| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Applied/modelling contexts |
| Difficulty | Standard +0.3 This is a standard integrating factor question with straightforward algebra. Part (a) follows the routine method (divide by coefficient, find IF = (t+2)^3, integrate both sides), part (b) is simple substitution to find k then evaluate, and part (c) requires basic interpretation of the limiting behavior. The working is longer than minimal examples but requires no novel insight—just careful execution of a well-practiced technique. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((t+2)\frac{dv}{dt} + 3v = k(t+2) - 3 \Rightarrow \frac{dv}{dt} + \frac{3v}{t+2} = k - \frac{3}{t+2}\), \(I = e^{\int\frac{3}{t+2}dt} = (t+2)^3\) | M1 | Correct process to find integrating factor; must achieve form \(A(t+2)^3\) |
| \(v(t+2)^3 = \int\left(k(t+2)^3 - 3(t+2)^2\right)dt\) | M1 | Attempts solution using correct method with their IF |
| \(v(t+2)^3 = \frac{k}{4}(t+2)^4 - (t+2)^3 (+c)\) | A1 | Correct solution; \(+c\) may be missing for this mark |
| \(t=0, v=0 \Rightarrow c = 8-4k\) | M1 | Interprets initial conditions; must have constant of integration |
| \(v(t+2)^3 = \frac{k}{4}(t+2)^4 - (t+2)^3 + 8 - 4k\) \(\Rightarrow v = \frac{k}{4}(t+2) - 1 + \frac{4(2-k)}{(t+2)^3}\) | A1* | Obtains printed answer with no errors; \(8-4k\) must be factorised |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(v=4, t=2 \Rightarrow 4 = k-1+\frac{4(2-k)}{64} \Rightarrow k = \ldots\ (\approx 5.2)\) | M1 | Uses given conditions to establish value of \(k\) |
| \(v = \frac{\text{"5.2"}}{4}(5+2) - 1 + \frac{4(2-\text{"5.2"})}{(5+2)^3} = \ldots\) | dM1 | Uses their \(k\) with \(t=5\) to find \(v\) |
| Velocity is \(8.06\ \text{ms}^{-1}\) (awrt) | A1 | Correct value including units; accept exact answer \(5531/686\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| E.g. model suggests velocity increases indefinitely which is unlikely; raindrop will reach ground so model not valid for all \(t\); raindrop reaches terminal velocity after finite time | B1 | Must evaluate model with suitable comment on validity; not just a statement about what happens |
# Question 5:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $(t+2)\frac{dv}{dt} + 3v = k(t+2) - 3 \Rightarrow \frac{dv}{dt} + \frac{3v}{t+2} = k - \frac{3}{t+2}$, $I = e^{\int\frac{3}{t+2}dt} = (t+2)^3$ | M1 | Correct process to find integrating factor; must achieve form $A(t+2)^3$ |
| $v(t+2)^3 = \int\left(k(t+2)^3 - 3(t+2)^2\right)dt$ | M1 | Attempts solution using correct method with their IF |
| $v(t+2)^3 = \frac{k}{4}(t+2)^4 - (t+2)^3 (+c)$ | A1 | Correct solution; $+c$ may be missing for this mark |
| $t=0, v=0 \Rightarrow c = 8-4k$ | M1 | Interprets initial conditions; must have constant of integration |
| $v(t+2)^3 = \frac{k}{4}(t+2)^4 - (t+2)^3 + 8 - 4k$ $\Rightarrow v = \frac{k}{4}(t+2) - 1 + \frac{4(2-k)}{(t+2)^3}$ | A1* | Obtains printed answer with no errors; $8-4k$ must be factorised |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $v=4, t=2 \Rightarrow 4 = k-1+\frac{4(2-k)}{64} \Rightarrow k = \ldots\ (\approx 5.2)$ | M1 | Uses given conditions to establish value of $k$ |
| $v = \frac{\text{"5.2"}}{4}(5+2) - 1 + \frac{4(2-\text{"5.2"})}{(5+2)^3} = \ldots$ | dM1 | Uses their $k$ with $t=5$ to find $v$ |
| Velocity is $8.06\ \text{ms}^{-1}$ (awrt) | A1 | Correct value **including units**; accept exact answer $5531/686$ |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| E.g. model suggests velocity increases indefinitely which is unlikely; raindrop will reach ground so model not valid for all $t$; raindrop reaches terminal velocity after finite time | B1 | Must evaluate model with suitable comment on validity; not just a statement about what happens |
---\begin{enumerate}
\item A raindrop falls from rest from a cloud. The velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downwards, of the raindrop, $t$ seconds after the raindrop starts to fall, is modelled by the differential equation
\end{enumerate}
$$( t + 2 ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 v = k ( t + 2 ) - 3 \quad t \geqslant 0$$
where $k$ is a positive constant.\\
(a) Solve the differential equation to show that
$$v = \frac { k } { 4 } ( t + 2 ) - 1 + \frac { 4 ( 2 - k ) } { ( t + 2 ) ^ { 3 } }$$
Given that $v = 4$ when $t = 2$\\
(b) determine, according to the model, the velocity of the raindrop 5 seconds after it starts to fall.\\
(c) Comment on the validity of the model for very large values of $t$
\hfill \mbox{\textit{Edexcel CP1 2024 Q5 [9]}}