2610 MEI Differential Equations
1(i)
I = exp
( ∫ 2 x dx )
Mark Scheme
M1
2
= ex
2 dy
2
2
2
ex
+ 2 x e x y = e x − ( x − 2)
dx
d x2
e y = e4 x − 4
dx
A1
M1
multiply
F1
follow their integrating factor
e x y = ∫ e 4 x − 4 dx
M1
integrate
= 14 e4 x − 4 + A
A1
( )
2
y = e− x
(ii)
January 2005
0=
(
2
(
e−1 14
1 4 x−4
e
+
4
A
)
F1
divide by their I (must divide constant also)
7
)
+A ⇒ A=−
1
4
M1
condition on y
y = 14 e− x e4 x − 4 − 1
A1
cao
x > 1 ⇒ 4 x − 4 > 0 ⇒ e4 x − 4 > 1
M1
attempt inequality for y
E1
fully justified
B1
through (1,0) and y > 0 for x > 1
B1
asymptotic to y = 0
2
and
1 −x
e
4
2
(
)
> 0 so y > 0
1
6
(iii)
dy
maximum ⇒
=0
dx
⇒ 2 xy = e − ( x − 2)
2
x = 2 ⇒ 4y = 1⇒ y =
1
4
= e −4
(
1 4
e +
4
M1
A
)
M1
1
4
⇒ A = 0 ⇒ y = 14 e − ( x − 2)
¼
must use DE
E1
2
M1
substitute into GS for y
A1
cao
B1
general shape consistent with their solution
B1
maximum labelled at (2, 14 )
2
7
2610 MEI Differential Equations
2(i)
January 2005
for x < 0, aux.eq. α 2 + 9 = 0 ⇒ α = ±3 j
M1
y = A cos 3x + B sin 3 x
A1
2
d y
− 16 y = 0
Mark Scheme
imaginary root or recognise SHM equation
B1
may be implied
y = C e −4 x + D e 4 x
M1
A1
F1
accept A,B again here but not in (ii)
(1) A = y0
C + D = y0
F1
F1
for x > 0,
dx 2
2
α − 16 = 0
α = ±4
6
(ii)
(2) x < 0,
dy
= −3 A sin 3 x + 3B cos 3 x
dx
dy
= −4C e −4 x + 4 D e 4 x
dx
3B = −4C + 4 D
x > 0,
M1
differentiate
M1
differentiate
(3) only e is unbounded so D = 0
hence C = y0
A1
B1
B1
B = − 43 y0
B1
4x
8
(iii)
B1
B1
B1
curve for x < 0
curve for x > 0
continuous gradient at x = 0 (must have
reasonable attempt at each of x < 0 and x > 0)
3
(iv)
2
d y
+ 4 y = 0 ⇒ y = E cos 2 x + F sin 2 x
dx 2
bounded so (3) provides no equation
hence insufficient (3 equations, 4 unknowns)
B1
M1
A1
3
2610 MEI Differential Equations
3(i)
rate of flow = area × speed ⇒
⇒
dV
= −0.0004 2 gx
dt
dV dx
= −0.0004 2 gx
dx dt
January 2005
Mark Scheme
M1
accept either in words or symbols for M1 but
need both for E1
M1
E1
use of chain rule
complete argument
3
(ii)
dV 5
dx
= ⇒ 53
= −0.0004 2 gx
dx 3
dt
∫ 53 x
10
3
x
− 12
1
2
dx = ∫ −0.0004 2 g dt
= −0.0004 2 g t + A
t = 0, x = 2 ⇒ A = 103 2
(
x = 2 1 − 0.00012 g t
)
2
x = 0 ⇒ t ≈ 2660
B1
M1
separate
M1
integrate
M1
condition on x
A1
cao
A1
6
(iii)
(1 + 2 x − x ) ddxt = −0.0004
2
∫ (x
− 12
+ 2 x 2 − x 2 )dx = ∫ −0.0004 2 g dt
1
2 x 2 + 43 x 2 − 52 x
1
2 gx
3
3
5
2
t = 0, x = 2 ⇒ B =
= −0.0004 2 g t + B
46
15
2 ≈ 4.337
M1
substitute for
M1
separate
M1
A1
integrate
M1
condition on x
dV
dx
x = 0 ⇒ t ≈ 2450
A1
x (0) = −0.0004( 4 g + 0) /1 = −0.00250
E1
must show working
x(0.1) = 2 − 0.00250 × 0.1
=1.99975
x (0.1) = −0.00251
M1
E1
B1
B1
use of algorithm
must show working
6
(iv)
x(0.2) = 1.99950
5
2610 MEI Differential Equations
4(i)
d2 x
dt
=8
2
January 2005
dx
dy
−3
dt
dt
dx
− 3(−2 x + 7 y )
dt
dx
21 ⎛
dx ⎞
= 8 + 6x − ⎜ 8x − ⎟
dt
3⎝
dt ⎠
=8
d2 x
dt
− 15
2
dx
+ 50 x = 0
dt
Mark Scheme
M1
A1
differentiate
M1
substitute for
M1
substitute for y
dy
dt
E1
5
(ii)
α 2 − 15α + 50 = 0
α = 5 or 10
x = Ae5t + Be10t
y = 13 ( 8 x − x )
(
(
= 13 8 Ae5t + 8Be10t − 5 Ae5t + 10 Be10t
))
= Ae5t − 23 Be10t
M1
A1
F1
auxiliary equation
CF for their roots
M1
rearrange equation (1)
M1
substitute for x and x
A1
cao
6
(iii)
−ae
−t
−t
−t
= 8ae − 3be + e
−t
−t
−t
−t
−be = −2ae + 7be + e
−9a + 3b = 1, 2a − 8b = 1
−t
M1
M1
M1
a = − 16
A1
− 16
A1
b=
substitute in (3)
substitute in (4)
compare coefficients and solve
5
(iv)
substituting f(t), g(t) into LHS gives zero
substituting solutions in (iii) gives e −t
as equations linear, substituting sums gives sum 0 + e −t = e −t
General solutions because two arbitrary constants as expected
for two first order equations.
B1
B1
E1
stated (or substitution)
(or verified by substitution)
E1
4
2610
1(i)
Mark Scheme
d2 x
dt
= −4
2
dx
dy
+6
dt
dt
dx
+ 6(−3 x + 2 y + 26)
dt
dx
12 ⎛ dx
⎞
= −4 + −18 x + ⎜ + 4 x − 28 ⎟ + 156
dt
6 ⎝ dt
⎠
= −4
d2 x
dt
2
+2
dx
+ 10 x = 100
dt
June 2005
M1
differentiate
M1
substitute for
M1
substitute for y
dy
dt
E1
4
(ii)
α 2 + 2α + 10 = 0
α = −1 ± 3 j
M1
A1
auxiliary equation
CF x = e−t ( A cos 3t + B sin 3t )
F1
CF for their solutions
PI x =
100
= 10
10
GS x = 10 + e−t ( A cos 3t + B sin 3t )
dx
= e −t (− A cos 3t − B sin 3t − 3 A sin 3t + 3B cos 3t )
dt
y = 16 ( x + 4 x − 28)
y = 2+
(iii)
1 −t
e
2
( ( A + B) cos 3t + ( B − A) sin 3t )
10 + A = 0 ⇒ A = −10
2 + 12 ( A + B) = 0 ⇒ B = 6
x = 10 + e−t (6sin 3t − 10 cos 3t )
y = 2 + e−t (8sin 3t − 2 cos 3t )
B1
F1
their CF + their PI
M1
differentiate
M1
rearrange and substitute
A1
cao
8
M1
M1
A1
use t = 0 with their x = 0
use t = 0 with their y = 0 (or their
x = 28 )
both (cao)
3
(iv)
B1
B1
B1
long-term values can be found by setting x = y = 0 in DE’s
and solving the resulting equations
initial condition and asymptote for one
graph
both generally correct (must start at
origin)
long-term values
M1
A1
5
2610
Mark Scheme
2(i)
June 2005
B1
B1
4
r starts at 56 and decreases, tending
to 4
I starts at 0, gradient is positive but
decreases
2
(ii)
M1
A1
dr
= − k (r − 4)
dt
dr
∫ r − 4 = − k ∫ dt
ln | r − 4 |= −kt + c1
M1
separate and integrate
A1
all correct
r = 4 + Ae
t = 0, r = 56 ⇒ A = 52
M1
M1
rearranging
use initial condition
r = 4 1 + 13e− kt
E1
− kt
(
)
7
(iii)
⎛ 13
⎞
I = ∫ rdt = 4 ⎜ t − e − kt ⎟ + c2
k
⎝
⎠
52
t = 0, I = 0 ⇒ c 2 =
k
52
1 − e − kt
I = 4t +
k
(
3000 = 4 × 620 + 52
k
)
(1 − e
−620 k
)
⇒ 10k = 1 − e−620k
unless k very small, e
−620 k
≈ 0 ⇒ 10k ≈ 1 ⇒ k ≈ 0.1
M1
integrate r
M1
use initial condition
A1
cao
M1
condition on their I
E1
must follow correct I
E1
independent of other marks
6
(iv)
α + 0.1 = 0 ⇒ α = −0.1 so CF r = B e
−0.1t
dr
for given PI
= −2b sin 2t + 2c cos 2t
dt
0.1a + (0.1c − 2b) sin 2t + (0.1b + 2c) cos 2t = 0.4 + 0.2 cos 2t
0.1a = 0.4
0.1c − 2b = 0
0.1b + 2c = 0.2
2
40
a = 4, b =
,c =
401
401
conditions ⇒ B = 51.995 , so
2
r = 51.995e −0.1t + 4 +
( cos 2t + 20sin 2t )
401
B1
M1
differentiate and substitute
M1
compare at least two coefficients and
solve
A1
A1
5
2610
3(a)(i)
Mark Scheme
dy cos x − 2sin y
=
dx
x +1
0.16
0.18
0.143637
0.20
0.155460
0.625629
0.591150
June 2005
M1
may be implied
M1
A1
M1
A1
use of algorithm
y(0.18)
use of algorithm
y(0.20)
5
(a)(ii)
dy/dx decreases
For each step, gradient used is greater than dy/dx over
interval, hence overestimates y.
B1
B1
sketch showing curve and step by
step solution
E1
convincing argument
3
(b)(i)
dy
2
cos x
+
y=
dx x + 1
x +1
⎛ 2
⎞
I = exp ⎜ ∫
dx ⎟ = ( x + 1) 2
+
x
1
⎝
⎠
d
y
( x + 1) 2
+ 2( x + 1) y = ( x + 1) cos x
dx
M1
rearrange
M1
A1
attempt integrating factor
M1
multiply
( x + 1)2 y = ∫ ( x + 1) cos xdx = ( x + 1) sin x − ∫ sin xdx
M1
attempt integration by parts
= ( x + 1) sin x + cos x + A
x = 0, y = 0 ⇒ A = −1
A1
M1
( x + 1) sin x + cos x − 1
( x + 1) 2
x = 0.2 ⇒ y = 0.1517
E1
y=
B1
9
(b)(ii)
dy cos x − 2sin y
Since sin y < y, replacing sin y by y in
=
dx
x +1
gives an underestimate for dy/dx.
Hence (while y > 0), the approx. DE will underestimate y.
M1
consider effect on dy/dx
A1
E1
3
2610
4(i)
Mark Scheme
α 3 + 2α 2 − α − 2 = 0
(α − 1)(α 2 + 3α + 2) = 0
α = 1,−1,−2
−t
CF y = A e + B e
−2t
+Ce
June 2005
M1
auxiliary equation
M1
factorise or demonstrate 1 is a root
A1
t
F1
PI y = a e−3t
B1
−3 t
−3 t
y = −3a e , y = 9a e , y = −27 a e
−27 a + 18a + 3a − 2a = 4
a = − 12
−3t
CF for their roots (must have 3
constants)
correct form
M1
M1
A1
differentiate and substitute
compare coefficients
y = A e −t + B e −2t + C et − 12 e −3t
F1
their CF + their PI
decays ⇒ C = 0
t = 0, y = − 32 ⇒ − 32 = A + B − 12
B1
M1
condition on y
M1
differentiate
M1
condition
9
(ii)
−t
y = − A e − 2 B e
−2 t
+
3 −3t
e
2
t = 0, y = ⇒ = − A − 2 B +
3
2
3
2
3
2
−t
A = −2, B = 1 so y = −2 e + e
−2 t
−
1 −3t
e
2
E1
5
(iii)
let u = e
−t
so y = −2u + u − u = − u (u − 2u + 4)
2
1
2
3
2
1
2
u = e−t ≠ 0 and u 2 − 2u + 4 = (u − 1) 2 + 3 > 0
hence y ≠ 0 for all t
M1
E1
consider quadratic (any valid method)
if discriminant used, value or working
must be shown
must indicate u non-zero
+ 89 > 0
M1
consider quadratic (any valid method)
hence no turning points
E1
(
y = 2 e−t − 2 e−2t + 32 e−3t = 32 u u 2 − 43 u + 43
((
= 32 u u − 32
)
2
)
)
B1
B1
starts at − 32 and asymptote y = 0
Shape (increasing)
6
1(i)
λ 2 + 6λ + 8 = 0
λ = −2 or − 4
M1
A1
F1
B1
M1
dM1
A1
F1
CF I = A e−2t + B e−4t
PI I = a e−t
a e −t − 6a e −t + 8a e −t = 6 e −t
3a = 6
a=2
I = 2 e −t + A e−2t + B e−4t
differentiate and substitute
compare
CF + PI
8
(ii)
λ 2 + 6λ + 9 = 0
λ = −3 (repeated)
M1
A1
F1
B1
M1
A1
CF I = (C + Dt ) e−3t
PI I = b e−t
b e−t − 6b e−t + 8b e −t = 6 e −t
b = 32
I = 32 e−t + (C + Dt ) e −3t
1.5 = 32 + C ⇒ C = 0
I = − 32 e−t − 3(C + Dt ) e−3t +
0 = − 32 − 3C + D ⇒ D = 32
I=
3
2
(e
−t
+ t e −3t
D e −3t
)
as t → ∞, I → 0
substitute and compare
F1
M1
M1
M1
CF + PI
condition on I
differentiate
condition on I
A1
cao
F1
recognise e−t → 0
12
(iii
)
λ 2 + 6λ + k = 0 ⇒ λ = −3 ± 9 − k
0 < k < 9 ⇒ 9 − k < 3 ⇒ two negative roots
hence I = A e−λ1t + B e−λ2t → 0
k > 9 ⇒ λ = −3 ± β j
e −3t ( A cos β t + B sin β t ) → 0
M1
E1
M1
complex roots with negative real part
(or CF)
E1
4
2(i)
1 dy 1
=
1 − 3 y dx x 2
1
1
∫ 1 − 3 y dy = ∫ x 2 dx
M1 separate
M1 integrate
A1
A1
M1
M1
A1
A1
1
− 13 ln 1 − 3y = − + c
x
1− 3y = Ae
3
x
y = 0, x = 1 ⇒ A = e−3
(
y = 13 1 − exp( 3x − 3)
)
x = 2 ⇒ y ≈ 0.259
± LHS
± RHS
rearrange
condition
from correct solution
8
(ii)
dy 3
1
+ y = 2 cos x
dx x
x
I = exp( ∫ 3x dx)
M1 divide
M1 attempt integrating factor
A1
=x
d 3
x y = x cos x
dx
3
( )
F1
x3 y = ∫ x cos x dx = x sin x − ∫ sin x dx
= x sin x + cos x + B
y = x −2 sin x + x −3 (cos x + B )
x = 1, y = 0 ⇒ B = − cos1 − sin1
y = x −2 sin x + x −3 (cos x − cos1 − sin1)
x = 2 ⇒ y ≈ 0.00258
(iii
)
(
(
y ′ = cos x − 3 x y + 0.1 y
2
)) x
M1 integrate (by parts)
RHS (or multiple) constant not
A1
required here
M1 divide to get y
M1 use condition
A1
M1 substitute x = 2
A1 cao
11
−2
y
y
0.034411 –0.12767
0.021644 –0.12380
0.00926(
3…)
Using smaller h would give greater
accuracy
x
1.8
1.9
2.0
follow their I
B1
seen or implied by correct numerical
value
M1 use algorithm
A1 y(1.9)
A1
y(2.0)
B1
5
3(i)
F = ma ⇒ mv
dv
= mg − 0.001mv 2
dx
dv
= g − 0.001v 2
dx
down positive so weight positive and resistance
negative as it opposes motion
⇒v
M1
N2L (accept just ma for M1)
E1
B1
3
(ii)
−0.002v
∫ g − 0.001v 2 dv = ∫ −0.002 dx
M1
separate
M1
integrate
ln g − 0.001v = −0.002 x + c
A1
LHS (or multiple)
M1
rearrange
M1
use condition
A1
cao
E1
must follow correct work
2
(
v = 1000 g − A e
2
−0.002 x
)
x = 0, v = 0 ⇒ A = g
(
v = 1000 g 1 − e
−0.002 x
)
x = 50 ⇒ v = 30.54
7
(iii)
dv
= mg − 2mv
dx
dv
v = g − 2v
dx
v
∫ g − 2v dv = ∫ dx
mv
M1
A1
M1
separate
A1
x+c
M1
attempt to integrate LHS
g ln g − 2v = x + c
x = 50, v = 30.54 ⇒ c = −74.91...
v = 5 ⇒ x = 76.36
so 26.4 m deep
A1
M1
M1
A1
use condition (correct value of v at least)
terminal velocity when acceleration zero
⇒ v = 4.9
M1
F1
1
2
⎛
g
⎞
∫ ⎜⎝ −1 + g − 2v ⎟⎠ dv = x + c
− 12 v − 14
9
(iv)
B1
B1
B1
follow their DE
increasing from (0,0)
decreasing to asymptote at 4.9 (or follow
their value)
cusp/max at (50, 30.54) (both coordinates
shown)
4.9
5
4(i)
x = x − 2 y + cos t
= x − 2(4 x − 3 y + cos t ) + cos t
= x − 8 x + 6 y − cos t
M1
M1
differentiate
substitute
y = 12 ( x + sin t − x )
M1
M1
E1
y in terms of x, x
substitute
x = x − 8 x + 3( x + sin t − x ) − cos t
x + 2 x + 5 x = 3sin t − cos t
5
(ii)
λ + 2λ + 5 = 0
λ = −1 ± 2 j
2
CF x = e−t ( A cos 2t + B sin 2t )
M1
M1
A1
F1
PI x = a sin t + b cos t
x = a cos t − b sin t , x = − a sin t − b cos t
−a − 2b + 5a = 3
−b + 2a + 5b = −1
a = 12 , b = − 12
x = e−t ( A cos 2t + B sin 2t ) + 12 ( sin t − cos t )
(iii)
( x + sin t − x )
M1
−t
−t
x = − e ( A cos 2t + B sin 2t ) + e ( −2 A sin 2t + 2 B cos 2
M1
+ 12 ( cos t + sin t )
y=
1
2
y = e−t ( ( A − B) cos 2t + ( A + B) sin 2t ) + 12 ( sin t − cos t )
(iv)
B1
M1
M1
M1
A1
F1
x~
y~
1
2
1
2
( sin t − cos t )
( sin t − cos t )
hence for large t , x ≈ y
but unless
A = B = 0,
A − B ≠ A or A + B ≠ B so x ≠ y
M1
A1
A1
auxiliary equation
solve to get complex roots
CF for their roots (for complex roots
must be in exp/trig form, not
complex exponentials)
differentiate twice and substitute
compare
solve
CF + PI
10
y in terms of x, x
differentiate x
substitute for x, x
CF part
PI part
5
F1
F1
B1
B1
must follow correctly from their
solutions
must follow correctly from their
solutions
4
4758
1(i)
Mark Scheme
λ =0
June 2006
B1
x = A cos 5t + B sin 5t
M1
cos 5t or sin 5t or A cos ωt + B sin ωt seen
or GS for their λ
A1
3
(ii)
2
(2λ ) − 4 ⋅ 5 < 0
0<λ < 5
M1
A1
A1
Use of discriminant
Correct inequality
Accept lower limit omitted or − 5
α 2 + 2α + 5 = 0
α = −1 ± 2 j
M1
A1
Auxiliary equation
x = e −t ( C cos 2t + D sin 2t )
F1
CF for their roots
3
(iii)
3
(iv)
x0 = C
x& = − e ( C cos 2t + D sin 2t ) + e
0 = −C + 2 D
D = 12 x0
−t
(
x = x0 e −t cos 2t + 12 sin 2t
−t
( −2C sin 2t + 2 D cos 2t )
)
M1
Condition on x
M1
M1
Differentiate (product rule)
Condition on x&
A1
cao
4
(v)
cos 2t + 12 sin 2t = 0
tan 2t = −2
t = 1.017
M1
M1
A1
cao
3
(vi)
α 2 + 6α + 5
α = −1, −5
x = E e −t + F e−5t
x0 = E + F
M1
A1
F1
M1
CF for their roots
Condition on x
x& = − E e −t − 5 F e−5t
0 = − E − 5F
M1
Condition on x&
−5 t
A1
cao
−4t
M1
Attempt complete method
E1
Fully justified (only ≠ 0 required)
E=
5
x ,F
4 0
=−
x=
1
x
4 0
−t
x = 14 x0
1
x
4 0
( 5e − e )
e (5 − e )
−t
Auxiliary equation
t > 0 ⇒ 5 > e −4t , x0 > 0, e −t > 0 ⇒ x > 0 i.e. never zero
8
4758
2(i)
Mark Scheme
λ + 2 = 0 ⇒ λ = −2
CF x = A e
PI x = at + b
a + 2(at + b) = t + 1
2a = 1, a + 2b = 1
M1
A1
B1
M1
M1
a = 12 , b =
A1
−2 t
x=
1
t
2
1
4
+ + Ae
1
4
−2t
June 2006
Differentiate and substitute
Compare
F1
CF + PI
M1
Condition on x
F1
Follow a non-trivial GS
M1
A1
Integrating factor
dx
e 2t
+ 2 e2t x = e2t ( t + 1)
dt
B1
Multiply DE by their I
e2t x = ∫ e2t ( t + 1) dt
M1
Attempt integral
M1
Integration by parts
t = 0, x = 1 ⇒ 1 = + A
1
4
3 −2t
e
4
x = 12 t + 14 +
Alternatively:
I = exp
( ∫ 2 dt ) = e
2t
dt
( t + 1) − ∫
e2t x = 12 e2t ( t + 1) − 14 e2t + A
=
1 2t
e
2
1 2t
e
2
x = 12 t + 14 + A e
A1
−2t
t = 0, x = 1 ⇒ 1 = + A
1
4
x=
1
t
2
3 −2t
e
4
+ +
1
4
F1
Divide by their I (must also divide constant)
M1
Condition on x
F1
Follow a non-trivial GS
9
(ii)
2 dy 1
=
y dx x
2
1
∫ y dy = ∫ x dx
M1
Separate
M1
Integrate
M1
Make y subject, dealing properly with constant
M1
Condition
F1
y = 4√(their x in terms of t)
2 ln y = ln x + c
y=B x
( t = 0 ) , x = 1, y = 4 ⇒ y = 4
y=4
1
t
2
+ 14 + 43 e−2t
x
5
(iii)
dz 2
+ z=6
dx x
M1
Divide DE by x
M1
Attempt integrating factor
= x2
d 2
x z = 6 x2
dx
A1
Simplified
F1
Follow their integrating factor
x 2 z = 2 x3 + C
z = 2 x + Cx −2
( t = 0 ) , x = 1, z = 3 ⇒ C = 1
A1
I = exp
(∫
2
x
dx
)
( )
−2
z = 2x + x
t = 1 ⇒ x = 0.852
y = 3.69
z = 3.08
F1
Divide by their I (must also divide constant)
M1
Condition on z
A1
cao (in terms of x)
B1
B1
Any 2 values (at least 3sf)
All 3 correct (and 3sf)
10
4758
3(i)
Mark Scheme
dv 1
dv
→ ±∞
= f( x) so ( unless f( x) = 0 ) , v → 0 ⇒
dx v
dx
i.e. gradient parallel to v-axis (vertical)
dv
1
1
−
=0
x = 4000 ⇒ v =
2
dx 5000
50002
so if v ≠ 0 then gradient parallel to x-axis (horizontal)
June 2006
Consider
M1
E1
M1
E1
M1
A1
dv
dx
or
when v = 0, but not if
dx
dv
dv
=0
dx
Must conclude about direction
dv
Consider
when x = 4000
dx
Must conclude about direction
Add to tangent field
Several vertical direction indicators on x-axis
6
(ii)
V0 = 0.05 ⇒ probe reaches B
V0 = 0.025 ⇒ probe returns to A
(iii)
∫ v dv = ∫ ( (9000 − x)
1 2
v
2
=
−2
− (1000 + x)
−2
1
1
+
+c
9000 − x 1000 + x
1
V2
2 0
=
v2 =
2
2
1
+
+ V0 2 − 450
9000 − x 1000 + x
1
9000
1
+ 1000
+c
) dx
M1
A1
M1
A1
Attempt one curve
B1
B1
Must be consistent with their curve
Must be consistent with their curve
N.B. Cannot score these if curve not drawn
Attempt second curve
6
M1
Separate
M1
B1
A1
Integrate
LHS
RHS
M1
Condition
A1
6
(iv)
minimum when x = 4000
2
2
1
vmin 2 = 5000
+ 5000
+ V0 2 − 450
need vmin 2 > 0
vmin 2 > 0 if V0 2 >
V0 > 0.0377
1
450
4
− 5000
B1
M1
F1
M1
Clearly stated
Substitute their x into v or v 2
Their v 2 or v when x = 4000
For vmin 2 > 0
M1
Attempt inequality for V0 2
A1
cao
6
4758
4(i)
Mark Scheme
June 2006
&&
x = 2 x& − y&
= 2 x& − (5 x − 4 y + 18)
y = 2 x + 3 − x&
&&
x = 2 x& − 5 x + 4(2 x + 3 − x& ) − 18
&&
x + 2 x& − 3 x = −6
M1
M1
M1
M1
E1
E1
λ 2 + 2λ − 3 = 0
λ = 1 or − 3
M1
A1
F1
B1
B1
F1
M1
Auxiliary equation
M1
Differentiate x and substitute
A1
Constants must correspond with those in x
x = 2 + 2 e −3t
M1
M1
M1
F1
Condition on x
Condition on y
Solve
Follow their GS
y = 7 + 10 e −3t
F1
Follow their GS
B1
B1
Sketch of x starts at 4 and decreases
Asymptote x = 2
B1
B1
Sketch of y starts at 17 and decreases
Asymptote y =7
Differentiate first equation
Substitute for y&
y in terms of x, x&
Substitute for y
LHS
RHS
6
(ii)
CF x = A e−3t + B et
PI x = a
−3a = −6 ⇒ a = 2
x = 2 + A e−3t + B et
y = 2 x + 3 − x&
= 4 + 2 A e −3t + 2 B et + 3 − (−3 A e −3t + B et )
y = 7 + 5Ae
(iii)
−3t
+ Be
4 = 2+ A+ B
17 = 7 + 5A + B
A = 2, B = 0
t
CF for their roots
Constant PI
PI correct
Their CF + PI
y in terms of x, x&
9
7
9
4758
1(i)
Mark Scheme
λ2 − λ − 2 = 0
λ = −1 or 2
M1
A1
F1
CF y = A e −t + B e2t
PI y = a e
y = −2a e
−2t
−2 t
Jan 2007
Auxiliary equation
CF for their roots
B1
, y = 4a e
−2 t
M1
Differentiate twice
M1
Substitute
4a = 1
a = 14
M1
A1
Compare and solve
y = A e−t + B e2t + 14 e−2t
F1
Their CF with 2 constants + their PI
4a e
−2t
(
− −2 a e
−2 t
) − 2a e
−2t
=e
−2t
9
(ii)
0 = A+
B + 14
−t
M1
Use initial condition
M1
Use asymptotic condition
A1
cao
y = 0 ⇔ e − t = e −2 t ⇔ e t = 1 ⇔ t = 0
M1
E1
B1
B1
Valid method to establish 0 is only root
Complete argument
Curve satisfies both conditions
y ≠ 0 for t > 0 and consistent with their
solution
CF y = C e−t + D e 2t
F1
Correct or same as in (i)
PI y = bt e −t
B1
t →∞⇒e
y=
1
4
(e
−2 t
→ 0, e
− e−t
−2t
)
2t
→ 0, e → ∞ so y → 0 ⇒ B = 0
7
(iii)
y = b e−t − bt e−t , y = −2b e−t + bt e −t
(
)
−2b e−t + bt e−t − b e−t − bt e−t − 2b e−t = e−t
M1
Differentiate (product) and substitute
⇒ −2b − b = 1 ⇒ b = − 13
A1
cao
GS y = C e−t + D e2t − 13 t e−t
F1
Their CF + their non-zero PI
y = 0, t = 0 ⇒ C + D = 0
y→0⇒ D=0
M1
M1
Use condition
Use condition
y = − 13 t e−t
A1
cao
8
40
4758
2(i)
Mark Scheme
d
1
cos x = cot x
( ln sin x ) =
dx
sin x
E1
Jan 2007
Differentiate (chain rule)
1
(ii)
1 dy
= −2 cot 2 x
y dx
M1
Rearrange
∫ y dy = ∫ −2 cot 2 x dx
M1
Integrate
ln y = − ln sin 2 x + c
A1
A1
M1
A1
One side correct (ignore constant)
All correct, including constant
Rearrange, dealing properly with constant
1
y = A cosec 2 x
6
(iii)
dy
+ 2 y cot 2 x = k
dx
I = exp
( ∫ 2 cot 2 x dx )
M1
Attempt integrating factor
M1
A1
Integrate
Simplified form of IF
M1
Multiply by their IF
y sin 2 x = ∫ k sin 2 x dx
M1
Integrate both sides
= − k cos 2 x + A
A1
cao
y = A cosec 2 x − k cot 2 x
E1
= exp ( ln sin 2 x )
= sin 2 x
dy
sin 2 x + 2 y cos 2 x = k sin 2 x
dx
1
2
1
2
7
(iv)
x = 14 π , y = 0 ⇒ 0 = A
M1
y = − k cot 2 x
A1
1
2
(v)
y=
A − k cos 2 x
1
2
sin 2 x
A = 12 k ⇒ y =
1
2
=
(
2
A − k 1 − 2sin x
1
2
Use condition
B1
Increasing and through
B1
Asymptote x = 0
( 14 π , 0 )
4
)
B1
2sin x cos x
k sin x
cos x
which tends to zero as x → 0
Both double angle formulae correct (or small
angle approximations or series expansion)
Use expressions in general solution
M1
A1
M1
E1
Identify value of A
Correct solution, fully justified
B1
Must be from correct solution
6
41
4758
3(i)
Mark Scheme
dv
= mg − R
dt
dv
= g − k1v
dt
1
∫ g − k1v dv = ∫ dt
m
−
1
ln g − k1v = t + c1
k1
g − k1v = A e− k1t
Alternatively
B1
N2L equation (accept ma, allow sign errors)
E1
Must follow from correct N2L
M1
Separate and integrate
A1
LHS
M1
Rearrange (dealing properly with constant)
M1
M1
Attempt integrating factor
d k1t
e v = g e k1t
dt
Integrate
Auxiliary equation
CF A e − k1t
Constant PI ( g / k1 )
M1
Use condition
A1
Alternatively
t = 0, v = 0 ⇒ A = g
v=
(
g
1 − e− k1t
k1
)
Jan 2007
M1
M1
A1
(
)
E1
7
(ii)
x = ∫ v dt =
⎞
g⎛
1 −k t
⎜t + e 1 + B⎟
k1 ⎝ k1
⎠
t = 0, x = 0 ⇒ B = −
x=
1
k1
g⎛
1 −k t 1 ⎞
⎜t + e 1 − ⎟
k1 ⎝ k1
k1 ⎠
M1
Integrate v
A1
cao (including constant)
M1
Use condition
A1
cao
B1
dv
N2L with mk2 v 2 (accept ma or m )
dt
E1
Must follow from correct N2L
M1
Integrate
A1
LHS
M1
Rearrange (dealing properly with constant)
M1
Use condition
A1
cao
B1
M1
First line
Use algorithm
A1
0.98
M1
Use algorithm
A1
1.84116 (accept 3sf or better)
4
(iii)
dv
mv = mg − mk2 v 2
dx
v
dv
=1
2
g − k2 v dx
v
∫ g − k v 2 dv = ∫ dx
2
−
1
ln g − k2 v 2 = x + c2
2k 2
g − k 2 v 2 = C e −2 k 2 x
x = 0, v = 0 ⇒ C = g
v=
(
g
1 − e −2 k 2 x
k2
)
7
(iv)
t
0
0.1
v
0
0.98
0.2
1.8411
6
v
9.8
8.6115
7
5
42
4758
(v)
Mark Scheme
3
g − k3v 2 = 0 when v = 4 ⇒ k3 =
g
4
3
2
= 1.225
E1
Jan 2007
Deduce or verify value (must relate to
resultant force or acceleration being zero)
1
43
4758
Mark Scheme
Jan 2007
4(i)
subtracting ⇒ −5 x + 5 = 0
x =1
y=7
M1
A1
A1
Solve simultaneously
(ii)
x = −3x − y
= −3 x − (2 x − y + 5)
M1
M1
M1
M1
E1
Differentiate
Substitute for y
y in terms of x, x
Substitute
3
= −3 x − 2 x + ( − x − 3 x + 10 ) − 5
x + 4 x + 5 x = 5
5
(iii)
2
λ + 4λ + 5 = 0
λ = −2 ± j
M1
M1
A1
Auxiliary equation
Solve to get complex roots
CF x = e−2t ( A cos t + B sin t )
F1
CF for their roots
PI x = 55 = 1
B1
GS x = e ( A cos t + B sin t ) + 1
y = − x − 3 x + 10
F1
Their CF with 2 constants + their PI
M1
y in terms of x, x
( − A sin t + B cos t ) + 2 e ( A cos t + B sin t )
− 3e ( A cos t + B sin t ) − 3 + 10
= e −2t ( ( − A − B ) cos t + ( A − B ) sin t ) + 7
M1
Differentiate their x
M1
Substitute
A1
cao
M1
M1
Use condition on x
Use condition on y
A1
Both correct
B1
B1
B1
Through origin
Positive gradient at t = 0
Asymptote x = 1, or their non-zero constant
PI (accept oscillatory or non-oscillatory)
−2 t
= −e
−2 t
−2t
−2t
(iv)
10
t = 0, x = 0 ⇒ A + 1 = 0
t = 0, y = 0 ⇒ − A + B + 7 = 0
A = −1, B = 8
x = e−2t ( 8sin t − cos t ) + 1
y = − e−2t ( 7 cos t + 9sin t ) + 7
3
(v)
NB Oscillates about x = 1 , but not apparent at this
scale due to small amplitude
3
44
4758
1(i)
Mark Scheme
λ 2 + 4λ + 29 = 0
λ = −2 ± 5 j
M1
M1
A1
CF y = e −2t ( A cos 5t + B sin 5t )
F1
PI y = a cos t + b sin t
y = − a sin t + b cos t , y = − a cos t − b sin t
B1
M1
CF for their roots (if complex, must
be exp/trig form)
Correct form for PI
Differentiate twice
M1
Substitute
− a cos t − b sin t + 4 ( − a sin t + b cos t )
+29 ( a cos t + b sin t ) = 3cos t
4b + 28a = 3
−4a + 28b = 0
a = 0.105
b = 0.015
y = e −2t ( A cos 5t + B sin 5t ) + 0.105cos t + 0.015sin t
(ii)
June 2007
M1
M1
A1
F1
Auxiliary equation
Solve for complex roots
Compare coefficients (both sin and
cos)
Solve for two coefficients
Both
GS = PI + CF (with two arbitrary
constants)
11
t = 0, y = 0 ⇒ 0 = A + 0.105
⇒ A = −0.105
y = −2 e −2t ( A cos 5t + B sin 5t )
M1
F1
Use condition on y
+ e −2t ( −5 A sin 5t + 5B cos 5t ) − 0.105sin t + 0.015cos t
t = 0, y = 0 ⇒ 0 = −2 A + 5B + 0.015
⇒ B = −0.045
M1
Differentiate (product rule)
M1
Use condition on y
y = − e−2t ( 0.105cos 5t + 0.045sin 5t ) + 0.105cos t + 0.015sin t
A1
cao
For large t, y ≈ 0.105cos t + 0.015sin t
M1
Ignore decaying terms
Calculate amplitude from solution of
this form
cao
2
2
amplitude ≈ 0.105 + 0.015 ≈ 0.106
M1
A1
8
(iii)
y (10π ) ≈ 0.105
B1
y (10π ) ≈ 0.015
B1
Their a from PI, provided GS of
correct form
Their b from PI, provided GS of
correct form
2
(iv)
y=e
−2t
( C cos 5t + D sin 5t )
F1
oscillations
with decaying amplitude (or tends to zero)
B1
B1
Correct or follows previous CF
Must not use same arbitrary
constants as before
Must indicate that y approaches
zero, not that y ≈ 0 for t > 10π
3
46
4758
2(i)
Mark Scheme
dy 2
1
− y = + x n −1
x
dx x
⎛ 2 ⎞
I = exp ⎜ ∫ − dx ⎟
x ⎠
⎝
= exp ( −2ln x )
−2
=x
d
yx −2 = x −3 + x n −3
dx
(
)
June 2007
M1
Rearrange
M1
Attempt IF
M1
Integrate to get k ln x
A1
Simplified form of IF
yx −2 = − 12 x −2 + n −1 2 x n − 2 + A
M1
A1
Multiply both sides by IF and recognise
derivative
Integrate
RHS including constant
y = − 12 + n −1 2 x n + Ax 2
F1
Their integral (with constant) divided by IF
From solution, x → 0 ⇒ y → − 12
B1
Limit consistent with their solution
From DE, x = 0 ⇒ 0 − 2 y = 1
M1
Use DE with x = 0
⇒ y=−
E1
Correctly deduced
M1
8
(ii)
1
2
3
(iii)
y = − 12 , x = 1 ⇒ − 12 = − 12 + n −1 2 + A
⇒ A=−
1
n−2
(
y = − 12 + n −1 2 x n − x 2
)
M1
Use condition
F1
Consistent with their GS and given
condition
n = 1, y = − 12 − x + x 2
B1
B1
–½
Shape for x > 0 consistent with their
solution (provided not y = constant)
Through (1, − 12 ) or (0, their value from (ii))
(1, –½)
4
(iv)
y = − 12 + x 2 ln x + Bx 2
M1
y (1) = − 12 + B
M1
Use result from (i) or attempt to solve from
scratch
Follow work in (i)
Integrate
RHS (accept repeated error in first term
from (i))
Divide by IF, including constant (here or
later)
Use condition at x = 1
M1
Use condition at x = 2
M1
Equate and solve
A1
cao
(
)
d
yx −2 = x −3 + x −1
dx
yx
−2
=− x
1
2
−2
M1
F1
M1
+ ln x + B
A1
y(2) =
− 12
+ 4 ln 2 + 4 B
y (1) = y (2) ⇒ 3B = −4 ln 2 ⇒ B = −
y=− +x
1
2
2
( ln x −
4
ln 2
3
)
4
ln 2
3
9
47
4758
3(i)
Mark Scheme
∫y
− 12
dy = ∫ − k (1 + 0.1cos 25t ) dt
2 y = −k ( t + 0.004sin 25t ) + c
1
2
t = 0, y = 1 ⇒ c = 2
(
y = 1 − 12 k ( t + 0.004 sin 25t )
)
2
June 2007
M1
Separate
M1
A1
A1
M1
F1
M1
A1
Integrate
LHS
RHS (condone no constant)
Use condition (must have constant)
Rearrange, dealing properly with constant
cao
8
(ii)
t = 1, y = 0.5 ⇒ 2 ( 0.5) = −k (1 + 0.004sin 25) + 2
M1
Substitute
⇒ k ≈ 0.586
E1
M1
A1
Calculate k (must be from correct solution)
Substitute
cao
M1
A1
A1
F1
Reasonable attempt at curve
From (0,1) and decreasing
Curve broadly in line with tangent field
Answer must be consistent with their curve
1
2
t = 2 ⇒ y = (1 −
1
× 0.586
2
( 2 + 0.004sin 50 ) )
2
≈ 0.172
4
(iii)
solution curve on insert
tank empty after 3.0 minutes
(iv)
4
x(0.1) = 1 + 0.1( −0.6446 )
M1
A1
E1
M1
A1
A1
= 0.93554
x(0.2) = 0.93554 + 0.1( −0.51985)
= 0.88356
–0.6446
Must be clearly shown
–0.51985
awrt 0.884
6
(v)
y < 0.01 ⇒ y < 0.1 ⇒ y + 0.1cos 25t < 0 for
some t
⇒
M1
dy
> 0 for some values of t
dt
E1
Consider size of
y and sign of
y + 0.1cos 25t
Complete argument
2
48
4758
4(i)
Mark Scheme
x = −5 x + 4 y − 2 e−2t
(
) − 2e
( x + 5x − e ) + 10 e
= −5 x + 4 −9 x + 7 y + 3e
= −5 x − 36 x + 28
4
−2t
−2t
−2t
−2 t
x − 2 x + x = 3e −2t
June 2007
M1
Differentiate
M1
Substitute for y
M1
M1
E1
y in terms of x, x
Substitute for y
5
(ii)
2
λ − 2λ + 1 = 0
λ = 1 (repeated)
M1
A1
Auxiliary equation
CF x = ( A + Bt ) et
F1
CF for their roots
B1
Correct form for PI
M1
Differentiate twice
M1
Substitute and compare
−2 t
PI x = a e
x = −2a e −2t , x = 4 a e −2 t
4a e
−2t
a=
1
3
(
− 2 −2 e
y=
=
1
4
1
4
)+ ae
−2 t
= 3e
−2t
A1
GS x =
(iii)
−2 t
1 −2 t
e +
3
( A + Bt ) e
t
F1
8
( x + 5x − e )
(−
−2 t
2 −2 t
e +
3
GS = PI + CF (with two arbitrary
constants)
B et + ( A + Bt ) et + 53 e −2t + 5 ( A + Bt ) et − e−2t
y = 14 et ( 6 A + B + 6 Bt )
)
M1
y in terms of x, x
M1
Differentiate x
x follows their x (but must use
product rule)
F1
A1
cao
4
(iv)
1
3
+A=0
M1
Condition on x
1
4
(6A + B) = 0
M1
Condition on y
A1
Both solutions correct
B1
Both values correct
B1
x through origin and consistent with
their solution for large t (but not
linear)
y through origin and consistent with
their solution for large t (but not
linear)
Gradient of both curves at origin
consistent with their values of x, y
A=
− 13 , B
=2
(
)
x = 13 e −2t + 2t − 13 et
y = 3t et
t = 0 ⇒ x = 1, y = 3
B1
B1
7
49
4758
Mark Scheme
4758
1(i
)
January 2008
Differential Equations
α 2 + 2α + 1 = 0
M1
Auxiliary equation
α = −1 (repeated)
A1
CF y = ( A + Bt ) e
F1
CF for their roots
PI y = a
in DE ⇒ y = 2
B1
B1
y = 2 + ( A + Bt ) e −t
F1
t = 0, y = 0 ⇒ 0 = 2 + A ⇒ A = −2
M1
Constant PI
PI correct
Their PI + CF (with two
arbitrary constants)
Condition on y
y = ( B − A − Bt ) e
M1
−t
−t
t = 0, y = 0 ⇒ 0 = B − A ⇒ B = −2
M1
y = 2 − ( 2 + 2t ) e
A1
−t
Differentiate (product rule)
Condition on y
10
(ii)
Both terms in CF hence will give zero if substituted in
LHS
PI y = bt 2 e −t
(
y = 2bt − bt
(
2
)e
−t
(
)e
+ 2 ( 2bt − bt ) + bt ) e
, y = 2b − 4bt + bt
2
E1
B1
−t
M1
Differentiate twice and
substitute
PI correct
Their PI + CF (with two
arbitrary constants)
Condition on y
t = 0, y = 0 ⇒ 0 = D − C ⇒ D = 0
M1
Condition on y
y = 12 t 2 e −t
A1
t > 0 ⇒ 12 t 2 > 0 and e −t > 0 ⇒ y > 0
E1
y = t − 12 t 2 e −t so y = 0 ⇔ t − 12 t 2 = 0 ⇔ t = 0 or 2
M1
Solve y = 0
Maximum at t = 2, y = 2 e −2
A1
B1
B1
B1
Maximum value of y
Starts at origin
Maximum at their value of y
y>0
in DE ⇒ 2b − 4bt + bt 2
⇒b=
(
1
2
2
2
−t
= e−t
A1
)
y = C + Dt + 12 t 2 e −t
t = 0, y = 0 ⇒ 0 = C
(
y = D + t − C − Dt −
M1
1 2
t
2
F1
)e
−t
8
(iii)
(
)
6
27
4758
2(i
)
Mark Scheme
dv
3
3
+
v=g−
dt 1 + t
1+ t
January 2008
M1
Rearrange
M1
A1
A1
Attempt integrating factor
Correct
Simplified
F1
Multiply DE by their I
)
(1 + t ) v = ∫ ( g (1 + t ) − 3 (1 + t ) ) dx
M1
Integrate
= 14 g (1 + t ) − (1 + t ) + A
A1
RHS
F1
Divide by their I (must also divide constant)
M1
Use condition
E1
Convincingly shown
I = exp
(1 + t )3
(
(∫
3
1+ t
)
3
3ln 1+ t
dt = e ( ) = (1 + t )
dv
2
3
2
+ 3 (1 + t ) v = g (1 + t ) − 3 (1 + t )
dt
d
(1 + t )3 v = g (1 + t )3 − 3 (1 + t )2
dt
3
3
4
2
3
v = 14 g (1 + t ) − 1 + A (1 + t )
−3
t = 0, v = 0 ⇒ 0 = g − 1 + A
1
4
(
)
v = 14 g (1 + t ) − 1 + 1 − 14 g (1 + t )
−3
10
(ii)
dv
(1 + t ) + 5v = (1 + t ) g
dt
dv
5
+
v=g
dt 1 + t
M1
Rearrange
M1
A1
Attempt integrating factor
Simplified
F1
Multiply DE by their I
(1 + t )5 v = ∫ g (1 + t )5 dx
M1
Integrate
= 16 g (1 + t ) + B
A1
RHS
F1
Divide by their I (must also divide constant)
M1
Use condition
F1
Follow a non-trivial GS
I = exp
(1 + t )5
(
(∫
5
1+ t
)
5
5ln 1+ t
dt = e ( ) = (1 + t )
dv
4
5
+ 5 (1 + t ) v = g (1 + t )
dt
)
d
(1 + t )5 v = g (1 + t )5
dt
6
v = 16 g (1 + t ) + B (1 + t )
−5
t = 0, v = 0 ⇒ 0 = g + B
1
6
(
v = 16 g 1 + t − (1 + t )
−5
)
9
(iii)
First model:
dv 1
−4
= g − 3 1 − 14 g (1 + t )
dt 4
As t → ∞, (1 + t )
(
−4
)
→0
Hence acceleration tends to
(
1
4
g
dv 1
−6
= g 1 + 5 (1 + t )
dt 6
Hence acceleration tends to 16 g
Second model
M1
Find acceleration
B1
Identify term(s) → 0 in their solution for either
model
A1
)
M1
Find acceleration
A1
5
28
4758
3(i)
Mark Scheme
P = A e0.5t
t = 0, P = 2000 ⇒ A = 2000
January 2008
M1
M1
A1
Any valid method
Use condition
CF P = A e0.5t
PI P = a cos 2t + b sin 2t
P = −2a sin 2t + 2b cos 2t
−2a sin 2t + 2b cos 2t = 0.5 ( a cos 2t + b sin 2t ) + 170sin 2t
F1
B1
M1
Correct or follows (i)
Differentiate
M1
Substitute
−2a = 0.5b + 170
2b = 0.5a
solving ⇒ a = −80, b = −20
M1
M1
A1
Compare coefficients
Solve
GS P = A e0.5t − 80 cos 2t − 20sin 2t
F1
Their PI + CF (with one arbitrary
constant)
M1
F1
Use condition
Follow a non-trivial GS
P = 2000 e0.5t
3
(ii)
(iii)
8
t = 0, P = 2000 ⇒ A = 2080
P = 2080 e0.5t − 80 cos 2t − 20sin 2t
(iv)
(v)
t
0
0.1
0.2
2
P
1000
1082.58
P
2000
2100
2208
(A) Limiting value ⇒ P = 0
1
2
⎛
P ⎞
⇒ P ⎜1 −
⎟ =0
⎝ 12 000 ⎠
(as limit non-zero) limiting value = 12000
M1
A1
A1
A1
Use of algorithm
2100
1082.5…
2208
M1
Set P = 0
M1
Solve
4
A1
3
(B) Growth rate max when
1
⎛
P ⎞2
f ( P ) = P ⎜1 −
⎟ max
⎝ 12 000 ⎠
1
⎛
⎛
P ⎞2
P ⎞
1
P ⎜1 −
f ′ ( P ) = ⎜1 −
⎟ −
⎟
×
12
000
2
12000
12
000 ⎠
⎝
⎠
⎝
⎛
P ⎞
1
P=0
f ′ ( P ) = 0 ⇔ ⎜1 −
⎟−
×
12
000
2
12000
⎝
⎠
⇔ P = 8000
− 12
M1
Recognise expression to maximise
M1
Reasonable attempt at derivative
M1
Set derivative to zero
A1
4
29
4758
4(i)
Mark Scheme
x = −3x + y
January 2008
M1
M1
M1
Differentiate first equation
Substitute for y
y in terms of x, x
M1
E1
Substitute for y
λ + 2λ + 2 = 0
λ = −1 ± j
M1
A1
Auxiliary equation
CF x = e −t ( A cos t + B sin t )
M1
F1
B1
B1
CF for complex roots
CF for their roots
Constant PI
PI correct
Their CF + PI (with two arbitrary
constants)
= −3 x + (−5 x + y + 15)
y = 3 x − 9 + x
x = −3x − 5 x + ( 3 x − 9 + x ) + 15
x + 2 x + 2 x = 6
5
(ii)
(iii)
2
PI x = a
2a = 6 ⇒ a = 3
GS x = 3 + e −t ( A cos t + B sin t )
F1
y = 3 x − 9 + x
M1
y in terms of x, x
M1
Differentiate x and substitute
A1
Constants must correspond with
those in x
x = 3 + 3e −t ( 2sin t − cos t )
M1
M1
F1
Condition on x
Condition on y
Follow their GS
y = 15e −t sin t
F1
Follow their GS
B1
B1
Sketch of x starts at origin
Asymptote x = 3
B1
Sketch of y starts at origin
Decaying oscillations (may
decay rapidly)
Asymptote y = 0
7
= 9 + 3e−t ( A cos t + B sin t ) − 9
− e−t ( A cos t + B sin t ) + e−t ( − A sin t + B cos t )
y = e −t ( ( 2 A + B ) cos t + ( 2 B − A ) sin t )
(iv)
3
0 = 3 + A ⇒ A = −3
0 = 2A + B ⇒ B = 6
4
(v)
B1
B1
5
30
4758
Mark Scheme
June 2008
4758 Differential Equations
1
(i)
2 x = 2 g − 8 ( x + 0.25 g ) − 2kv
M1
Weight positive as down, tension negative as
up.
Resistance negative as opposes motion.
⇒ x + kx + 4 x = 0
B1
E1
x = A cos 2t + B sin 2t
t = 0, x = 0.1 ⇒ A = 0.1
x = −2 A sin 2t + 2 B cos 2t so t = 0, x = 0 ⇒ B = 0
x = 0.1cos 2t
B1
M1
M1
A1
Find the coefficient of cos
Find the coefficient of sin
cao
α 2 + 2α + 4 = 0
α = −1 ± 3 j
M1
Auxiliary equation
N2L equation with all forces using given
expressions for tension and resistance
B1
Must follow correct N2L equation
4
(ii)
4
(iii)
(
x = e −t C cos 3 t + D sin 3 t
t = 0, x = 0.1 ⇒ C = 0.1
)
(
x = − e−t C cos 3 t + D sin 3 t
+e
−t
(−
)
3 C sin 3 t + 3 D cos 3 t
)
0 = −C + 3 D
D=
A1
M1
CF for complex roots
F1
CF for their roots
M1
Condition on x
M1
Differentiate (product rule)
M1
Condition on x
A1
cao
B1
B1
B1
Curve through (0,0.1) with zero gradient
Oscillating
Asymptote x = 0
M1
A1
A1
B1
Use of discriminant
Correct inequality
Accept k < −4 in addition (but not k > –4)
Curve through (0,0.1)
Decays without oscillating (at most one
intercept with positive t axis)
0.1
3
(
x = 0.1e−t cos 3 t +
1
3
sin 3 t
)
11
(iv)
k 2 − 4 ⋅1 ⋅ 4 > 0
(As k is positive) k > 4
B1
5
38
4758
2
(i)
Mark Scheme
x = A e −2t
t = 0, x = 8 ⇒ A = 8
x = 8e−2t
June 2008
M1
Any valid method
M1
A1
Condition on x
M1
M1
A1
Substitute for x
Auxiliary equation
3
(ii)
y + y = 16 e −2t
α + 1 = 0 ⇒ α = −1
CF y = B e −t
PI y = a e −2t
−2t
B1
−2t
−2t
−2a e + a e = 16 e
a = −16
GS y = −16 e −2t + B e −t
F1
t = 0, y = 0 ⇒ B = 16
M1
Differentiate and substitute
cao
Their PI + CF (with one arbitrary
constant)
Condition on y
F1
Follow a non-trivial GS
M1
M1
A1
B1
M1
A1
F1
Substitute for x
Attempt integrating factor
IF correct
(
−t
y = 16 e − e
−2t
M1
A1
)
Alternative mark scheme for first 7 marks:
I = et
d(y e t )/dt = 16e –t
y e t = –16e –t + B
y = –16e –2t + Be –t
(iii)
(
y = 16 e −t 1 − e −t
16 e
−t
)
M1
> 0 and t > 0 ⇒ e
−t
E1
B1
< 1 hence y > 0
B1
B1
Integrate
cao
Divide by their I (must divide constant)
Or equivalent (NB e –t > e –2t needs
justifying)
Complete argument
Starts at origin
General shape consistent with their
solution and y > 0
Tends to zero
9
5
(iv)
d
( x + y + z ) = ( −2 x ) + ( 2 x − y ) + ( y ) = 0
dt
⇒ x+ y+z =c
Hence initial conditions ⇒ x + y + z = 8
z = 8− x− y
(
−t
z = 8 1− 2e + e
−2t
) = 8 (1 − e )
−t 2
M1
Consider sum of DE’s
E1
E1
M1
Substitute for x and y and find z
E1
Convincingly shown (x, y must be correct)
5
(v)
(
0.99 × 8 = 8 1 − e
)
−t 2
t = −0.690638 or 5.29581
99% is Z after 5.30 hours
B1
Correct equation (any form)
B1
Accept value in [5.29, 5.3]
2
39
4758
3
(i)
Mark Scheme
y+
k
y =1
t
M1
Divide by t (condone LHS only)
M1
Attempt integrating factor
A1
F1
Integrating factor
Multiply DE by their I
d
yt k = t k
dt
M1
LHS
yt = ∫ t dt
M1
Integrate
I = exp
( ∫ dt ) = exp ( k ln t ) = t
k
t
k
t k y + kt k −1 y = t k
( )
k
=
k
1 t k +1
k +1
y=
1
t
k +1
+A
A1
cao (including constant)
+ At − k
F1
Divide by their I (must divide constant)
M1
Use condition
F1
Follow a non-trivial GS
t = 1, y = 0 ⇒ 0 =
(ii)
y=
1
k +1
y=
1
3
(
June 2008
(t − t )
1
+
k +1
A ⇒ A = − k 1+1
−k
t − t −2
10
)
B1
B1
B1
Shape consistent with their solution for t > 1
Passes through (1, 0)
Behaviour for large t
1
3
(iii)
yt −1 = ∫ t −1 dt
M1
Follow their (i)
= ln t + B
y = t ( ln t + B )
A1
F1
cao
Divide by their I (must divide constant)
t = 1, y = 0 ⇒ B = 0 ⇒ y = t ln t
A1
cao
4
(iv)
dy
= 1 + t −1 sin y
dt
t
1
1.1
1.2
y
0
0.1
0.2091
dy/dt
1
1.0908
M1
Rearrange DE (may be implied)
M1
A1
A1
Use algorithm
y(1.1)
y(1.2)
B1
Must give reason
M1
Identify effect of decreasing step length
A1
Convincing argument
M1
Identify derivative increasing
A1
Convincing argument
4
(v)
0.2138 as smaller step size
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.
Hence underestimates.
Alternative mark scheme for last 2 marks:
dy/dt seems to be increasing, hence Euler’s
method
will underestimate true value + sketch (or
explanation).
3
40
4758
4
(i)
Mark Scheme
June 2008
x = 4 x − 6 y − 9 cos t
M1
Differentiate first equation
= 4 x − 6(3x − 5 y − 7 sin t ) − 9 cos t
M1
Substitute for y
y=
1
6
( 4 x − x − 9sin t )
M1
y in terms of x, x
x = 4 x − 18 x + 5(4 x − x − 9sin t ) + 42sin t − 9 cos t
x + x − 2 x = −3sin t − 9 cos t
M1
E1
E1
Substitute for y
LHS
RHS
α2 +α − 2 = 0
α = 1 or − 2
Auxiliary equation
CF x = A et + B e−2t
PI x = a cos t + b sin t
( −ac − bs ) + ( −as + bc ) − 2 ( ac + bs ) = −3s − 9c
M1
A1
F1
B1
M1
−a + b − 2a = −9
M1
−b − a − 2b = −3
⇒ a = 3, b = 0
M1
A1
x = 3cos t + A et + B e−2t
F1
Their PI + CF (with two arbitrary
constants)
M1
y in terms of x, x
M1
Differentiate x and substitute
A1
Constants must correspond with
those in x
6
(ii)
(iii)
y=
=
1
6
1
6
9
( 4 x − x − 9sin t )
(12 cos t + 4 A e + 4B e
t
1
2
CF for their roots
PI of this form
Differentiate twice and substitute
Compare coefficients (2
equations)
Solve (2 equations)
t
−2t
+ 3sin t − A et + 2 B e −2t − 9sin t
y = 2 cos t − sin t + A e + B e
−2t
)
3
(iv)
(v)
x bounded ⇒ A = 0
M1
⇒ y bounded
E1
t = 0, y = 0 ⇒ 0 = B + 2 ⇒ B = −2
M1
x = 3cos t − 2 e −2t , y = 2 cos t − sin t − 2 e −2t
F1
x = 3cos t
y = 2 cos t − sin t
A1
A1
Identify coefficient of exponentially
growing term must be zero
Complete argument
2
Condition on y
Follow their (non-trivial) general
solutions
cao
cao
4
41
4761
Mark Scheme
January 2009
4758 Differential Equations
1(i)
α 3 + 2α 2 − α − 2 = 0
B1
E1
M1
A1
M1
F1
(−2) + 2(−2) − (−2) − 2 = 0
(α + 2)(α 2 − 1) = 0
α = −2, ± 1
y = Ae −2 x + Be − x + Ce x
3
2
Or factorise
Solve
Attempt CF
CF for their three roots
6
(ii)
PI y =
2
= −1
−2
GS y = −1 + Ae
M1 Constant PI
−2 x
+ Be
−x
+ Ce
A1
F1
x
Correct PI
GS = PI + CF
3
(iii)
e → ∞ as x → ∞
so finite limit ⇒ C = 0
x = 0, y = 0 ⇒ 0 = -1 + A + B
x = ln 2, y = 0 ⇒ 0 = −1 + 14 A + 12 B
Solving gives A = -2, B = 3
y = −2e −2 x + 3e − x − 1
M1
F1
M1
M1
M1
E1
x
Consider as x → ∞
Must be shown, not just stated
Use condition
Use condition
Convincingly shown
6
(iv)
−x
−x
y = −(2e − 1)(e − 1)
y = 0 ⇔ e − x = 12 or 1
⇔ x = ln 2 or 0
dy
= 4e − 2 x − 3e − x = e − x (4e − x − 3)
dx
dy
= 0 ⇔ e − x = 34 as e − x ≠ 0
dx
⇔ x = ln 43
Stationary point at (ln 43 , 18 )
M1 Solve
E1 Convincingly show no other roots
M1 Solve
E1
A1
Show only one root
B1
B1
Through (0, 0)
Through (ln 2, 0)
Stationary point at their answer to
(iv)
y → −1 as x → ∞
5
(v)
y
(ln(4/3),1/8)
ln2
x
B1
B1
—1
4
29
4761
2(i)
Mark Scheme
dy
+ y tan x = x cos x
dx
I = exp ∫ tan xdx
M1 Rearrange
M1 Attempt IF
= exp ln sec x
= sec x
d
( y sec x) = x
dx
y sec x = 12 x 2 + A
A1
A1
M1 Integrate
A1 RHS
Divide by their IF (must divide
F1
constant)
M1 Use condition
F1 Follow their non-trivial GS
x = 0, y = 1 ⇒ A = 1
y = ( 12 x 2 + 1) cos x
y
B1
B1
1
Shape correct for − π < x < π
Through (0,1)
1
2
10
1
2
π/2
—π/2
x
(iii)
Correct IF
Simplified
M1 Multiply and recognise derivative
y = ( 12 x 2 + A) cos x
(ii)
January 2009
2
x cos x sin x − y sin x
cos x
′
y (0) = 0
y (0.1) = 1
y ′(0.1) = −0.090351
y (0.2) = 1 + 0.1 × −0.090351 = 0.990965
y′ =
M1 Rearrange
B1
B1
B1
M1 Use of algorithm for second step
A1 3sf or better
6
(iv)
Same IF as in (i) or attempt from
scratch
I = sec x
M1
d
( y sec x) = x tan x
dx
A1
[ y sec x]xx==00.2 = ∫0
M1 Integrate
0.2
x tan xdx
A1 Accept no limits
M1 Substitute limits (both sides)
A1 Awrt 0.983
y (0.2) sec(0.2) − 1 × sec 0 ≈ 0.002688
⇒ y (0.2) ≈ 0.982701
6
30
4761
3(i)
Mark Scheme
60v
dv
= 60 g − 14 v 2
dx
January 2009
M1 N2L
1
v
dv
=
2
240 g − v dx 240
v
1
∫ 240 g − v 2 dv = ∫ 240 dx
1
1
− ln 240 g − v 2 =
x+c
240
2
A1
Correct N2L equation
E1
Convincingly shown
M1 Integrate
A1
ln 240 g − v 2 seen
A1
RHS
Rearrange, dealing properly
M1
with constant
M1 Use condition
x
120
240 g − v 2 = Ae
x = 0, v = 0 ⇒ A = 240 g
v = 240 g (1 − e
2
(ii)
(iii)
−
x
120
)
−
10
x = 10 ⇒ v = 240 g (1 − e 120 ) ≈ 13.71
dv
60 = 60 g − 60v − 90 g
dt
1
dv
1
dv
= − g − v or
+v = − g
2
dt
2
dt
Cao
E1
Convincingly shown
A1
Correct DE
M1 Separate
ln v + 12 g = −t + k
M1 Integrate
A1 LHS
Rearrange, dealing properly
M1
with constant
v + 12 g = Ae −t
α + 1 = 0 ⇒ α = −1
CF Ae
1
M1 N2L
Solving DE (three alternative methods):
dv
∫ v + 12 g = ∫ − dt
or
9
A1
M1
M1
M1
PI
A1
−t
PI − 12 g
v = Ae −t − 12 g
Solve auxiliary equation
CF for their root
Attempt to find constant
All correct
or
M1 Attempt integrating
factor
I = et
d t
1
(e v) = − ge t
dt
2
1
et v = − get + A
2
1
v = Ae −t − g
2
v = 13.71, t = 0 ⇒ 13.71 = A −
M1 Multiply
M1 Integrate
A1 All correct
1
2
g ⇒ A = 18.61 M1 Use condition
v = 18.61e −t − 4.9
E1
31
Complete argument
8
4761
Mark Scheme
(iv) At greatest depth, v = 0
⇒ e −t =
4.9
⇒ t = 1.3345
18.61
1.3345
Depth = ∫
0
[
M1
M1 Integrate
]
1.3345
A1
Ignore limits
Use limits (or evaluate constant
M1
and substitute for t)
A1 All correct
0
= 7.17 m
4(i)
− 3 x − y + 7 = 0⎫
x =1
⎬⇔
2x − y + 2 = 0 ⎭
y=4
2
M1
M1
M1
M1
E1
M1
A1
M1
F1
−2 t
( A cos t + B sin t )
5
PI x = = 1
5
F1
5
CF for complex roots
CF for their roots
GS = PI + CF with two
arbitrary constants
6
y = −3x + 7 − x
y in terms of x, x
Differentiate their x (product
M1
rule)
A1 Constants must correspond
M1 Use condition on x
M1 Use condition on y
M1
x = −2e −2t ( A cos t + B sin t ) + e −2t (− A sin t + B cos t )
(v)
Differentiate
Substitute for y
y in terms of x, x
Substitute for y
Complete argument
Auxiliary equation
B1
GS x = 1 + e −2t ( A cos t + B sin t )
(iv)
6
B1
B1
(ii) x = −3 x − y
= −3x − (2 x − y + 2)
y = −3x + 7 − x
x = −3 x − 2 x − 3 x + 7 − x − 2
⇒ x + 4 x + 5 x = 5
(iii) α 2 + 4α + 5 = 0
⇒ α = −2 ± i
CF e
Set velocity to zero and attempt
to solve
A1
(18.61e −t − 4.9)dt
= − 18.61e − t − 4.9t
January 2009
y = 4 + e −2t (( A − B) sin t − ( A + B) cos t )
1+A=4
4–A–B=0
A = 3, B = 1
x = 1 + e −2t (3 cos t + sin t )
y = 4 + e −2t (2 sin t − 4 cos t )
(vi)
A1
B1
B1
Both solutions
(0, 4)
→1
B1
B1
(0, 0)
→4
B1
Must refer to gradients
3
3
4
1
As the solutions approach the asymptotes, the
gradients approach zero.
5
32
4758
Mark Scheme
June 2009
4758 Differential Equations
1(i)
α 2 + 25 = 0
α = ±5j
CF y = Acos5t + Bsin5t
PI y = at cos5t + bt sin5t
y = a cos5t − 5at sin5t + b sin5t + 5bt cos5t
y = −10a sin5t − 25at cos5t + 10b cos5t − 25bt sin5t
In DE ⇒ 10b cos5t − 10a sin5t = 20cos5t
coefficients
⇒ b = 2, a = 0
PI y = 2t sin5t
GS y = 2t sin5t + Acos5t + Bsin5t
M1
A1
F1
B1
Auxiliary equation
CF for their roots
M1
M1
Differentiate twice
Substitute and compare
A1
F1 8
(ii)
t = 0, y = 1 ⇒ A = 1
y = 2sin5t + 10t cos5t − 5Asin5t + 5Bcos5t
t = 0, y = 0 ⇒ B = 0
y = 2t sin5t + cos5t
B1
M1
M1
A1 4
(iii)
Curve through (0,1)
Curve with zero gradient at (0,1)
Oscillations
Oscillations with increasing amplitude
B1
B1
B1
B1 4
(iv)
y = 2sin5t , y = 10cos5t , y = −50sin5t
y + 2y + 25y = −50sin5t + 20cos5t + 50sin5t
= 20cos5t
α 2 + 2α + 25 = 0
α = −1 ± 24 j
M1
E1
M1
A1
Substitute into DE
CF e−t (Ccos 24t + Dsin 24t )
F1
CF for their complex roots
F1
Their PI + their CF with two
GS y = 2sin5t + e−t (Ccos 24t + Dsin 24t )
arbitrary
From correct GS
Differentiate
Use condition on y
Auxiliary equation
constants
6
(v)
Oscillations of amplitude 2
oscillate
Compared to unbounded oscillations in first model
B1
or bounded oscillations; or both
B1
2
44
or equivalent;
or one bounded, one
unbounded
4758
Mark Scheme
dy 3
sin x
+ y= 2
dx x
x
3
I = exp dx
x
= exp(3lnx)
= x3
d 3
x y = x sin x
dx
derivative
June 2009
M1
Rearrange
M1
Attempting integrating factor
A1
A1
Correct and simplified
M1
Multiply and recognise
x3y = x sin xdx = − x cos x + cos xdx
M1
Integrate
= − cos x + sin x + A
y = (− x cos x + sin x + A) / x3
A1
A1
F1
All correct
Must include constant
2(i)
∫
( )
∫
∫
9
(ii)
⎛ ⎛ 1 ⎞
⎞
1
y ≈ ⎜ − x ⎜ 1 − x2 ⎟ + x − x3 + A⎟ / x3
6
⎝ ⎝ 2 ⎠
⎠
M1
Substitute given approximations
F1
1 A
= + 3
3 x
A=0
y = (sin x = x cos x) / x3
1
lim y =
x→ 0
3
M1
Use finite limit to deduce A
A1
B1
Correct particular solution
B1
Correct limit
6
y = 0 ⇒ sin x − x cos x = 0
(iii)
get tanx
⇒ tanx = x
M1
Equate to zero and attempt to
E1
Convincingly shown
2
(iv)
dy 3
1 1
+ y = − x, multiply by I = x3
dx x
x 6
d 3
1
(x y) = x2 − x4
dx
6
M1
Rearrange and multiply by IF
B1
Same IF as in (i) or correct IF
A1
Recognise derivative and RHS
M1
Integrate
A1
cao
M1
Use condition to find constant
correct
1
1
x3y = x3 − x5 + B
3
30
1 1
B
y = − x2 + 3
3 30
x
Finite limit ⇒ B = 0
45
4758
Mark Scheme
lim y =
x→ 0
1
3
June 2009
E1
Show correct limit (or same limit
as (ii)
7
3(a)(i) 2α + 4 = 0 ⇒ α = −2
CF Ae−2t
PI I = a cos2t + b sin2t
I = −2a sin2t + 2b cos2t
−4a sin2t + 4b cos2t + 4a cos2t + 4b sin2t = 3cos2t
3
−4a + 4b = 0,4b + 4a = 3 ⇒ a = b =
8
3
PI I = (cos2t + sin2t )
8
3
GS I = Ae−2t + (cos2t + sin2t )
8
arbitrary
M1
A1
B1
M1
M1
Find root of auxiliary equation
M1
Compare coefficients and solve
Differentiate
Substitute
A1
F1 8
Their PI + their CF with one
constant
(ii)
(iii)
3
3
t = 0, I = 0 ⇒ 0 = A + ⇒ A = −
8
8
3
I = (cos2t + sin2t − e−2t )
8
3
For large t, I ≈ (cos2t + sin2t )
8
M1
Use condition
A1 2
cao
M1
Consider behaviour for large t
(may be implied)
(b)(i)
(ii)
4(i)
3 2 2 3
Amplitude =
1 +1 =
2
8
8
Curve with oscillations with constant amplitude
Their amplitude clearly indicated
dy
(A) t = 0, y = 0 ⇒
= 2 − 2(0) + e0
dt
Gradient =3
dy
9
(B) At stationary point,
= 0, y =
dt
8
9
1
−t
−t
⇒ 0 = 2 − 2( ) + e ⇒ e =
8
4
⇒ t = ln4
dy
→ 0,e−t → 0
(C)
dt
Giving 0 = 2 − 2y + 0, so y → 1
A1
B1
B1 4
M1
Substitute into DE
A1
M1
Substitute into DE
M1
Solve for t
A1
M1
Substitute into DE
A1 7
Curve through origin with positive gradient
With maximum at (ln4, 9/8)
With y → 1 as x → ∞
B1
B1
B1 3
Follow their ln4
Follow their (C)
x = 7x + 6y − 6e−3t
M1
Differentiate
M1
Substitute for y
M1
y in terms of x, x ,t
−3t
= 7x + 6(−12x − 10y + 5sint ) − 6e
1
y = (x − 7x − 2e−3t )
6
46
4758
Mark Scheme
x = 7x − 72x − 10(x − 7x − 2e−3t ) + 30sint − 6e−3t
x + 3x + 2x = 14e−3t + 30sint
June 2009
M1
Substitute for y
E1
Complete argument
5
(ii)
x = ae−3t − 9cost + 3sint
x = −3ae−3t + 9sint + 3cost
x = 9ae−3t + 9cost − 3sint
In DE gives
9ae−3t + 9cost − 3sint
+3( −3ae−3t + 9sint + 3cost )
+2( ae−3t − 9cost + 3sint )
= 2ae−3t + 30sint
So PI with 2a =14
⇒a=7
AE α 2 + 3α + 2 = 0
α = −1, −2
CF Ae−t + Be−2t
GS x = Ae−t + Be−2t + 7e−3t − 9cost + 3sint
arbitrary
M1
Differentiate twice
M1
Substitute
E1
A1
M1
A1
Correct form shown
F1
F1
CF for their roots
Their PI + their CF with two
Auxiliary equation
8
(iii)
(iv)
(v)
1
x = (x − 7x − 2e−3t )
6
x = − Ae−t − 2Be−2t − 21e−3t + 9sint + 3cost
constants
M1
y in terms of x, x ,t
M1
F1
Differentiate GS for x
Follow their GS
4
3
y = − Ae−t − Be−2t − 12e−3t + 11cost − 2sint
3
2
x ≈ 3sint − 9cost
y ≈ 11cost − 2sint
x = y ⇒ 11cost − 2sint ≈ 3sint − 9cost
⇒ 20cost ≈ 5sint ⇒ tant ≈ 4
A1 4
cao
B1
B1
M1
A1 4
Follow their x
Follow their y
Equate
Complete argument
Amplitude of x ≈ 32 + 92 = 3 10
M1
Attempt both amplitudes
A1
One correct
A1 3
cao (accept reciprocal)
2
2
Amplitude of y ≈ 11 + 2 = 5 5
5
Ratio is
2
6
47
2
2
(b)
(i)
y
2
1
–1
0
1
x
–1
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4758/01 Ins Jan10
4758
Mark Scheme
January 2010
4758 Differential Equations
1(i)
α 2 + 6α + 9 = 0
α = −3 (repeated)
y = e-3t ( A + Bt )
PI y = a sin t + b cos t
y = a cos t − b sin t

y = −a sin t − b cos t
M1
A1
F1
Auxiliary equation
CF for their roots
B1
−a sin t − b cos t + 6 ( a cos t − b sin t )
+9 ( a sin t + b cos t ) = 0.5 sin t
8a − 6b = 0.5
8b + 6a = 0
Solving gives a = 0.04, b = −0.03
M1
Differentiate twice and substitute
M1
M1
A1
Compare coefficients
Solve
GS y = e −3t ( A + Bt ) + 0.04 sin t − 0.03 cos t
F1
PI + CF with two arbitrary constants
9
t = 0, y = 0  A = 0.03
M1
M1
F1
M1
Use condition
A1
Cao
(iii)
For large t , the particle oscillates
With amplitude constant ( ≈ 0.05)
B1
B1
Oscillates
Amplitude approximately constant
(iv)
t = 20π  e −60π very small
y ≈ −0.03
y ≈ 0.04
M1
A1
A1
(ii)
y = e
−3t
( B − 3 A − 3Bt ) + 0.04 cos t + 0.03 sin t
t = 0, y = 0  0 = B − 3 A + 0.04
y = 0.01( e −3t ( 3 + 5t ) + 4 sin t − 3 cos t )
Differentiate
Follows their GS
Use condition
5
2
(v)
y=e
−3t
3
( C + Dt )
M1
A1
CF of correct type or same type as in (i)
Must use new arbitrary constants
20π
B1√
B1
B1
y ≈ −0.03 at t = 20π
Gradient at 20π consistent with (iv)
Shape consistent
-0.03
5
29
4758
2(a)(i)
Mark Scheme
I = exp  − tan x dx
M1 Attempt IF
= exp ( −ln sec x )
A1
Correct IF
= ( sec x ) = cos x
A1
Simplified
−1
cos x
dy
− y sin x = sin x
dx
M1 Multiply by IF
d
( y cos x ) = sin x
dx
M1 Recognise derivative
M1 Integrate
A1 RHS (including constant)
A1 LHS
y cos x = − cos x + A
( y = A sec x − 1 )
dy
= (1+ y ) tanx
dx
ln(1+y)=lnsecx + A
(ii)
January 2010
x = 0, y = 0  0 = A − 1
y = sec x − 1
y
M1
A1
M1
A1
M1
A1
M1
A1
M1
A1
B1
B1
Rearrange equation
M1
A1
M1
A1
Attempt one curve
Reasonable attempt at one curve
Attempt second curve
Reasonable attempt at both curves
8
Separate variables
RHS
LHS
8
Use condition
Shape and through origin
Behaviour at ±½π
1
– π/2
π/2
x
4
(b)(i)
(ii)
4
y ' = (1 + y ) tan x
2
M1 Rearrange
x = 0, y = 1  y ' = 0
y ( 0.1) = 1 + 0.1 × 0 = 1
x = 0.1, y = 1  y ' = 0.201
M1 Use of algorithm
y ( 0.2 ) = 1 + 0.1 × 0.201
= 1.0201
M1 Use of algorithm for second step
E1
A1
5
(iii)
tan
π
2
undefined so cannot go past
π
M1
2
So approximation cannot continue to1.6 >
π
2
A1
2
(iv)
Reduce step length
B1
1
30
4758
3(i)
Mark Scheme
x = Ae − kt
M1
A1
x =  v1e dt
− kt
x=
M1 Integrate
v1 − kt
e +B
k
t = 0, x = 0  B =
Any valid method (or no method
shown)
A1
M1 Use condition
t = 0, x = v1  A = v1
x = v1e − kt
=−
January 2010
A1
v1
k
M1 Use condition
v1
1 − e− kt )
(
k
E1
8
(ii)

dy
= − kdt
y + g / k 
M1 Separate and integrate
g

In  y +  = − kt + C
k

A1
LHS
RHS
Rearrange, dealing properly with
M1
constant
A1
y +
g
= De − kt
k
t = 0, y = v2  D = v2 +
g
k
M1 Use condition
g
g

y =  v2 +  e − kt −
k
k


g
g
y =    v2 +  e − kt − dt
k
k

1
g
g
= −  v2 +  e − kt − t + E
k
k
k
1
g
t = 0, y = 0  0 = −  v2 +  + E
k
k
1
g
y = 2 ( kv2 + g ) (1 − e − kt ) − t
k
k
A1
M1 Integrate
A1
M1 Use condition
E1
1
0
(iii)
1 − e − kt =
kx
v1
M1
1  kx 
t = − ln  1 − 
k 
v1 
 kv + g 
g  kx 
y = 2
 x + 2 ln  1 − 
k
v1 
 kv1 

(iv)
A1
M1 Substitute
E1
Convincingly shown
4
x = 8  y = 4.686
Hence will not clear wall
M1
A1
2
31
4758
4(i)
Mark Scheme
4 y = −3x + 23 − x
4 y = −3x − 
x
1
1
( −3x − x ) = 2 x + ( −3x + 23 − x ) − 7
4
4
−3x − 
x = 8 x − 3x + 23 − x − 28
 
x + 2 x + 5 x = 5
January 2010
M1
M1
y or 4 y in terms of x, x
Differentiate
M1
Substitute for y
M1
E1
Substitute for y
5
(ii)
α + 2α + 5 = 0
 α = −1 ± 2i
2
CF e
−t
PI x =
( A cos 2t + B sin 2t )
5
=1
5
GS x = 1 + e − t ( A cos 2t + B sin 2t )
M1
A1
M1
F1
Auxiliary equation
B1
Constant PI
B1
Correct PI
PI + CF with two arbitrary
constants
F1
CF for complex roots
CF for their roots
7
(iii)
y=
=
1
( −3x + 23 − x )
4
M1
1
− 3 − 3e − t ( A cos 2t + B sin 2t ) + 23
4
+e − t ( A cos 2t + B sin 2t )
M1
F1
Differentiate and substitute
Expression for x follows their GS
−e − t ( −2 A sin 2t + 2 B cos 2t )
1
y = 5 − e − t ( ( A + B ) cos 2t + ( B − A ) sin 2t )
2
(iv)
4
t = 0, x = 8  1 + A = 8  A = 7
1
t = 0, y = 0  5 − ( A + B ) = 0  B = 3
2
−t
x = 1 + e ( 7 cos 2t + 3sin 2t )
y =5−e
(v)
A1
−t
M1
Use condition
M1
Use condition
A1
( 5cos 2t − 2sin 2t )
A1
4
For large t, e-t tends to 0
y→5
x→1
y>x
M1
B1
B1
E1
Complete argument
4
32
GCE
Mathematics (MEI)
Advanced GCE 4758
Differential Equations
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4758
1(i)
Mark Scheme
α2  4α  8  0
α  2  2 j
M1
A1
M1
F1
B1
CF e2 x ( A cos 2 x  B sin 2 x)
PI y  ax 2  bx  c
y  2ax  b, 
y  2a
(ii)
(iv)
Auxiliary equation
CF for complex roots
CF for their roots
Differentiate twice and
substitute
Compare coefficients
2a  4(2a x  b)  8(a x 2  bx  c)  32 x 2
M1
8a  32
8a  8b  0
2a  4b  8c  0
a  4, b  4, c  1
M1
GS y  4 x 2  4 x  1  e2 x ( A cos 2 x  B sin 2 x)
F1
x  0, y  0  A  1
y  8 x  4  e2 x (2 A sin 2 x  2 B cos 2 x
2 A cos 2 x  2 B sin 2 x)
x  0, y   0  0  4  (2 B  2 A)  B  1
M1
PI + CF with two arbitrary
constants
Use condition
M1
Differentiate (product rule)
M1
Use condition
A1
Cao
4
B1
B1
B1
Oscillates
Amplitude growing
2
B1
Minimum point at origin
Oscillates for x<0 with
growing amplitude
Approximately parabolic for
x>0
M1
A1
2 x
y  4 x  4 x  1  e (sin 2 x  cos 2 x)
x    y oscillates
With (exponentially) growing amplitude
y  (2 x  1)2 or 4 x 2  4 x  1
2
(iii)
June 2010
Solve
10
1
(v)
B1
B1
3
(vi)
At stationary point
dy
0
dx
M1
d2 y
 32 x 2  8 y
dx 2
d2 y
y 0 2 0
dx
minimum
Set first derivative (only) to
zero in DE
A1
So
M1
E1
1
Deduce sign of second
derivative
Complete argument
4
4758
2(a)(i)
Mark Scheme
IF  exp  2dt
M1
 e 2t
dy
e2t
 2e 2t y  1
dt
d 2t
(e y )  1
dx
e2t y  t  A
A1
M1*
June 2010
Attempt IF
Multiply by IF
A1
*M1A1
Integrate both sides
6
2 t
[ y  e (t  A)]
Alternative method:
CF y  Ee 2t
B1
B1
PI y  Fte 2t
M1
In DE: e2t ( F  2 Ft )  2 Fte2t  e2t
F=1
y  e2t (t  E )
(ii)
dz
 2 z  e 2t (t  A)
dt
I  e2t
d 2t
(e z )  t  A
dt
e2t z  12 t 2  At  B
2 t
(b)(i)
(ii)
(iii)
M1A1
F1
B1
Correct or follows (i)
M1
Multiply by IF and integrate
A1
z  e ( t  At  B)
t  0, z  1  1  B
z  2et ( 12 t 2  At  B)  e2t (t  A)
t  0, z  0  0  2 B  A  A  2
M1
M1
M1
z  e2t ( 12 t 2  2t  1)
A1
1 2
2
Alternative method:
PI x  ( Pt  Qt 2 )e 2t
P = A and Q = 0.5
z  e 2t ( 12 t 2  At  B)
Then as above
α  2  0  α  2
CF x  Ce 2t
PI x  a sin t  b cos t
x  a cos t  b sin t
In DE: a cos t  b sin t  2a sin t  2b cos t  sin t
a  2b  0, b  2a  1
 a  52 , b   15
Use condition
Differentiate (product rule)
Use condition
7
B1
Correct form of PI
M1A1
Complete method
B1
B1
CF correct
Correct form of PI
M1
M1
A1
Differentiate and substitute
Compare and solve
GS x  15 (2sin t  cos t )  Ce 2t
x  0, t  0  x  0 (from DE)
0   15  C
F1
Their PI + CF
M1
M1
Or differentiate
Use condition
x  15 (2sin t  cos t  e 2t
A1
For large t, x  15 (2sin t  cos t ) 
So x varies between 
1
5
5 and
5 sin (t-)
1
5
1
5
5
2
6
3
M1
Complete method
A1
Accept
x  15 5
2
4758
3(i)
Mark Scheme
y
dy    kdt
 12
1
2
2 y   kt  B
t  0, y  1  2  B
t  2, y  0.81  1.8  2k  2
 k  0.1
June 2010
M1
Separate and integrate
A1
A1
M1
M1
A1
LHS
RHS
Use condition
Use condition
1
y 2  1  0.05t
y  (1  0.05t )2
Valid for 1  0.05t  0 , i.e. t  20
y
1
20
(ii)
 πy dy   0.4dt
5
2
πy  0.4t  C
0.4 y
π (2 y  y 2 )
y
t
(iv)
1
0.987268
0.974614
Shape
Intercepts
M1
Separate and integrate
A1
A1
M1
LHS
RHS
Use condition
A1
y  
0
0.1
0.2
B1
B1
10
t  0, y  1  C  52 π
y  0.81  t  1.287
(iii)
√ on arithmetical error in k
t
3
2
2
5
A1
B1√
y
–0.12732
–0.12653
hy
–0.01273
–0.01265
If V  volume, v  velocity,
A  horizontal cross-sectional area,
dV
 k1v
then
dt
v  k2 y
dy dV

dt dt
dy
 A   k1k2 y
dt
dy

 k y
dt
A
M1
Rearrange (implied by
correct values)
M1
A1
M1
A1
Use algorithm
y (0.1) (awrt 0.987)
Use algorithm
y (0.2) (0.974 to 0.975)
M1
Rate of change of volume
M1
M1
E1
3
5
5
Relate rates of change of y
and volume
Eliminate volume and/or
velocity
Complete argument
4
4758
4(i)
Mark Scheme
5 y  2 x  9e2t  x
5 y  2 x  18e2t  
x
1
5
(2 x  18e
2 t
 
x)
(ii)
(iii)
M1
M1
Differentiate
Substitute for y
Substitute for y
E1
M1
A1
F1
B1
M1
 4a  4a  3a  e 2t  3e 2t
M1
a  1
A1
GS x  Aet  Be3t  e2t
F1
y  15 (2 x  9e 2t  x )
M1
M1
(2 Aet  2 Be 3t  2e 2t  9e 2t
( Aet  3Be3t  2e2t ))
(v)
y or 5y in terms of x, x
 
x  2 x  3 x  3e 2t
α 2  2α  3  0
 α  1, 3
CF Aet  Be 3t
PI x  ae 2t
x  2ae 2 t , 
x  4 a e 2 t
1
5
(iv)
M1
M1
 x  54 (2 x  9e2t  x )  3e 2t
June 2010
F1
y  15 Aet  Be 3t  e 2t
A1
t  0, x  0  0  A  B  1
t  0, y  2  2  15 A  B  1
 A  0, B  1
x  e 3t  e 2t
y  e3t  e2t
As t  , x  0, y  0
M1
M1
y (0)  2  A  0
M1
x, y   as t  
E1
5
Auxiliary equation
CF for their roots
PI of correct form
Differentiate and substitute
Compare coefficients and
solve
PI + CF with two arbitrary
constants
8
Differentiate and substitute
Expression for x follows
their GS
4
Use condition
Use condition
A1
A1
4
B1
4
Consider coefficient(s) of et
and mention of y < 2
Complete argument
3
GCE
Mathematics (MEI)
Advanced GCE
Unit 4758: Differential Equations
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4758
Mark Scheme
Question
1
(a) (i)
Answer
  2  5  0
  1  2 j
2
CF e  t ( A cos 2t  B sin 2t )
PI x  aet
x  aet , 
x  ae t
ae  2ae  5ae  4e
a  12
t
t
t
t
GS x  12 et  e t ( A cos 2t  B sin 2t )
Marks
M1
A1
M1
January 2011
Guidance
Auxiliary equation
CF for complex roots
F1
B1
M1
M1
A1
CF for their roots
F1
PI + CF with two arbitrary constants
Differentiate twice and substitute
Compare coefficients and solve
[9]
1
(a)
(ii)
t  0, x  0   A  0
1
2
M1
Use condition
x  e -e ( A cos 2t  B sin 2t )
M1
Differentiate (product rule)
e  t (2 A sin 2t  2 B cos 2t )
t  0, x  0  0  12  A  2 B
M1
Use condition
x  12 e - 12 e (cos 2t  sin 2t )
A1
cao
AE  3  4 2    6  0
13  4  12  1  6  0
(  1)( 2  5  6)  0
  1, 2 or  3
GS y  Ae x  Be 2 x  Ce 3 x
B1
E1
M1
Factorise (or solve by other means)
A1
F1
GS = CF with three arbitrary constants
1
2
t
t
t
t
[4]
1
1
(b)
(b)
(i)
(ii)
y0 A0
y (0)  1  B  C  1
y (0)  4  2 B  3C  4
B  1, C  2
y  2e 3 x  e 2 x
[5]
B1
M1
M1
Use condition
Use condition
A1
cao
[4]
4
4758
Mark Scheme
Question
1
(b) (iii)
Answer
y0e
3 x
Marks
Guidance
(2  e )  0
x
e 2
 x  ln 2
x
M1
A1
[2]
2
(a)
(i)
I  exp  2 xdx
M1
Attempt IF
 ex
2 dy
2
ex
 2 xe x y  sin x
dx
2
d
( ye x )  sin x
dx
A1
Correct IF
M1
Multiply by IF
M1
Recognise derivative
M1
Integrate
A1
M1
A1
F1
RHS (including constant)
Use condition
2
ye x   sin xdx
2
  cos x  A
x  0, y  0  1  1  A
 A2
y  e  x (2  cos x)
2
Divide by their IF, including constant
[9]
2
(a)
(ii)
 x2
e  0,cos x  1
 2  cos x  0  y  0
2
2
dy
 2 xe  x (2  cos x)  e  x (sin x)
dx
dy
x 0
0
dx
x  y0
M1
E1
Or use DE
E1
B1
B1
B1
January 2011
Through (0,1)
Shape consistent with results shown
[6]
5
4758
Mark Scheme
Question
2
(b) (i)
Answer
dy
 1  2 xy
dx
x
y
y
0
1
1
0.1
1.1 0.78
0.2 1.178
Marks
B1
M1
A1
A1
Guidance
First row
Use algorithm
1.1
1.178
[4]
2
(b)
(ii)
F1
I e
2
2
d
(ye x )  e x
dx
x2
x  0.2
M1
 ye x 
  e x dx

 x 0
0
A1
y (0.2)e0.2  1  0.2027
y (0.2)  1.15(55)
M1
  k  0    k
M1
A1
B1
M1
Differentiate
M1
Substitute and compare
2
0.2
2
2
3
(i)
 kx
CF Ae
PI y  a cos3 x  b sin 3 x
dy
 3a sin 3x  3b cos3 x
dx
3a sin 3x  3b cos3 x
 k (a cos3 x  b sin 3 x)  cos3 x
3a  kb  0
3b  ka  1
k
3
a
,b 
9  k2
9  k2
1
(k cos3 x  3sin 3 x)
y  Ae  kx 
9  k2
A1
[5]
Root of auxiliary equation
A1
A1
F1
January 2011
PI + CF with one arbitrary constant
[8]
6
4758
Mark Scheme
Question
3
(ii)
Answer
x  0, y   1  y  0 (from DE)
OR differentiate y
k
0  A
9  k2
1
( k cos3 x  3sin 3 x  ke  kx )
y
9  k2
Marks
M1A1
M1
A1
January 2011
Guidance
Use condition
[4]
3
(iii)
 kx
F1
B1
CF Be
PI y  cxe  kx
y   ce  kx  kcxe  kx
 kx
ce (1  kx)  kcxe
c2
y  Be  kx  2 xe  kx
 kx
 2e
 kx
M1
Differentiate
M1
Substitute and compare
A1
F1
PI + CF with one arbitrary constant
[6]
3
(iv)
d
(previous DE) with k  2
dx
y  Be 2 x  2 xe 2 x  C
x  0, y  0  0  B  C
y   2 Be 2 x  2e 2 x  4 xe 2 x
x  0, y   1  1  2 B  2
B   ,C 
1
2
M1
Recognise relationship
F1
B1
M1
Condition
Differentiate
M1
Use condition
NEED ALTERNATIVE SOLUTION
1
2
y   12 e 2 x  2 xe 2 x  12
A1
[6]
OR for first 2 marks
m 2  2m  0 ; CF y  Be 2 x  C
and PI y  pxe 2 x giving p  2
GS y  Be 2 x  2 xe 2 x  C
M1
A1
Complete method
7
4758
Mark Scheme
Question
4
(i)
4
(ii)
Answer
y  10 x  x
y  10 
x  x
10 
x  x  0.2 x  3x  0.3 x

x  0.4 x  0.05 x  0
 2  0.4  0.05  0
  0.2  0.1j
x  e0.2t ( A cos 0.1t  B sin 0.1t )
4
(iii)
x  0.2e0.2t ( A cos 0.1t  B sin 0.1t )
0.1e0.2t ( A sin 0.1t  B cos 0.1t )
y  10 x  x
e
0.2 t
(( A  B) cos 0.1t  ( B  A)sin 0.1t )
Marks
M1
M1
M1
M1
E1
[5]
M1
A1
M1
F1
[4]
M1
A1
M1
A1
Guidance
Eliminate y
Eliminate y
Auxiliary equation
CF for complex roots
CF for their roots
Differentiate (product rule)
Substitute to find y
[4]
4
(iv)
x0  A
y0  A  B
( x0 cos 0.1t  ( y0  x0 )sin 0.1t )
B1
M1
A1
( y0 cos 0.1t  ( y0  2 x0 )sin 0.1t )
A1
 y0
y  0 when tan 0.1t 
 1.25
y0  2 x0
So (for least positive t), t  22.5
 x0
1
x  0 when tan 0.1t 

y0  x0
9
So (for least positive t), t  30.3
Hence rabbits die out first
M1
F1
xe
0.2 t
ye
0.2 t
Use condition
[4]
4
(v)
January 2011
A1
M1
F1
Or compare values of tan 0.1t
A1
A1
Or compare values of tan 0.1t
Complete argument
[7]
8
GCE
Mathematics (MEI)
Advanced GCE
Unit 4758: Differential Equations
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4758
1(i)
Mark Scheme
June 2011
 2  4  3  0
  1 or -3
CF Ae  t  Be 3t
PI y  a cos 2t  b sin 2t
M1
A1
F1
B1
y  2a sin 2t  2b cos 2t
M1
Differentiate twice and
substitute
M1
Compare coefficients
Auxiliary equation
CF for their roots

y  4a cos 2t  4b sin 2t
4a cos 2t  4b sin 2t  8a sin 2t  8b cos 2t  3a cos 2t  3b sin 3t  13cos 2t
8b  a  13
b  8a  0
1
8
a , b
5
5
y  15 (8sin 2t  cos 2t )  Ae  t  Be 3t
GS
A1
A1
F1
PI + CF with two arbitrary
constants
M1
Use condition
M1
F1
M1
Differentiate
9
(ii)
t  0, y  0  0   15  A  B
y  15 (16 cos 2t  2 sin 2t )  Ae  t  3Be 3t
t  0, y  0  0  165  A  3B
13
3
, B=
10
2
13  t
1
y  5 (8sin 2t  cos 2t )  10
e  32 e 3t
 A
(iii)
(iv)
Use condition
A1
If z  y  c , differentiating (*) gives new DE
and has 3 arbitrary constants so must be GS
or
Integrating gives (*) with  k on RHS
PI will be previous PI  13 k , CF as before, so GS y  c
A1
Cao
M1
A1
Recognise derivative
6
M1
A1
SC1 for showing that correct y from (i) + c satisfies new DE
z  15 (8sin 2t  cos 2t )  De  t  Ee 3t  c
t  0, z  2  2   15  D  E  c
t
z  (16 cos 2t  2sin 2t )  De  3Ee
1
5
t  0, z  0  0 
16
5
3 t
 D  3E
2
M1
Use condition
F1
Derivative
M1
Use condition
Second derivative:
condone, for this mark
only, +c appearing
Use condition

z  15 (32sin 2t  4 cos 2t )  De  t  9 Ee 3t
F1
t  0, 
z  13  13  54  D  9 E
M1
13
3
,E ,c2
10
2
13  t
z  15 (8sin 2t  cos 2t )  10
e  32 e 3t  2
D
B1
A1
Cao
7
1
4758
2(a)(i)
Mark Scheme
I  exp(   2x dx)
M1
 exp(2 ln x )
M1
1
M1
A1
F1
y  2 x 2  Ax 2
3
(iii)
Multiply both sides by IF
M1
x 2 y  2 x  2  A
Integrate both sides
Must divide constant
8
0  2  A
y  2 x2  2 x
Attempt integrating factor
A1
A1
 x 2
dy
3
x 2
 2 x 3 y  x  2
dx
d 2
3
( x y)  x  2
dx
(ii)
June 2011
3
M1
A1
2
2
x  0, y  0
F1
dy
9
1
 4 x  3x 2  0  x 
(as x  0)
dx
16
dy
x  0,
0
dx
M1
F1
1
B1
B1
Behaviour at origin
and shape for x  1
Through
B1
Stationary point at
6
(b)(i)
Circle centre origin
Radius 1
B1
B1
2
(ii)
B1
B1
B1
One isocline correct
All three isoclines correct
Reasonably complete and accurate
direction indicators
3
(iii)
B1
Solution curve
(iv)
B1
B1
Solution curve
Zero gradient at origin
1
2
2
4758
3(a)(i)
Mark Scheme
N2L: ma  2k 2 x
dv
2v  2k 2 x
dx
dv
v
 k 2 x
dx
June 2011
M1
M1
Acceleration
E1
3
(ii)
 vdv   k
2
xdx
M1
Separate and integrate
x  a, v  0  A  12 k 2 a 2
A1
A1
M1
LHS
RHS
Use condition
v 2  k 2 (a 2  x 2 )
A1
1
2
v  k x A
2
1
2
2
So for v  0,
2
dx
 k a 2  x 2
dt
E1
6
(iii)
1

dx   kdt
a  x2
arcsin ax  B  kt
M1
Separate and integrate
x  a, t  0  B   12 
A1
A1
M1
LHS
RHS
Use condition
x  a sin( 12   kt )  a cos kt
A1
Either form
2
5
(b)(i)
 d   9sin  d
M1
Separate and integrate
  13  ,   0  C   92
A1
A1
M1
LHS
RHS
Use condition
So  2  9(2 cos   1)
A1
d
 3 2 cos   1 (decreasing)
dt
E1
1
2
  9 cos   C
2
6
(ii)
      0
1
3
M1
So estimate    0  
1
3
1
3
A1
The algorithm will keep giving   
1
3
B1
B1
but  is not constant so not useful
4
3
4758
4(i)
Mark Scheme
y   12 x  32 x  32 t
M1
y   
x  x 
M1
1
2
3
2
3
2
June 2011
 12 
x  32 x  32  2 x  ( 12 x  32 x  32 t )  t  2
M1
Eliminate
M1
E1
Eliminate

x  2 x  x  5t  1
M1
A1
F1
B1
Auxiliary equation
Root
CF for their root(s) (with two constants)
M1
Differentiate and substitute
M1
A1
F1
Compare and solve
5
(ii)
 2  2  1  0
  1 (repeated)
CF: ( A  Bt )e t
PI: x  at  b
x  a, 
x0
In DE: 0  2a  at  b  5t  1
a  5
2 a  b  1
a  5, b  9
GS: x  9  5t  ( A  Bt )e t
GS = PI + CF with two arbitrary constants
8
(iii)
y   12 x  32 x  32 t
M1
t
t
  [5  Be  ( A  Bt )e ]
1
2
t
 [9  5t  ( A  Bt )e ]  t
3
2
 9t  11  ( A  B  Bt )e
1
2
(iv)
3
2
t
M1
Differentiate (product rule)
M1
Substitute
A1
4
t  0, x  9  A  0
t  0, y  0  0  11  12 B  B  22
M1
M1
x  9  5t  22te  t
y  9t  11  (11  22t )e t
A1
A1
Use condition
Use condition
4
(v)
t
e 0
x  9  5t
y  9t  11
M1
F1
F1
3
4
4758
Mark Scheme
Question
1
(i)
June 2012
Marks
Answer
 2  6  9  0
  3 (repeated)
Guidance
M1
A1
F1
y   A  Bx  e 3 x
y  ax 2  bx  c
y   2ax  b, y   2a
B1
2a  6(2ax  b)  9(ax 2  bx  c)  x 2
9a  1
12a  9b  0
2a  6b  9c  0
1
4
2
a ,b ,c
9
27
27
1 2 4
2
y x 
x
 ( A  Bx) e3 x
9
27
27
M1
Differentiate and substitute
M1
M1
Compare at least two coefficients
Solve for at least two unknowns
A1
F1
Non-zero PI + CF with two arbitrary constants
[9]
1
(ii)
2
x  0, y  0  A  
27
dy 2
4
 x
 B e3 x  3( A  Bx) e 3 x
dx 9
27
4
x  0, y   0  0    B  3 A
27
2
B
27
1 2 4
2
2
y  x  x
 (1  x) e3 x
9
27
27 27
M1
Use condition
M1
F1
Differentiate using product rule
FT only c from (i)
M1
Use condition
A1
cao
[5]
1
(iii)
Both appear in CF
B1
[1]
1
(iv)
y   x 2 e 3 x
B1
dy
 2 x e3 x  3 x 2 e3 x
dx
M1
5
Or any clear statement or reason
Allow for y   x e   x e
Differentiate twice using product rule
3 3 x
2 3 x
4758
Mark Scheme
Question
June 2012
Marks
Answer
Guidance
2
d y
 2  e 3 x  12  x e 3 x  9  x 2 e 3 x
dx 2
in DE:

A1

e3 x 2  12 x  9 x 2  12 x  18 x 2  9 x 2  e3 x
1
2
 2  1   


y  A  Bx  x e
2
1
2
3 x
M1
Substitute into DE
M1
Compare coefficients
A1
Correct PI correct working only
F1
Non-zero PI + CF with two arbitrary constants
[7]
1
(v)
(A)
1
(v)
(B)
(e 3 x  0) , quadratic has +ve coefficient of x 2
so can have y  0 for all x for suitable A and B
would need quadratic < 0 for all x, but for large x it will
be positive
FT for any three term quadratic
B1
B1
Reasonably complete explanation or show for a specific example
FT for any three term quadratic
Reasonably complete explanation
For both marks to be awarded in (v), e 3 x  0 must be stated
[2]
2
(i)
dv
 mg  mkv 2
dx
dv
v  g  kv 2
dx
v
 g  kv 2 dv   dx
mv

M1
E1
1
ln g  kv 2  x  c1
2k
2 kx
g  kv  A e
x  0, v  0  A  g
2
v2 

g
1  e2 kx
k
N2L 3 terms, allow sign errors and any form for accn, including a

M1*
Separating variables. Integrating factor attempt gets zero
A1
LHS
A1
M1dep*
RHS (including constant on one side)
Rearrange, dealing properly with constant
M1dep*
Use condition
E1
[8]
2
(ii)
g
g
55   k 
3025
k
2
B1
[1]
6
Or 0.00324: cao
4758
Mark Scheme
Question
2
(iii)
June 2012
Marks
Answer
dv
m  mg  0.1mgv
dt
dv
 g (1  0.1v)
dt
EITHER
M1
Guidance
N2L 3 terms, allow sign errors and any form for accn, including a
A1
1
 1  0.1v dv   g dt
10 ln 1  0.1v  gt  c2
1  0.1v  B e0.1gt
t  0, v  54  B  4.4
v  10  44 e 0.1gt
M1
Separating variables
A1
A1
M1
M1
A1
[8]
LHS
RHS (including constant on one side)
Rearrange
Use condition
B1
FT their equation
M1
A1
M1
M1
A1
[8]
Multiply through by IF and integrate both sides
FT
Divide through by IF
Use condition
OR
e0.1gt
 10e0.1gt  A
Integrating factor
ve
0.1 gt
v  10  Ae
0.1 gt
t  0, v  54  A  44
v  10  44e
0.1 gt
OR
dv
 0.1gv  g
dt
Auxiliary equation m  0.1g  0
CF
M1
A1
B1
M1
M1
A1
[8]
Ae 0.1gt
PI v =10
GS v  10  A e 0.1gt
t  0, v  54  A  44
v  10  44e
0.1 gt
7
Use condition
4758
Mark Scheme
Question
2
(iv)
2
(v)
June 2012
Marks
Answer
1
 12  10 
t
ln 
  3.15 s
0.1g  44 
B1 FT
Guidance
FT for v obtained by a correct method
[1]
x   v dt  10t 
440 0.1gt
 c3
e
g
M1
F1
440
g
t  3.15  x  74.4
M1
A1 FT
t  0, x  0  c3 
3
(i)
dy 2
 y  x 2 sin x
dx x
 2 
I  exp    dx 
 x 
 e2ln x
 x 2
d 2
x y  sin x
dx

(ii)
FT for x obtained by a correct method
M1
A1
[6]
cao
M1
Divide both sides by x (NOTE that a MR (eg sinx missing) can earn 7/8)
M1
Allow 
A1
A1

x y   sin xdx
M1
B1
Multiply by their IF and attempt to integrate both sides
LHS must be their IF x y
x 2 y   cos x  A
A1
+ A is needed
  cos x  A 
F1
Divide every term by the multiplier of y (including a constant)
  cos   A 
[8]
M1
Use condition
A1
cao
2
3
FT attempt to integrate their answer to part (iii)
yx
2
0
2
 A  1
y   x 2 (cos x  1)
8
4758
Mark Scheme
Question
June 2012
Marks
Answer
B1
B1
B1
Guidance
Oscillations corresponding to their expression
Changing amplitude corresponding to their expression
y  0 with maxima on x-axis; cusps get B0; cao
Max B1 if MR has made graph simpler
[5]
3
(iii)
1 dy 2

y 2 dx x
1
 y2

dy  
2
dx
x
1
 2 ln x  B
y
1
y
2ln x  B
M1
Attempt to separate variables
Attempt at IF, M0 for this part
M1
Integrate both sides
A1
LHS
A1
RHS (including constant on one side)
A1
Any correct form of answer
[5]
3
3
(iv)
(v)
dy x sin x  2 y

dx
x
x
3.14
3.15
3.16
3
2
y
0
0.00015703
–0.000677
y’
0.015703
–0.083421
hy’
0.00015703
–0.00083421
Reduce step length (and carry out more steps)
9
B1
May be implied by correct values
M1
A1
A1
A1
[5]
B1
[1]
Use algorithm can get this even if B0
y(3.15) = 0.000157… Agreement to 3sf
–0.08342… Agreement to 3sf Award for -0.000834
–0.000676…Agreement to 3sf. ONLY award if this is the FINAL answer
No contradictions
4758
Question
4
(i)
Mark Scheme
Marks
Answer
y   x  2 x  6
y   
x  2 x
 
x  2 x  x  2 x  4 x  12  7

x  4 x  5 x  5
AE  2  4  5  0
  2  j
CF e
2 t
PI x 
June 2012
( A cos t  B sin t )
5
1
5
GS x  1  e2t ( A cos t  B sin t )
Guidance
M1
M1
M1
M1
A1
M1
A1
Substitute for y
Substitute for y
cao
M1
B1
B1
B1
CF for complex roots
CF for their roots
Appropriate form of PI
Correct PI
F1
Non-zero PI + CF with 2 arbitrary constants
cao
[12]
4
(ii)
dx
 2 e2t ( A cos t  B sin t )  e2t ( A sin t  B cos t )
dt
y   x  2 x  6  4  e2t ( A sin t  B cos t )
Alternative method:
dy
 1 e
2 t
M1
Differentiate using product rule
M1
A1
[3]
Substitute
cao
M1
Substitute for x in
 A cos t  B sin t   2 y  7
dt
dy
 2y  8  e
2 t
 A cos t  B sin t 
dt
2t
 (8e
y  4e
(iii)
 x  2 y  7 and arrange in correct IF form
dt
ye 
4
dy
2t
2 t
 A cos t  B sin t )dt
( A sin t  B cos t )
7  1 A
0  4B
x  1  e 2t (6 cos t  4sin t )
y  4  e 2t (6sin t  4 cos t )
M1
Find IF, multiply both sides by IF and show intention to integrate
A1
[3]
M1
M1
A1
Use condition on a GS
Use condition on a GS
cao
A1
[4]
10
cao
4758
Mark Scheme
Question
4
(iv)
Answer
Marks
y
 4 , so k  4
x
M1
y  4 x  6sin t  4 cos t  4(6 cos t  4sin t )
F1
M1
 10sin t  28cos t
 tan t  2.8
A1
which has infinitely many roots
E1
[5]
11
June 2012
Guidance
FT wrong constant term(s) and wrong form of expressions for x and y
The exponential term needs to have been cancelled
Or equivalent e.g. tan(t  0.59)  4
This can be awarded for a correct justification following a wrong value for
tant (i.e. can get A0 E1)
GCE
Mathematics (MEI)
Advanced GCE
Unit 4758: Differential Equations
Mark Scheme for January 2013
Oxford Cambridge and RSA Examinations
4758
Mark Scheme
Question
1
(i)
Answer
  2  5  6  0
3
2
23  2  2 2  5  2  6  0
(  2)(  1)(  3)  0
  (2), 1, 3
CF A e 2 x  B e  x  C e3 x
PI y  a sin x  b cos x
y   a cos x  b sin x
y    a sin x  b cos x
y   a cos x  b sin x
(a cos x  b sin x)  2( a sin x  b cos x)
5(a cos x  b sin x)  6(a sin x  b cos x)  sin x
6a  8b  0 
3
2
  a   25 , b  50
8a  6b  1
1
1
(ii)
(iii)
M1
Differentiate and substitute
M1
Compare coefficients and solve
A1
x
3
2
GS y  50
cos x  25
sin x  A e  B e  C e
2x
Marks
Guidance
M1
E1
Allow if implicit in factorisation of cubic
M1 Attempt roots (any method)
A1
F1
B1
3 x
F1
[10]
F1
F1
Use condition
3
2
y    50
sin x  25
cos x  B e  x  3C e3 x
M1
Differentiate
2  B  3C
x  0, y   0  0   25
M1
Use condition
bounded so A  0
3
x  0, y  1  1  50
 BC
B
29
,C
20

51
 100
y
3
29  x
51 3 x
2
cos x  25
sin x  20
e  100
e
50
3
2 sin x
y  50
cos x  25
1 32  4 2  1
amplitude  50
10
A1
F1
[6]
F1
M1
A1
Sketch showing oscillations with their
amplitude. More than one oscillation, ignore
origin
B1
[4]
5
January 2013
4758
Mark Scheme
Question
1 (iv)
Answer
bounded so B  C  0
3
47
x  0, y  1  1  50
 A  A  50
dy
47
2
  25
 2  50
0
dx
So no such solution
x 0
2
(i)
dv
dv
 9.8m  mkv 
 9.8  kv
dt
dt
1
EITHER 
dv   dt
9.8  kv
1
 ln 9.8  kv  t  c
k
9.8  kv  A e  kt
t  0, v  0  9.8  A
9.8
v
1  e  kt
k
OR Integrating factor ekt
N2L: m


ve kt   9.8e kt dt
ve kt 
9.8 kt
e A
k
t  0, v  0  A  


9.8
k
9.8
1  e  kt
k
OR Auxiliary equation   k  0
CF v  Ae  kt
9.8
PI v  b, v  0  b 
k
v
January 2013
Marks
B1
M1
Guidance
M1
Or A 
A1
[4]
www
1
25
E1
M1
Separate and integrate
A1
A1
M1
M1
LHS
RHS (including constant on one side)
Rearrange, dealing properly with constant
Use condition
A1
cao
M1
A1
Multiply both sides by IF and recognise
derivative on LHS
M1
Integrate both sides
A1
Must include constant
M1
Use condition
A1
cao
M1
A1
M1
6
4758
Mark Scheme
Question
Answer
GS v  Ae  kt 
9.8
k
t  0, v  0  A  
v

9.8
1  e  kt
k
Marks
January 2013
Guidance
A1
9.8
k

M1
Use condition
A1
cao
[7]
2
(ii)
2
(iii)
9.8
k
 1.4
7
B1
[1]
dv
 9.8m  0.2mv 2
dt
1
 9.8  0.2v 2 dv   dt
5
 49  v 2 dv  t  c2
5  1
1 


 dv  t  c2

14  7  v 7  v 
5
ln 7  v  ln 7  v   t  c2
14 
m
7  v 14
  t  c2 
7v 5
7v
 B e14t /5
7v
t  0, v  0  B  1
M1
M1
A1
RHS (including constant on one side)
M1
Integrate
A1
LHS
ln
M1
Rearrange into a form without ln, dealing
properly with constant
Use condition
7  v
M1
Rearrange to get v in terms of t
 e14t /5  1 
 1  e 14t /5 
v  7  14t /5   7 
14t /5 

1
e
1 e

A1
oe
7ve
14t /5
as t  , v  7  1100   7
M1
E1
[10]
7
4758
Mark Scheme
Question
2 (iv)
Answer
v  g  0.529v
t
v
0
0
0.1
0.98
0.2
1.9087
Marks
3/2
v
9.8
9.2868
2
(v)
v  0  g  0.529v3/2  v  7.00 (3 sf)
3
(a)
I  exp
hv
0.98
0.92868
   tan x dx 
 cos x
dy
cos x  y sin x  sin x cos x
dx
d
 y cos x   sin x cos x
dx
y cos x   sin x cos x dx
  12 sin 2 x dx
1
sin 2
2
xk)
x  0, y  1  1   14  c1
y
E1
M1
A1
A1
A1
[5]
E1
[1]
Guidance
Use algorithm
v(0.1)
v(0.1) at least 3d.p.
v(0.2) =1.91 to 3s.f.
Or 9.8  0.529  7
3
2
0
M1
 exp   ln sec x  or exp  ln cos x 
  14 cos 2 x  c1 (or
January 2013
sin 2 x  2
3  cos 2 x
5  cos 2 x
or y 
or y 
2cos x
2cos x
4cos x
A1
A1
M1
Multiply and recognise derivative
M1
Attempt integral
M1
Use identity, substitution or inspection on RHS
A1
oe (but must include constant)
M1
Use condition
A1
oe
[9]
8
4758
Mark Scheme
Question
3 (b)
3
(c)
(i)
Answer
p '( x)  f ( x)p( x)  g( x)
c '( x)  f ( x)c( x)  0
dy
 f ( x) y  p '( x)  Ac '( x)  f ( x)  p( x)  Ac( x) 
dx
 p '( x)  f ( x)p( x)  A  c '( x)  f ( x)c( x) 
 g( x)  A  0  g( x)
2
y  ex 
2
dy
 2x ex
dx
2
2 2
so LHS of DE  2 x e x  e x
x
2
2 
2  x 1
1
 2e x  x    2e x 

x

 x 
Marks
M1 Must be p( x ) oe
M1 Must be c( x ) oe
M1
Substitute in DE
M1
E1
[5]
Separate p and c terms
Complete argument
B1
M1
E1
[3]
3
(c)
(ii) dy
2y
1
2

  dy    dx
dx
x
y
x
ln y  2ln x  c2
M1
y  Ax 2
A1
A1
A1
OR Integrating factor = x 2
B1
d
 yx 2   0
dx
yx 2  A
y  Ax
2
LHS
RHS including constant
cao
M1
A1
A1
[4]
cao
9
January 2013
Guidance
4758
Mark Scheme
Question
3
(c) (iii)
Answer
y  e x  Ax 2
2
B1
1  e1  A
M1
y  e  1  e  x
x2
4
(i)
2
Guidance
Here A combines the arbitrary constants of (b)
and (c) (ii) into a single arbitrary constant.
Use condition
A1
  x  x  t 
y  23   
x  12 x  1
2
 
x  12 x  1  32 x  12  23   x  12 x  t   2t
3
y
2
3
Marks
1
2
2 
x  2 x  5 x  2  5t
AE 2 2  2  5  0
   12  32 j
[3]
M1
M1
Differentiate
M1
Substitute
A1
M1
oe
A1
 A cos 32 t  B sin 32 t 
M1
Correct form
PI x  a  bt
x  b, 
x  0  2b  5(a  bt )  2  5t
F1
B1
M1
FT wrong roots
2b  5a  2 
4
  a  5 , b  1
5b  5 
M1
A1
CF e
 12 t
GS x  54  t  e
 12 t
 A cos 32 t  B sin 32 t 
F1
[13]
10
Differentiate and substitute
Equate coefficients and solve
January 2013
4758
Mark Scheme
Question
4 (ii)
Answer
  x  12 x  t 
 t
x  1  12 e  A cos 23 t  B sin 23 t 
 t
 e   32 A sin 32 t  32 B cos 32 t 
 t
y  52  t  e  A sin 32 t  B cos 32 t 
y
2
3
1
2
1
2
1
2
4
(iii)
x  1, t  0  1   A  A 
4
5
2
5
M1
M1
F1
Must be using product rule
Must be GS from (i)
A1
M1
 15 cos 23 t  52 sin 23 t 
 t
y  52  t  e  15 sin 32 t  52 cos 32 t 
x  54  t  e
Guidance
[4]
M1
1
5
y  0, t  0  0  52  B  B 
Marks
January 2013
 12 t
1
2
A1
Both
[3]
4
(iv)
x y e
6
5
 12 t

3
sin 32 t
5

1
cos 23 t
5

M1
Adding and attempting the limit
 0  x  y  65
E1
FT for finite limit
x  y  65  53 sin 32 t  15 cos 32 t  0  tan 23 t  13
M1
which occurs (infinitely often)
E1
Establish equation and indicate method
Correctly investigate the existence of a
solution, but explicit solution for t not required.
t e
 12 t
[4]
11
4758
Mark Scheme
Question
1
(i)
Answer
2
2
 3  1  0
   1, 
CF
PI
Marks
M1
A1
F1
B1
1
2
x  Ae
t
 Be
t/2
x  a c o s t  b s in t
x   a s in t  b c o s t
,
 a  3b  1 
  a  
b  3a  0 
1
(ii)
x 
1
10
 3 s in t
x 
1
10
 3 cos t
1
10
 3 s in t
1
10
,b 
3
10
t

1
2
B
 A
 cos t  
3
10
t
 Be
t/2
 A B
 s in t   A e
x  0, t  0  0 
x 
1
10
 cos t   A e
x  0, t  0  0  
1
2
e
t

1
2
2
5
Be
e
t/2
t/2
Graph
1
1
(iii)
Guidance
Auxiliary equation
Correct roots
FT roots
x   a c o s t  b s in t
2 (  a c o s t  b s in t )  3(  a s in t  b c o s t )  ( a c o s t  b s in t )  c o s t
GS
June 2013
x 
1
10
 3 s in 1 0 
 c o s 1 0

 
x 
1
10
 3 c o s 1 0
 s in 1 0 


1
10
3
10
M1
M1
A1
Differentiate and substitute
Compare and solve
Correct values
F1
[8]
M1
FT CF with 2 arbitrary constants + PI
M1
Differentiate GS from (i)
M1
Use condition
A1
Correct expression
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]
M1
A1
A1
[3]
Use t  1 0  in either x or x soi
FT incorrect a and b from (i)
FT incorrect a and b from (i)
(iv)
3  4  2 1  1
M1
 0
A1
[2]
2
or state roots are real and distinct. Overdamped
5
Use condition
Consider auxiliary equation; either
discriminant or nature of roots found in
(i). Or state the CF again.
Must give a reason
4758
Mark Scheme
Question
1
(v)
Answer
x  C e

t
 De
10
 C e
1
10
x  C e
t
C  
10
1
2
e

 D e
1
2
Marks
B1
t/2
De
 5
t/2
,D 
2
5
e

3
10
 C e
10

1
2
D e
5
5
June 2013
Guidance
FT from (i)
M1
Use condition. Accept applied at
t  0
M1
Use condition. Accept applied at
t  0
A1
cao
 2 .2  1 0
x  
1
2
e
10  t

2
5
e
13
e
t
6
 2 .6 5  1 0 e
t/2
FT their C and D if correct t used
(1 0   t ) / 2
A1
sc Award for
x  
1
e
t
2

2
e
t/2
if new t
5
clearly defined.
[5]
2
(i)
EITHER:
500
dv
 10 000  500 g  kv
M1
dt
dv
 10 
k
500
dt
 10 

500
k
10 
1
k
500
k
500
A1
v
dv 
v
ln 1 0 
k
500
v  Ae
 dt
v  t c
 kt / 500
M1
Separate and integrate
A1
A1
LHS
RHS
Rearrange, dealing properly with
constant
Use condition
M1
t  0, v  0  A  10
M1
1  e
E1
v 
5000
k
N2L
 kt / 500

[8]
OR:
500
dv
 10 000  500 g  kv
M1
dt
dv
 10 
dt
k
500
A1
v
IF: e k t 5 0 0
ve
kt 500

e
N2L
B1
kt 500
M1
.1 0 d t
6
Multiply through by IF and recognise
LHS
4758
Mark Scheme
Question
Answer
5000

e
Marks
 A
kt 500
June 2013
Guidance
A1
Integrate RHS
M1
Use condition
M1
Rearrange
k
5000
t  0, v  0  A  
k
v 
5000
k
1  e
 kt / 500

E1
[8]
OR:
500
dv
 10 000  500 g  kv
M1
v
A1
dt
dv
 10 
dt
k
500
k
Auxiliary equation:  
 0
Use N2L
M1
500
CF: v  A e
kt

A1
500
5000
PI: v 
M1
Find PI
k
GS: v  A e

kt
500
5000

A1
k
5000
t  0, v  0  A  
M1
k
1  e
v 
5000
k
x 
 v dt
 kt / 500

Use condition
E1
[8]
2
(ii)

5000
k
t  0, x  0 
x 
5000
k
t 
t 
5000
k
500
k
e
500
k
e
 kt / 500
 5 0k 0   c 2
 kt /500
1

(  c2 )
 0

M1
Attempt to integrate both terms
M1
Use condition
A1
cao
[3]
7
4758
Mark Scheme
Question
2
(iii)
Answer
k  2 .5 , t  5  x 
v 
1  e
5 
e
500
2 .5
 1 2 .5 / 5 0 0
Marks
1

 1 2 .5 / 5 0 0
Guidance
M1
i.e. consistent
 1 2 4 .0 (1 d p )
5000
2 .5
5000
2 .5
June 2013
E1
  4 9 .4 to 1dp
B1
cao
[3]
2
(iv)
dv
500v
 1 0 0 0 0  5 0 0 g  0 .4 v
2
M1
dx
v
dv
 1 0  0 .0 0 0 8 v
2
N2L
E1
dx
[2]
2
(v)
v
 1 0  0 .0 0 0 8 v 2
dv 
 dx
M1
Separate variables: attempt to integrate
A1
A1
M1
LHS
RHS
Rearrange, dealing properly with
constant
Use correct condition
A1
oe
Graph
B1
B1
[8]
Increasing and condition shown
Asymptote at (awrt) 112
  2  0    2
M1
Auxiliary equation or IF :

1
0 .0 0 1 6
ln 1 0  0 .0 0 0 8 v
1 0  0 .0 0 0 8 v
2
 Be
2
 x  c3
 0 .0 0 1 6 x
M1
x  1 2 4 , v  4 9 .4  B  9 .8

v 
3
(a)
(i)
1 2 5 0 1 0  9 .8 e
 0 .0 0 1 6 x

d
Starting from (124, 49.4) FT
 ye  
2x
dx
CF
Ae
2 x
A1
B1
y  a c o s 2 x  b s in 2 x
y    2 a s in 2 x  2 b c o s 2 x
 2 a s in 2 x  2 b c o s 2 x  2 ( a c o s 2 x  b s in 2 x )  s in 2 x
M1
Differentiate and substitute
2a  2b  1 
  a  
2b  2 a  0 
M1
A1
Compare and solve
cao
F1
[7]
CF with one arbitrary constant + PI
y  
1
4
cos 2 x 
1
4
1
4
,b 
s in 2 x  A e
1
4
2 x
8
0
4758
Mark Scheme
Question
3
(a) (ii)
Answer
2  
y  
Marks
 A
1
4
cos 2 x 
1
4
1
4
s in 2 x 
9
4
e
June 2013
2 x
Graph
Guidance
M1
Use condition
A1
B1
cao
Starts from (0, 2) with negative gradient
Oscillations with constant amplitude and
Their function must have
frequency for large x. Scales not
exponential decay term
required
B1
[4]
3
(b)
(i)
I  exp
e
2x
dy
  2 dx   e
 2e
2x
y  e
2x
B1
x
dx
d
dx
e
e
2x
2x
y
e
e
y 
x
x
M1
dx
x
 e c
y  e
3
3
(b)
(c)
(ii)
x
 ce
2 x
x  0, y  2  2  1  c  c  1
y  e
x
I  e
2x
d
dx
e
2x
 e
y
2 x
e
x 1
1
e2x y 


 x0
2
y (1) e  2 e
0
2x
0 e
(i)
z  Ae
 2 .7 1 8 6 2
t
Conditions
 z  2e
A1
F1
[5]
M1
A1
[2]
M1
All correct
Divide through by IF
A1
A1
B1
A1
[6]
B1
B1
[2]
ta n x d x
y (1)  0 .6 3 8 6
4
Integrate
Use condition
cao
M1
ta n x
2x
M1
t
9
LHS with limits
RHS
LHS
0.639 or better
4758
Mark Scheme
Question
4
(ii)
Answer
x  x  2y  2e
t
y  x  3y  2e
t
y 
1
2
x  x  2e 
y 
1
2
x  x  2e 
1
2
x  4x  5x  4 e

x
3
2
x  x  2e  2e
t
Substitute
M1
Rearrange
M1
Differentiate
M1
Substitute
A1
M1
A1
M1
F1
cao
Auxiliary equation
Correct roots
Correct form of CF for their roots
FT their roots
B1
Correct form of PI
M1
A valid method to find a
A1
[12]
cao
t
t
 4  5  0
  2 j
CF
PI
e
 B cos t  C
2t
x  ae
x  a e
x 
GS
4
4
(iii)
(iv)
y 
t
1
2
2
5
s in t 
t
, x  ae
e
t
2t
 e
t
ae
 B cos t
t
 4a e
t
 5a e
t
 4e
t
 a 
 C s in t 
2
5
x  x  2e 
t
x  
2
5
e
y  
3
5
e
 2e
2t
 B cos t  C
t

2t
(B
1
2
x  1, t  0 
e
B 
3
5
,C  
x 
2
5
e
y  
3
5
e
s in t   e
2t
  B s in t
 C cos t 
M1
 C ) c o s t  ( B  C ) s in t 
A1
 B 1
2
5
y  0, t  0  
t
M1
t
3
5

1
2
B 
1
2
C  0
2t
3
5
e
t

3
5
e
 c o s t  3 s in t 
2t
 cos t 
Use condition
M1
Use condition
A1
2 s in t 
A1
[4]
10
Substitute x and x into expression for
y
Differentiate using product rule
cao aef, simplified to a maximum of 5
terms
[3]
M1
9
5

Guidance
M1
t
x  x  2e  
2
Marks
t
t
June 2013
FT if exponential term on
RHS of d.e.
4758
Question
4
(v)
Mark Scheme
Answer
x  y 
2
5
e
t
 
 e
t
 e
2t
3
5

e
3
5
t
e
2t

 3 s in t 
3
5
 cos t
e
2t
Marks
Guidance
 3 s in t 
 cos t 
 e
June 2013
3t
2 s in t 
 3 s in t
B1
M1
E1
[3]
3t
For t  0 , 0  e  1 so its graph
will meet the graph of 3 s in t infinitely often
11
Be convinced
Consider graphs oe
Complete argument