2y a x b x

2 sin 2 2 cos 2 2 ( cos 2 sin 2 ) sin 2a x b x a x b x x M1 Differentiate and substitute

1 1

4 4

2 2 1,

2 2 0

a ba b

b a

M1

A1

Compare and solve

cao

21 1

4 4co s 2 sin 2 e

xy x x A

F1 CF with one arbitrary constant + PI

[7]


9


3 (a) (ii) 1

42 A M1 Use condition

291 1

4 4 4co s 2 sin 2 e

xy x x

A1 cao

Graph B1 Starts from (0, 2) with negative gradient

B1

Oscillations with constant amplitude and

frequency for large x. Scales not

required

Their function must have

exponential decay term

[4]

3 (b) (i) 2

ex p 2 d ex

I x B1

2 2de 2 e e

d

x x xyy

x

2de e

d

x xy

x M1

2e e d

x xy x M1 Integrate

ex

c A1 All correct

2e e

x xy c

F1 Divide through by IF

[5]

3 (b) (ii) 0 , 2 2 1 1x y c c M1 Use condition

2e e

x xy

A1 cao

[2]

3 (c) 2e

xI M1

2 2de e ta n

d

x xy x

x M1

1 12 2

00

e e ta n dx

x x

x

y x x

A1

A1

LHS with limits

RHS

2 0(1) e 2 e 2 .7 1 8 6 2y B1 LHS

(1) 0 .6 3 8 6y A1 0.639 or better

[6]

4 (i) et

z A

B1

Conditions 2 et

z

B1

[2]


10


4 (ii) 2 2 et

x x y

3 2 et

y x y

M1 Substitute

1

22 e

ty x x

M1 Rearrange

1

22 e

ty x x

M1 Differentiate

31

2 22 e 2 e 2 e

t t tx x x x x

M1 Substitute

4 5 4 et

x x x

A1 cao

24 5 0 M1 Auxiliary equation

2 j A1 Correct roots

CF 2

e co s s int

B t C t M1 Correct form of CF for their roots

F1 FT their roots

PI e

tx a

B1 Correct form of PI

FT if exponential term on

RHS of d.e.

e , et t

x a x a

2

5e 4 e 5 e 4 e

t t t ta a a a

M1 A valid method to find a

GS 22

5e e co s s in

t tx B t C t

A1 cao

[12]

4 (iii) 1

22 e

ty x x

M1

Substitute x and x into expression for

y

2 22

5e 2 e co s s in e s in co s

t t tx B t C t B t C t

M1 Differentiate using product rule

23 1

5 2e e ( ) co s ( ) s in

t ty B C t B C t

A1

cao aef, simplified to a maximum of 5

terms

[3]

4 (iv) 2

51, 0 1x t B M1 Use condition

3 1 1

5 2 20 , 0 0y t B C M1 Use condition

3 9

5 5,B C

232

5 5e e co s 3 s in

t tx t t

A1

23 3

5 5e e co s 2 s in

t ty t t

A1

[4]


11


4 (v)

232

5 5

23 3

5 5

e e c o s 3 s in

e e c o s 2 s in

t t

t t

x y t t

t t

2

e e 3 s int t

t

3

e 3 sint

t

B1 Be convinced

For 0t , 30 e 1

t so its graph M1 Consider graphs oe

will meet the graph of 3 s in t infinitely often E1 Complete argument

[3]


1

1. Annotations and abbreviations

Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning

E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working


2

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


3

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.)

We are usually quite flexible about the accuracy to which the final answer is expressed and we do not penalise over-specification.

When a value is given in the paper Only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case.

When a value is not given in the paper


4

Accept any answer that agrees with the correct value to 2 s.f.

ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions. There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given.

There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.

i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct


5

answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.


6

Question Answer Marks Guidance 1 (i) Auxiliary equation

2 2 2 0m m M1

1m i A1

CF e cos sint A t B t

F1 Their roots

PI sin 2 cos2x P t Q t

B1 cao

M1 Differentiate twice ( 4 4 ) 2(2 2 ) 2( ) 30Ps Qc Pc Qs Ps Qc c

M1 Substitute 4 4 2 0

4 4 2 30

P Q P

Q P Q M1 Compare coefficients and solve

6sin 2 3cos2x t t A1 Implied by P = 6 and Q = – 3 cwo

GS: e cos sin 6sin 2 3cos2tx A t B t t t

F1 Their PI + their CF with 2 arbitrary constants. Must be t on RHS

[9] 1 (ii) 0, 0x t 3A M1 Use given condition e sin cos e cos sin

12cos2 6sin 2

t tx A t B t A t B t

t t

M1 Differentiate, product rule

10, 0x t 1B M1 Use given condition e 3cos sin 6sin 2 3cos2tx t t t t

A1 cao

[4] 1 (iii) 6sin 2 3cos2x t t B1 FT Use large t

Amplitude =

2 26 3= 3 5

6.71

B1 FT

[2] 1 (iv) Auxiliary equation

3 22 2 0m m m :

m = 0 is a root B1 Some working must be shown

1m i B1 FT roots from (i) [2]


7

Question Answer Marks Guidance 1 (v)

= e cos sintx P t Q t C

B1 Follow CF from (i), or new CF if different roots in (iv), plus a non-zero

constant.

0, 0x t 0 P C M1 Use condition, dependent on CF including the non-zero constant 10, 0x t 10 Q P M1 Use condition and use product rule 4, 0x t 4 2P Q P

2, 12, 12Q P C

M1 Use condition and use product rule

e 12cos 2sin 12tx t t

A1 cao

[5] 1 (vi) Starts at origin with positive gradient B1 Ignore negative t Tends to 12 for large values of t B1 FT their value of C if correct form of solution in (v) [2] 2 (i) ln 0.25P t A or

0.25e tP B M1 Separate and integrate (need ln on LHS)

0, 100t P ln100A or 100B M1 Use condition

0.25100e tP A1 cao Not suitable as tends to infinity as t increases

B1

Allow if constant of integration not found

'Increases for ever/always' gets B1; 'exponential growth' gets B1

'Suitable' gets B0; 'Suitable for small t ' gets B1

Alternatives for first M1:

Auxiliary eqn gives 0.25m : 0.25e tP B

Integrating factor 0.25e t

: 0.25e tP B :

0.25e tP B

[4] 2 (ii)

CF: 0.25Ae t

B1

PI:

0.5e tP k B1 Correct form

0.50.5 e tP k =0.5 0.50.25 e 18et tk M1 Differentiate and substitute

M1 Compare coefficients and solve 24k A1 GS:

0.25 = Ae tP +0.524e t

F1 Their PI + their CF with one arbitrary constant

[6]


8

Question Answer Marks Guidance 2 (iii) 0, 100t P 76A M1 Use condition 0.25 = 76e tP +

0.524e t

A1 cao

Unsuitable, not bounded (or equivalent)

F1

'Increases for ever/always' gets B1; 'exponential growth' gets B1

'Suitable for small t ' gets B1'

'Better/more suitable than previous one' gets F0;

[3] 2 (iv) M1 Separate variables 1 1

400P P M1 Use partial fractions with correct denominators

41 1 1

d 6 10 d400 400

P tP P

A1

1

ln400 400

P

P M1 Integrate LHS (dependent on previous M1)

41

ln 6 10 ( )400 400

Pt B

P A1

0, 100t P

1 1ln

400 3B

M1 Use condition (not dependent on partial fractions)

[ 25 3

ln6 400

Pt

P ] M1 Rearrange (dependent on partial fractions)

6

25

400

1 3et

P

or

6

25

6

25

400e

3 e

t

tP

A1 cao aef

[8] 2 (v)

, 200t T P 25 600 25

ln ln36 200 6

T E1 AG

[1]

2 (vi) Tends to a limit B1 Dependent on correct form of solution in (iv) limit of 400 B1 cao (Note that 400P gets B1B1) [2]


9

Question Answer Marks Guidance 3 (a) (i) 2e

1 1

xxy y

x x B1 Divide by x + 1

IF

11 dd

11e ex

xxxx M1 Splitting into

11

1x

A1 ln( 1)e x x

B1 Seen ln(1 )e ex x e 1x x

E1 Show answer

de (1 ) e

d

x xy xx

M1 Multiply both sides by given IF

A1 RHS must be simplified e (1 ) ex xy x C

M1 Integrate both sides, including + C

A1 e e

1

x x Cy

x

M1

Make y the subject (need to divide all terms by coeff of y on LHS)

A1 cao aef [11] 3 (a) (ii) 0, 2x y

3C M1 e e 3

1

x x

yx

A1 cao

[2]


10

Question Answer Marks Guidance 3 (b) (i) Circle, centre O B1

Radius 1 1

4 2

B1

[2] 3 (b) (ii) One correct isocline B1 All 3 correct isoclines B1 Ratios of radii must be correct Correct direction indicators B1 At least four on each of the three circles [3] 3 (b) (iii) Attempt at a solution curve B1 Correct solution curve B1 [2] 3 (b) (iv) One correct use of algorithm M1 1.05 A1 1.09525 A1 1.095 or better 1.137

A1 1.14 or better. Do not penalise lack of accuracy in intermediate values

if final answer correct

[4]


11

Question Answer Marks Guidance 4 (i) 24 2 2e tx y x M1 Differentiate with respect to t

212 e

4

ty x x

M1 Rearrange

2 24 5 2e 3 2 2et tx y x x

M1 Substitute for y

2 2 22 5 2 e 8e 12 2et t tx x x x x

M1 Substitute for y

23 2 e tx x x A1 Auxiliary equation 2 3 2 0 M1

1,2 A1

CF 2e et tA B F1

PI 2e tx a B1 Correct form for their RHS

2 22 e ; 4 e :t tx a x a 12a = 1 M1 Differentiate and equate

1

12a

A1

2 21

e e e12

t t tx A B

F1 Their PI + their CF with two arbitrary constants

[12] 4 (ii) 21

2 2e4

ty x x

M1 Rearrange first given equation (all terms must be present)

2 2 21 1 1 1

( e 2 e e ) e4 6 2 4

t t t ty A B x

M1 Substitute for x

M1 Substitute for x

2 23 1

e e e4 4

t t ty A B

A1 cwo in (i) and (ii)

[4]


12

Question Answer Marks Guidance 4 (iii) 3 1 2

4 4 3A B

5

3 4 03

A B

M1

Use given condition

3 12 0

4 2A B

3 8 2 0A B

M1

Use given condition

4 1;

9 12A B

M1 Solve

2 24 1 1

e e e9 12 12

t t tx

A1 cao

2 21 1 1

e e e3 12 4

t t ty

A1 cao

[5] 4 (iv) 2 21 1 1

e e e3 12 4

t t t

>

2 24 1 1e e e

9 12 12

t t t

21 1e e

9 3

t t

M1 Use condition and simplify

3e 3t

M1 Take logs

1

ln33

t

A1 cao

[3]


GCE

Mathematics (MEI)


Advanced GCE


OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2015


3

Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning



4

Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.


c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.


5

Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.


e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader.

Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.




When a value is not given in the paper Accept any answer that agrees with the correct value to 2 s.f.

ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions.


6

There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests. If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.


i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.

j If in any case the scheme operates with considerable unfairness consult your Team Leader.


7


1 (i) Auxiliary equation: 2 8 25 0m m M1

4 3m i A1

CF: 4e cos3 sin3tx A t B t F1 From their roots

0A M1 Use condition

4 4e 3 cos4 4e . sin3t tx B t B t M1 Differentiate, product rule

0.25 3B :

1

12B

M1

A1 Use condition (must use 0.25)

41

e sin312

tx t F1

[8]

(ii) d0

d

x

t : 4e 3cos3 4sin3 0t t t M1 Equate their x to zero

3tan 3

4t : 0.2145t M1 Obtain expression for tan and attempt to solve

A1 Dependent on using correct answer to (i)

0.0212x A1 cao

[4]

(iii) Oscillations with decreasing amplitude B1

Accept amplitude 0 ; accept 0x . accept "oscillations with small

amplitude" Do not accept "oscillations"

[1]


8


(iv) CF: 4e cos3 sin3tx C t D t F1

PI: sin5 cos5x P t Q t B1 Correct form

5 cos5 5 sin5x P t Q t

25 sin5 25 cos5x P t Q t

M1

M1

A1

F1

M1

M1

M1

A1

[10]

Differentiate twice and substitute

25 40 25 5P Q P

25 40 25 0Q P Q : Compare coefficients and solve

10,

8P Q

4 1e cos3 sin3 cos5

8

tx C t D t t

Their CF with 2 arbitrary constants + their PI

0, 0x y :

1

8C Use condition

4

4

e 3 sin 3 3 cos3

54e cos3 sin 3 sin 5

8

t

t

x C t D t

C t D t t

Differentiate using product rule

0.25, 0x t : 0.25 3 4D C

(1

4D )

Use condition (must use 0.25)

4 1 1 1

e cos3 sin3 cos58 4 8

tx t t t

cao

(v) Oscillations of approximately constant

amplitude 1

8

B1 FT their amplitude

[1]


9


2 (i) 2 1 12

n xy y

x x x

B1 Divide through by x

IF =

d

en

xx

B1

(=lne n x

) = nx B1

1d

2d

n n nyx x xx

M1 Multiply both sides by their IF

12

1

n nn x x

yx An n

M1

A1

Integrate both sides

Must include arbitrary constant

M1 Divide both sides, including an arbitrary constant, by their IF

2 1

1

nxy Ax

n n

A1 cao

[8]

(ii) 0, 2, 1y x n : ( 2A ) M1 Use condition to find a value for A

21y x

x A1

Curve in 1st and 4th quadrants through (2,0) B1

Must use correct form of solution. Sketch going through (2, 0) with positive

gradient at (2, 0)

Correct behaviour as 0x and x B1 Ignore curve for x < 0

[4]

(iii) 1 1 2d

2d

yx x xx

M1 Find and multiply by IF

A1

1 12lnyx x x B M1 Integrate both sides

A1 Must include arbitrary constant

2 ln 1y x x Bx A1 cao

[5]


10


(iv) 1B M1 Use condition 1, 0x y

2ln 2 1 0y x M1 Differentiate, equate to zero

1ln

2x : One solution E1

[3]

(v) yvalues: 1 M1 NB the DE is used in the form

d 1 1

d 2 1

yy

x x x

0.92079 or 0.192079 A1 Agree to 3 s.f.

0.86436 A1 Agree to 3 s.f.

y value: 0.2785 A1 0.279 (or better)

[4]

3 (i) 2d

20 100 4d

vv v

x M1* Use N2L with accn in terms of v and x

2

5 dd

25

v vx

v

M1dep

* Separate variables

25

ln 252

v x A M1dep

* Integrate both sides

A1 lhs

A1 rhs, including +A

5ln 25

2A

M1dep

* Use condition

2

2 525 25ex

v

M1dep

* rearrange

2

2 525 1 ex

v

E1 cao

When x = 10, 4.95(4)v B1

[9]


11


(ii) 2d

20 100 4d

vv

t M1* Use N2L with accn in terms of v and t

2

5dd

25

vt

v

M1dep

* Separate variables

1 5

ln2 5

vt B

v

M1dep

*

Integrate both sides. The integral may be quoted from MF2 or PF

1 1 1

2 5 5v v

used.


B = 0

M1dep

* Use condition

1 5ln

2 5

vt

v

A1 aef

When v =4.954, t = 2.689 M1 Use answer from (i)

[7]

(iii) d20 100 2

d

vv

t

M1 Use N2L with accn in terms of v and t

10ln(50 )v t C M1 Separate and integrate

10ln50C M1 Use condition

5010ln

50t

v

A1

1

1050 1 et

v

M1

Make v subject


Terminal velocity = 150ms B1

[7]


12


OR: d

20 100 2d

vv

t

M1 Use N2L with accn in terms of v and t

[ IF: 0.1e t : 0.1 0.1d

e 5ed

t tvt

]

0.1 0.1e 50et tv A M1

A1

Multiply through by IF and integrate

50A M1 Use condition

1

1050 1 et

v

M1

Make v subject



[7]

OR: d

20 100 2d

vv

t

M1

Use N2L with accn in terms of v and t

Auxiliary eqn 10 1 0m : CF: 0.1e tv A M1

PI: v B : 50B M1

GS: 0.150 e tv A A1

50A M1 Use condition

1

1050 1 et

v

A1

Correct expression


[7]

(iv) When v = 4.954, 10ln1.11 1.04t s B1

[1]


13


4 (i) 2

2

d d d2 2

d d d

x x y

t t t M1 Differentiate

1

2d 3

2 60ed 4

txx

t

M1 Substitute for dy

dt

12

22

d d 32 60e

d d 4

tx xx

t t

A1 AG Rearrange

Auxiliary equation

2 32 0

4m m M1

1 3,

2 2m A1

CF:

3 1

2 2e et t

x A B F1

PI:

1

2et

x P

B1 Correct form for their CF

1 1

2 21 1

e , e2 4

t t

x P x P

M1 Differentiate and substitute

30P A1 Solve

3 1 1

2 2 2e e 30et t t

x A B

F1 PI + CF with 2 arb const

[10]

(ii) 1 d

2 d

xy x

t M1 Rearrange

Substitute for x and

d

d

x

t M1

3 1 1

2 2 21 3 75

e e e4 4 2

t t t

y A B

A1 cao. As final answer

[3]


14


(iii) 40, 0 40 30x t A B M1 Use condition

1 3 7550, 0 50

4 4 2y t A B

M1 Use condition

10, 20A B

3 1 1

2 2 210e 20e 30et t t

x

A1 cao

3 1 1

2 2 25 75

e 15e e2 2

t t t

y

A1 cao

[4]

(iv) When 0x

220e 10e 30 0T T M1 Multiply by

1

2eT

2e 2e 3 0T T M1 Attempt to solve as a quadratic

e 3T (or -1) A1

ln3T (= 1.10) A1 cao

M1 Substitute T in expression for y

34.64y A1 cao

[6]

(v) Unsuitable, X is negative B1

[1]

Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2015

OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: [email protected] www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored

mailto:[email protected]

http://www.ocr.org.uk/


GCE

Mathematics (MEI)


Advanced GCE


OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2016


3

Annotations and abbreviations

Annotation in scoris Meaning

and

Benefit of doubt

Follow through

Ignore subsequent working

, Method mark awarded 0, 1

, Accuracy mark awarded 0, 1

, Independent mark awarded 0, 1

Special case

Omission sign

Misread

Highlighting Other abbreviations in mark scheme

Meaning



4

1. Subject-specific Marking Instructions for GCE Mathematics (MEI) Mechanics strand

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.


c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


5

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.


e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.





6

When a value is not given in the paper Accept any answer that agrees with the correct value to 2 s.f.

ft should be used so that only one mark is lost for each distinct error made in the accuracy to which working is done or an answer given. Refer cases to your Team Leader where the same type of error (e.g. errors due to premature approximation leading to error) has been made in different questions or parts of questions. There are some mistakes that might be repeated throughout a paper. If a candidate makes such a mistake, (eg uses a calculator in wrong angle mode) then you will need to check the candidate’s script for repetitions of the mistake and consult your Team Leader about what penalty should be given.

There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.



7

i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader.

j If in any case the scheme operates with considerable unfairness consult your Team Leader.


8

1. (i) AE: 24 8 3 0m m M1

1 1 31 ,

2 2 2m

A1

CF:

1 3

2 2e et t

x A B

F1

PI:

1

2et

x k

B1

1

2k

M1

A1

Differentiate and substitute to find k

1 3 1

2 2 21

e e e2

t t t

x A B

F1

PI + CF with 2 arbitrary constants

[7]

(ii) 0, 6t x

16

2A B

M1

Use condition

0,t

d4

d

x

t

173

2A B

M1

M1

Differentiate and use condition

Solve

34,

2A B

1 3 1

2 2 23 1

4e e e2 2

t t t

x

A1

[4]

(iii) Minimum when

1 3 1

2 2 29 1

2e e e 04 4

t t t

M1

Differentiate and equate to 0

2e 8e 9 0t t

M1 Multiply through by 3

2et

e 1 e 9 0t t M1 Solve

ln9t = 2ln3 A1 cao o.e (2.19722…)

[4]


9

(iv) CF:

1 3

2 2e et t

x C D

FT from (i)

PI: 0.5e tx Qt B1

0.5(1 0.5 )e tx Q t : 0.5( 1 0.25 e )tx Q t M1 Differentiate using product rule and substitute

4 4Q M1 Compare coefficients and solve

1Q A1

0.5e tx t +

1 3

2 2e et t

C D

A1

cao

1 3

0.52 21 3

e e e 1 0.52 2

t ttx C D t

M1

Differentiate using product rule

t = 0, x = 6: 6C D M1 Use condition correctly

0,t

d4

d

x

t : 3 10C D

M1

Use condition correctly (independent of use of product rule) and solve

4, 2C D

0.5e tx t +

1 3

2 24e 2et t

A1

cao

[9]

2. (a)(i) Divide through by x B1 Both sides

IF:

3d

e x

x

M1

Attempt to find integrating factor

= 3lne x M1 Integrate and simplify log term

= 3x A1

3d

cosd

yx x x

x

M1

Multiply both sides by IF and express LHS as differential o.e.

RHS = sin cos x x x c M1 Integrate by parts

A1 Including + c

3sin cos y x x x x c F1 Divide through by their integrating factor

[8]


10

(ii) y = 0 when

1

2x :

3

02 2

c

M1

Use condition

2

c : 3

sin cos2

y x x x x

A1

cao

[2]

(b)(i) 2

d 3d

sec

y x

y x

B1

Separate the variables

M1 Attempt to integrate

1

y = 3cos d x x

A1

LHS

= 3sin x c A1 Integral of RHS including + c

1y when

1

2x : 4 c

M1

Use condition

M1 Make y the subject

1

4 3sin

y

x

A1

cao Final answer

[7]

(ii) Max value when sin 1x M1 Correct statement for their solution to (b)(i)

Or any other valid method e.g. differentiation, second derivative not

required

Max value = 1/(4-3) = 1 AG E1 Correctly shown; 1/(4-3) must be seen

[2] SC2 for stating BOTH max value = 1 and min value = 1/7

(c) 2d

(3 )cosd

y

y x xx

B1

May be implied by correct values

x y y hy

1 0 0.5403023 0.005403023

1.01 0.005403023 0.5372259 0.005372259

1.02 0.010775282

M1

A1

A1

Use algorithm

y(1.01) = 0.005403.....Agreement to 3 sf

y’(1.01) = 0.5372259…Agreement to 3 sf

0.0108 A1 0.011 or better

[5]


11

3. (i) 2d

90 90 0.36d

v

v g vx

M1

Use N2L, 3 terms, allow sign errors, allow any form for accn, including

a

2d

9.8 0.004 .d

v

v vx

E1

[2]

(ii) 2

dd

9.8 0.004

v vx

v

M1*

Separate variables

21ln

0.0089.8 0.004 x Av

A1

Integrate to obtain LHS

A1 RHS including constant on one side

Use v = 0, x = 0:

1ln 9.8

0.008 A

M1dep*

Use condition

2 0.0082450 1 e

xv M1dep* Rearrange

E1

[6]

(iii) Increasing graph through (0, 0) B1

Asymptote 2450v B1 Allow 49.5

[2]

(iv) 353 m M1A1 Accept 2.s.f.

[2]

(v) d90 90 72

d

vg v

t

M1

Use N2L, 3 terms, allow sign errors

d0.8

d

vg v

t

A1

dd

9.8 0.8

vt

v

M1

Separate variables

1ln

0.89.8 0.8 t Bv

A1

Integrate to obtain LHS

A1 RHS, including constant on one side


12

Use t = 0, v = 48:

1ln 28.6

0.8 B

M1

Use condition

0.812.25 35.75e

tv M1 Rearrange

A1 cao

Alternative method 1:

d90 90 72

d

vg v

t

d0.8

d

vg v

t

Integrating factor: e0.8t

0.8 0.8de

det t

v gt

0.8 0.8e 12.25et tv A 0.812.25 e tv A

Use t = 0, v = 48: 143

35.754

A

0.812.25 35.75e

tv

Alternative method 2:

d90 90 72

d

vg v

t

d0.8

d

vg v

t

CF: A 0.8e t

PI: v = 12.25

GS: 0.8e 12.25tv A

Use t = 0, v = 48: 35.75A 0.8

12.25 35.75e

t

v

[8]

M1

A1

B1

M1

A1

M1

M1

A1

[8]

M1

A1

M1A1

M1

A1

M1

A1

[8]


Multiply both sides by IF and express LHS as differential o.e.

Rearrange

Use condition

cao


Use condition


13

(vi) 0.835.75

12.25 e0.8

tx t D

M1

Integrate

A1 cao

Use t = 0, x = 0:

35.75

0.8D

M1

Use condition

When t = 5, distance = 105 m A1 cao

[4]


4. (i) 3sinx x y t M1 Differentiate

5 12sin 3sinx x x y t t M1 Substitute for y

3cosy x t x M1 Substitute for y

4 9sin 3cosx x t t A1 oe

AE 2 4 0m M1

2m j A1

CF cos2 sin2x A t B t F1

PI sin cosx P t Q t B1

cos sin sin cosx P t Q t x P t Q t M1 Differentiate twice and substitute

3 9 3 3P Q M1 Equate coefficients and solve

3, 1P Q M1

cos2 sin2 3sin cosx A t B t t t A1 cao

[12]

(ii) 2 sin2 2 cos2 3cos sinx A t B t t t M1 Differentiate x

Substitute x and x in 3cosy x t x M1

( 2 )cos2 (2 )sin2 4sin cosy A B t A B t t t A1

[3]


14


(iii) 0, 0: 2 1 0t y A B M1 Use condition

dy0, 5: 5 5 5 ( 2 ) 1

dt A A B

t M1 Use condition

Solve to give 1

0,2

A B

1sin 2 3sin cos

2x t t t

1cos2 sin 2 4sin cos

2y t t t t

A1

A1

[4]

(iv) M1 Equate their expressions for x and y

cos2 sint t A1 cao

22sin sin 1 0t t : 1

sin sin 12

t t M1 Use double angle formula and attempt to solve quadratic equation

5 or

6 6t

A1 One correct value

2

3

A1 cao

[5]

Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2016

OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: [email protected] www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored

mailto:[email protected]

http://www.ocr.org.uk/

Documents

2610 MEI Differential Equations January 2005 Mark Scheme packs/MS_MEI_DE.pdf · 2018-04-04 · 2610 MEI Differential Equations January 2005 Mark Scheme 1(i) I =exp 2 d()∫ xx M1