| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2020 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a standard second-order linear ODE with constant coefficients requiring complementary function (complex roots), particular integral, and application of initial conditions. While methodical, it involves multiple techniques (auxiliary equation with complex roots, finding particular solution, applying two initial conditions, analyzing maximum via calculus) and contextual interpretation across three parts, making it moderately challenging but still within typical A-level Further Maths scope. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral4.10g Damped oscillations: model and interpret |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(100m^2 + 60m + 13 = 0 \Rightarrow m = -0.3 \pm 0.2\text{i}\) | M1 | Uses model to form and solve auxiliary equation |
| \(x = e^{-0.3t}(A\cos 0.2t + B\sin 0.2t)\) | A1 | Correct CF, does not need \(x =\) |
| PI: \(x = 2\) | B1 | Correct PI |
| \(x = e^{-0.3t}(A\cos 0.2t + B\sin 0.2t) + 2\) | A1ft | Deduces correct GS (CF + PI). Must have \(x = f(t)\) and PI not 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 0\), \(x = 0 \Rightarrow A = -2\) | M1 | Uses model and initial conditions to establish value of \(A\) |
| \(\frac{\text{d}x}{\text{d}t} = -0.3e^{-0.3t}(-2\cos 0.2t + B\sin 0.2t) + e^{-0.3t}(0.4\sin 0.2t + 0.2B\cos 0.2t)\), \(t=0\), \(\frac{\text{d}x}{\text{d}t} = 10 \Rightarrow B = 47\) | M1 | Differentiates using product rule and uses initial conditions. Must use \(x=0\) and \(\frac{\text{d}x}{\text{d}t} = 10\) |
| \(x = e^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + 2\) | A1 | Correct particular solution |
| \(-0.3e^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + e^{-0.3t}(9.4\cos 0.2t + 0.4\sin 0.2t) = 0 \Rightarrow t = \ldots\) or \(x = \sqrt{2213}e^{-0.3t}\sin(0.2t - 0.0425) + 2\), \(\frac{\text{d}x}{\text{d}t} = -0.3\sqrt{2213}e^{-0.3t}\sin(0.2t - 0.0425) + 0.2\sqrt{2213}e^{-0.3t}\cos(0.2t - 0.0425)\), \(t = \ldots\) | M1 | Uses solution with correct strategy to obtain required value of \(t\), e.g. differentiates, sets equal to zero and solves |
| \(\tan 0.2t = \frac{100}{137} \Rightarrow 0.2t = 0.630\ldots\) or \(\tan(0.2t - 0.0425) = \frac{2}{3} \Rightarrow 0.2t = 0.630\) | M1 | Correct trigonometric approach leading to value for \(t\) |
| \(t = 3.15\ldots\) weeks | A1 | Correct value for \(t\) |
| \(x = e^{-0.3 \times 3.15\ldots}(47\sin(0.2 \times 3.15\ldots) - 2\cos(0.2 \times 3.15\ldots)) + 2\) | M1 | Uses model and their value of \(t\) to find maximum concentration |
| \(= \text{awrt } 12.1\ \{\mu\text{g/ml}\}\) | A1 | Correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 10 \Rightarrow x = e^{-3}(47\sin(2) - 2\cos(2)) + 2 = 4.16\ldots\) | M1 | Uses model to find concentration when \(t = 10\) |
| The model suggests it would be safe to give the second dose | A1ft | Makes suitable comment consistent with calculated value. Special case: if candidate's max value is less than 5, M1: never reaches 5 as maximum is… or max is less than 5; A1: yes, it is safe |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $100m^2 + 60m + 13 = 0 \Rightarrow m = -0.3 \pm 0.2\text{i}$ | M1 | Uses model to form and solve auxiliary equation |
| $x = e^{-0.3t}(A\cos 0.2t + B\sin 0.2t)$ | A1 | Correct CF, does not need $x =$ |
| PI: $x = 2$ | B1 | Correct PI |
| $x = e^{-0.3t}(A\cos 0.2t + B\sin 0.2t) + 2$ | A1ft | Deduces correct GS (CF + PI). Must have $x = f(t)$ and PI not 0 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 0$, $x = 0 \Rightarrow A = -2$ | M1 | Uses model and initial conditions to establish value of $A$ |
| $\frac{\text{d}x}{\text{d}t} = -0.3e^{-0.3t}(-2\cos 0.2t + B\sin 0.2t) + e^{-0.3t}(0.4\sin 0.2t + 0.2B\cos 0.2t)$, $t=0$, $\frac{\text{d}x}{\text{d}t} = 10 \Rightarrow B = 47$ | M1 | Differentiates using product rule and uses initial conditions. Must use $x=0$ and $\frac{\text{d}x}{\text{d}t} = 10$ |
| $x = e^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + 2$ | A1 | Correct particular solution |
| $-0.3e^{-0.3t}(47\sin 0.2t - 2\cos 0.2t) + e^{-0.3t}(9.4\cos 0.2t + 0.4\sin 0.2t) = 0 \Rightarrow t = \ldots$ or $x = \sqrt{2213}e^{-0.3t}\sin(0.2t - 0.0425) + 2$, $\frac{\text{d}x}{\text{d}t} = -0.3\sqrt{2213}e^{-0.3t}\sin(0.2t - 0.0425) + 0.2\sqrt{2213}e^{-0.3t}\cos(0.2t - 0.0425)$, $t = \ldots$ | M1 | Uses solution with correct strategy to obtain required value of $t$, e.g. differentiates, sets equal to zero and solves |
| $\tan 0.2t = \frac{100}{137} \Rightarrow 0.2t = 0.630\ldots$ or $\tan(0.2t - 0.0425) = \frac{2}{3} \Rightarrow 0.2t = 0.630$ | M1 | Correct trigonometric approach leading to value for $t$ |
| $t = 3.15\ldots$ weeks | A1 | Correct value for $t$ |
| $x = e^{-0.3 \times 3.15\ldots}(47\sin(0.2 \times 3.15\ldots) - 2\cos(0.2 \times 3.15\ldots)) + 2$ | M1 | Uses model and their value of $t$ to find maximum concentration |
| $= \text{awrt } 12.1\ \{\mu\text{g/ml}\}$ | A1 | Correct value |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 10 \Rightarrow x = e^{-3}(47\sin(2) - 2\cos(2)) + 2 = 4.16\ldots$ | M1 | Uses model to find concentration when $t = 10$ |
| The model suggests it would be safe to give the second dose | A1ft | Makes suitable comment consistent with calculated value. Special case: if candidate's max value is less than 5, M1: never reaches 5 as maximum is… or max is less than 5; A1: yes, it is safe |
---\begin{enumerate}
\item A scientist is investigating the concentration of antibodies in the bloodstream of a patient following a vaccination.\\
The concentration of antibodies, $x$, measured in micrograms ( $\mu \mathrm { g }$ ) per millilitre ( ml ) of blood, is modelled by the differential equation
\end{enumerate}
$$100 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 60 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 26$$
where $t$ is the number of weeks since the vaccination was given.\\
(a) Find a general solution of the differential equation.
Initially,
\begin{itemize}
\item there are no antibodies in the bloodstream of the patient
\item the concentration of antibodies is estimated to be increasing at $10 \mu \mathrm {~g} / \mathrm { ml }$ per week\\
(b) Find, according to the model, the maximum concentration of antibodies in the bloodstream of the patient after the vaccination.
\end{itemize}
A second dose of the vaccine has to be given to try to ensure that it is fully effective. It is only safe to give the second dose if the concentration of antibodies in the bloodstream of the patient is less than $5 \mu \mathrm {~g} / \mathrm { ml }$.\\
(c) Determine whether, according to the model, it is safe to give the second dose of the vaccine to the patient exactly 10 weeks after the first dose.
\hfill \mbox{\textit{Edexcel CP2 2020 Q3 [14]}}