| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.3 This is a standard second-order differential equation problem with straightforward steps: converting weight to mass (trivial), finding a particular integral using the method of undetermined coefficients (routine for cos t forcing), finding complementary function (standard auxiliary equation), and applying initial conditions. While it has a real-world context and multiple parts, each step follows textbook procedures with no novel insight required. Slightly easier than average due to the guided structure and simple arithmetic. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Weight \(= \text{mass}\times g \Rightarrow m = \frac{30000}{g} = 3000\); but mass is in thousands of kg, so \(m=3\) | M1 | Correct explanation that in the model, \(m=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = 40\cos t + 20\sin t,\quad \frac{d^2x}{dt^2} = -40\sin t + 20\cos t\) | M1 | Differentiates the given PI twice |
| \(3(-40\sin t + 20\cos t)+4(40\cos t+20\sin t)+40\sin t - 20\cos t = \ldots\) | M1 | Substitutes into the given differential equation |
| \(= 200\cos t\), so PI is \(x = 40\sin t - 20\cos t\) | A1* | Reaches \(200\cos t\) and makes a conclusion |
| Or alternative method: Let \(x = a\cos t + b\sin t\) | M1 | Uses the correct form for the PI and differentiates twice |
| \(4b - 2a = 200,\quad -2b-4a = 0 \Rightarrow a=\ldots, b=\ldots\) | M1 | Substitutes into the given differential equation and attempts to solve |
| \(x = 40\sin t - 20\cos t\) | A1* | Correct PI |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3\lambda^2 + 4\lambda + 1 = 0 \Rightarrow \lambda = -1,\, -\frac{1}{3}\) | M1 | Uses the model to form and solve the auxiliary equation |
| \(x = Ae^{-t} + Be^{-\frac{1}{3}t}\) | A1 | Correct complementary function |
| \(x = PI + CF\) | M1 | Uses the correct notation for the general solution by combining PI and CF |
| \(x = Ae^{-t} + Be^{-\frac{1}{3}t} + 40\sin t - 20\cos t\) | A1 | Correct General Solution for the model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0,\, x=0 \Rightarrow A+B = 20\) | M1 | Uses the initial conditions of the model, \(t=0\) at \(x=0\), to form an equation in \(A\) and \(B\) |
| \(x=0,\, \frac{dx}{dt} = -Ae^{-t} - \frac{1}{3}Be^{-\frac{1}{3}t}+40\cos t+20\sin t = 0\) \(\Rightarrow A + \frac{1}{3}B = 40\) | M1 | Uses \(\frac{dx}{dt}=0\) at \(x=0\) in the model to form an equation in \(A\) and \(B\) |
| \(x = 50e^{-t} - 30e^{-\frac{1}{3}t} + 40\sin t - 20\cos t\) | A1 | Correct particular solution |
| \(t = 9 \Rightarrow x = 33\text{m}\) | A1 | Obtains 33m using the assumptions made in the model |
## Question 9:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Weight $= \text{mass}\times g \Rightarrow m = \frac{30000}{g} = 3000$; but mass is in thousands of kg, so $m=3$ | M1 | Correct explanation that in the model, $m=3$ |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 40\cos t + 20\sin t,\quad \frac{d^2x}{dt^2} = -40\sin t + 20\cos t$ | M1 | Differentiates the given PI twice |
| $3(-40\sin t + 20\cos t)+4(40\cos t+20\sin t)+40\sin t - 20\cos t = \ldots$ | M1 | Substitutes into the given differential equation |
| $= 200\cos t$, so PI is $x = 40\sin t - 20\cos t$ | A1* | Reaches $200\cos t$ and makes a conclusion |
| **Or alternative method:** Let $x = a\cos t + b\sin t$ | M1 | Uses the correct form for the PI and differentiates twice |
| $4b - 2a = 200,\quad -2b-4a = 0 \Rightarrow a=\ldots, b=\ldots$ | M1 | Substitutes into the given differential equation and attempts to solve |
| $x = 40\sin t - 20\cos t$ | A1* | Correct PI |
### Part (a)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3\lambda^2 + 4\lambda + 1 = 0 \Rightarrow \lambda = -1,\, -\frac{1}{3}$ | M1 | Uses the model to form and solve the auxiliary equation |
| $x = Ae^{-t} + Be^{-\frac{1}{3}t}$ | A1 | Correct complementary function |
| $x = PI + CF$ | M1 | Uses the correct notation for the general solution by combining PI and CF |
| $x = Ae^{-t} + Be^{-\frac{1}{3}t} + 40\sin t - 20\cos t$ | A1 | Correct General Solution for the model |
**(8 marks)**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0,\, x=0 \Rightarrow A+B = 20$ | M1 | Uses the initial conditions of the model, $t=0$ at $x=0$, to form an equation in $A$ and $B$ |
| $x=0,\, \frac{dx}{dt} = -Ae^{-t} - \frac{1}{3}Be^{-\frac{1}{3}t}+40\cos t+20\sin t = 0$ $\Rightarrow A + \frac{1}{3}B = 40$ | M1 | Uses $\frac{dx}{dt}=0$ at $x=0$ in the model to form an equation in $A$ and $B$ |
| $x = 50e^{-t} - 30e^{-\frac{1}{3}t} + 40\sin t - 20\cos t$ | A1 | Correct particular solution |
| $t = 9 \Rightarrow x = 33\text{m}$ | A1 | Obtains 33m using the assumptions made in the model |
**(4 marks)**
**(12 marks total)**
The image appears to be a blank page (page 42) from a Pearson Edexcel Level 3 Advanced GCE in Further Mathematics Sample Assessment Materials document. There is no mark scheme content visible on this page to extract.\begin{enumerate}
\item A company plans to build a new fairground ride. The ride will consist of a capsule that will hold the passengers and the capsule will be attached to a tall tower. The capsule is to be released from rest from a point half way up the tower and then made to oscillate in a vertical line.
\end{enumerate}
The vertical displacement, $x$ metres, of the top of the capsule below its initial position at time $t$ seconds is modelled by the differential equation,
$$m \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = 200 \cos t , \quad t \geqslant 0$$
where $m$ is the mass of the capsule including its passengers, in thousands of kilograms.\\
The maximum permissible weight for the capsule, including its passengers, is 30000 N .\\
Taking the value of $g$ to be $10 \mathrm {~ms} ^ { - 2 }$ and assuming the capsule is at its maximum permissible weight,\\
(a) (i) explain why the value of $m$ is 3\\
(ii) show that a particular solution to the differential equation is
$$x = 40 \sin t - 20 \cos t$$
(iii) hence find the general solution of the differential equation.\\
(b) Using the model, find, to the nearest metre, the vertical distance of the top of the capsule from its initial position, 9 seconds after it is released.\\
\hfill \mbox{\textit{Edexcel CP1 Q9 [12]}}