Mathematics · Oct / Nov 2025 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. A container initially contains 6 litres of water. Water starts to flow into the container. The rate at which the water flows into the container is modelled by the differential equation V d t 12 t e t d where V litres is the volume of water in the container, t minutes after water starts flowing into the container.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[5]
−
t
−
t
−
t
−
t
−
t
= −
+
= −
−
t
e
d
t
t
e
e
d ,
t
t
e
e
d
V
−
t
−
t
−
t
=
= −
−
+
12 e
t
V
12 e
t
12e
c
d
t
=
=
=
−
+
=
t
0,
V
6
6
0 12
c
c
...
−
t
−
t
=
−
−
V
18 12 e
t
12e
8(b)[1]
No, because e.g. upper limit of volume is 18 (litres)
(a)
Note that differentiating using the product rule scores 0 marks
−
t
−
t
−
t
M1:
Uses integration by parts to achieve
t
e
d
t
= α
t
e
±β
e
(d )
t
where
αβ≠
but may
be equal to each other (you do not need to be
,
0
concerned about how they arrive at this) May
be implied by further
work.
dM1: Uses a correct method to integrate an expression of the form
−
t
−
t
−
t
t
e
(d )
t
= α
t
e
±γ
e
where ,
βγ≠
. It is dependent on the previous
0
method mark.
A1:
Solves the differential equation with or without a constant of integration
d
V
−
−
−
e.g.
t
t
t
(
)
o.e. May be written in an
=
12 e
t
V
= −
12 e
t
−
12e
+
c
d
t
(
)(
−
t
−
t
)
equivalent form e.g.
=
+
+
...
V
... e
t
e
c
Watch out for the DI method
D
I
12t
−
+
e t
−
−
e t
−
12
+
0
e t
−
−
t
−
t
−
t
= −
−
t
e
d
t
12 e
t
12e
Giving correct integration e.g.
M1:
Ether states
and states a value for c or uses the initial conditions
t
=
0,
V
=
6
for their integral to find the value of c. You do not need to be concerned
by the mechanics of the rearrangement. Whilst not dependent on the
earlier method marks, there must have been some attempt to solve the
differential equation however poor (but not clearly differentiation)
May alternatively substitute in limits V and 6 and t and 0 onto each side of
their integrated expression so the constant of integration is within the
calculation.
−
t
−
t
A1:
V
=
18 12 e
−
t
−
12e
oe Condone if seen in (b)
(b)
B1ft: No, because upper limit of volume is 18 (litres) oe Ft on their
−
t
−
t
V
=
"18"
−
At
e
−
B
e
where A, B > 0
May set
V =
and show that the equation cannot be solved.
20
t t
−
e.g. 2
−=
+
and explains that e
−>
t
and
so cannot be
12e (
1)
0
t + >
1
0
negative
If their "18"
>
then they need to conclude yes with a reason.
20
d
y
−
1
, d
x
2
d
y
2
t
+
5
=
=
= −
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