Question 1:
1 | (i) | dx
=−1(1+e−t)−2.−e−t
dt
e−t
=
(1+e−t)2
1
1−x=1−
1+e−t
1+e−t−1 e−t
1−x= =
1+e−t 1+e−t
1 e−t e−t
x(1−x)= =

1+e−t 1+e−t (1+e−t)2
dx
= x(1−x)

dt
1
x= =0.5
When t = 0,
1+e0 | M1
A1
M1
A1
E1
B1
[6] | chain rule
substituting for x(1 – x)
1+e−t−1 e−t
1−x= =
1+e−t 1+e−t
[OR,M1 A1 for solving
differential equation for t, B1
use of initial condition, M1
A1 making x the subject, E1
required form]
(ii) | 1 3
=
(1+e−t) 4
 e–t = 1/3
 t = –ln 1/3 = 1.10 years | M1
M1
A1
[3] | correct log rules
(iii) | 1 A B C
= + +
x2(1−x) x2 x 1−x
 1 = A(1 – x) + Bx(1 – x) + Cx2
x = 0  A = 1
x = 1  C = 1
coefft of x2: 0 = –B + C  B = 1 | M1
M1
B(2,1,0)
[4] | clearing fractions
substituting or equating
coeffs for A,B or C
A = 1, B = 1, C = 1 www
(iv) | dx
 dx=dt
x2(1−x)
1 1 1
 t=( + + )dx
x2 x 1−x
= –1/x + ln x – ln(1 – x) + c
When t = 0, x = ½  0 = –2 + ln ½ – ln ½ + c
 c = 2.
 t = –1/x + ln x – ln(1 – x) + 2
x 1
=2+ln − *
1−x x | M1
B1
B1
M1
E1
[5] | separating variables
-1/x + …
ln x – ln(1 – x) ft their A,B,C
substituting initial conditions
(v) | 3/4 1 2
t=2+ln − =ln3+ =1.77yrs
1−3/4 3/4 3 | M1A1
[2]
1
1−x=1−
1+e−t
1+e−t−1 e−t
1−x= =
1+e−t 1+e−t
1 e−t e−t
x(1−x)= =
1+e−t 1+e−t (1+e−t)2
dx
= x(1−x)
dt
dx
 dx=dt
x2(1−x)
1 1 1
t=( + + )dx
x2 x 1−x
2(i) (A) 9 / 1.5 = 6 hours
(B)1 /1.5 = 12 hours | B1
B1
[2]
dθ
(ii) =−k(θ−θ)
dt 0
dθ
⇒ ∫ =∫−kdt
θ−θ
0
⇒ ln(θ−θ)=−kt+c
0
θ−θ =e−kt+c
0
θ=θ +Ae−kt*
0 | M1
A1
A1
M1
E1
[5] | separating variables
ln(θ−θ)
0
–kt + c
anti-logging correctly(with c)
A=ec
(iii) 98 = 50 + Ae0
⇒ A = 48
dθ
Initially =−k(98−50)=−48k=−1.5
dt
⇒ k = 0.03125* | M1
A1
M1
E1
[4]
(iv) (A) 89=50+48e−003125t
⇒ 39/48 = e–003125t
⇒ t = ln(39/48)/(–0.03125) = 6.64 hours
(B) 80=50+48e−003125t
⇒ 30/48 = e–003125t
⇒ t = ln(30/48)/(–0.03125) = 15 hours | M1
M1
A1
M1
A1
[5] | equating
taking lns correctly for either
(v)Models disagree more for greater temperature loss | B1
[1]
3 (i) x = a(1 + kt)−1
⇒ dx/dt = −ka(1 + kt)−2
= −ka(x/a)2
= −kx2/a *
OR kt=a/x – 1, t= a/kx – 1/k
dt/dx= -a/kx²
⇒dx/dt= -kx²/a | M1
A1
E1
[3]
M1
A1
E1
[3] | Chain rule (or quotient rule)
Substitution for x
(ii) When t = 0, x = a ⇒ a = 2.5
When t = 1, x = 1.6 ⇒ 1.6 = 2.5/(1 +
k)
⇒ 1 + k = 1.5625
⇒ k = 0.5625 | B1
M1
A1
[3] | a = 2.5
(iii) In the long term, x → 0 | B1
[1] | or, for example, they die out.
(iv) 1 1 A B
= = +
2y−y2 y(1−y) y 2−y
⇒ 1 = A(2 − y) + By
y = 0 ⇒ 2A = 1 ⇒ A = ½
y = 2 ⇒ 1 = 2B ⇒ B = ½
⇒ 1 1 1
= +
2y−y2 2y 2(2−y) | M1
M1
A1
A1
[4] | partial fractions
evaluating constants by substituting
values, equating coefficients or
cover-up
(v) 1
∫ dy=∫dt
2y−y2
⇒ 1 1
∫[ + ]dy=∫dt
2y 2(2−y)
⇒ ½ ln y − ½ ln(2 − y) = t + c
When t = 0, y = 1 ⇒ 0 − 0 = 0 + c ⇒ c = 0
⇒ ln y − ln(2 − y) = 2t
⇒ y *
ln =2t
2−y
y
=e2t
2−y
⇒ y = 2e2t − ye2t
⇒ y + ye2t = 2e2t
⇒ y(1 + e2t) = 2e2t
⇒ 2e2t 2 *
y= =
1+e2t 1+e−2t | M1
B1 ft
A1
E1
M1
DM1
E1
[7] | Separating variables
½ ln y − ½ ln(2 − y) ft their A,B
evaluating the constant
Anti-logging
Isolating y
(vi) As t →∞ e−2t→ 0 ⇒ y → 2
So long term population is 2000 | B1
[1] | or y = 2
x 2y
4(i) cosθ= ,sinθ=
k k
cos2θ+sin2θ=1
⎛x⎞ 2 ⎛2y⎞ 2
⇒ ⎜ ⎟ +⎜ ⎟ =1
⎝k⎠ ⎝ k ⎠
x2 4y2
⇒ + =1
k2 k2
⇒ x2 + 4y2 = k2 * | M1
M1
E1
[3] | Used
substitution
dx dy 1
(ii) =−ksinθ, = kcosθ
dθ dθ 2
1
kcosθ
dy dy/dθ 2
= =−
dx dx/dθ ksinθ
= – ½ cot θ
x 2kcosθ 1 dy
− =− =− cotθ=
4y 4ksinθ 2 dx
or, by differentiating implicitly
2x + 8y dy/dx = 0
⇒ dy/dx = –2x/8y = –x/4y* | M1
A1
E1
M1 A1
E1
[3] | oe
(iii)k = 2 | B1
[1]
y
(iv)
k = 1
k = 2
1 2 3 4 x
k = 3
k = 4 | B1
B1
B1
[3] | 1 correct curve –shape and position
2 or more curves correct shape- in concentric
form
all 3 curves correct
(v)grad of stream path = –1/grad of contour
dy 1 4y
⇒ =− = *
dx (−x/4y) x | M1
E1
[2]
dy 4y dy 4dx
(vi) = ⇒∫ =∫
dx x y x
⇒ ln y = 4 ln x + c = ln ecx4
⇒ y = Ax4 where A = ec.
When x = 2, y = 1 ⇒ 1 = 16A ⇒ A = 1/16
⇒ y = x4/16 * | M1
A1
M1
M1
A1
E1
[6] | Separating variables
ln y = 4 ln x (+c)
antilogging correctly (at any stage)
substituting x = 2, y = 1
evaluating a correct constant
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