| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Systems of differential equations |
| Type | Complex roots giving oscillatory solution |
| Difficulty | Challenging +1.2 This is a structured multi-part question on coupled differential equations with a modeling context. Part (a) is routine algebraic manipulation (differentiate and substitute), part (b) requires solving a standard second-order DE with constant coefficients, part (c) applies initial conditions, and part (d) interprets the steady-state. While it involves several steps and techniques, each part follows standard procedures taught in Core Pure 2 with clear signposting. The coefficients are messy but the mathematical pathway is straightforward, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2y}{dt^2} = 0.032\frac{dx}{dt} - 0.025\frac{dy}{dt}\) and \(\frac{dx}{dt} = \frac{1}{0.032}\left(0.025\frac{dy}{dt} + \frac{d^2y}{dt^2}\right)\) | B1 | Differentiates the second equation with respect to \(t\) correctly. May have rearranged to make \(x\) the subject first. Dot notation allowed. |
| Substitutes for \(x\): \(\frac{d^2y}{dt^2} = 0.0008y - \frac{0.00144}{0.032}\left(\frac{dy}{dt} + 0.025y\right) + 0.064 - 0.025\frac{dy}{dt}\) | M1 | Uses the second equation to eliminate \(x\) to achieve an equation in \(y\), \(\frac{dy}{dt}\), \(\frac{d^2y}{dt^2}\) |
| \(40000\frac{d^2y}{dt^2} + 2800\frac{dy}{dt} + 13y = 2560\) | A1\* | Achieves the printed answer with no errors. Allow dot notation. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(40000m^2 + 2800m + 13 = 0 \Rightarrow m = \ldots\) | M1 | Uses the model to form and attempt to solve the auxiliary equation |
| CF: \(y = Ae^{m_1 t} + Be^{m_2 t}\) | M1 | Forms the complementary function correct for their roots. Must be in terms of \(t\) only (not \(x\)) |
| CF: \(y = Ae^{\frac{-t}{200}} + Be^{\frac{-13t}{200}}\) | A1 | Correct CF |
| PI: Try \(y = k \Rightarrow 13k = 2560 \Rightarrow k = \frac{2560}{13}\) | M1 | Chooses correct form of PI and uses complete method to find PI |
| GS: \(y = Ae^{-0.005t} + Be^{-0.065t} + \frac{2560}{13}\) | A1ft | Combines CF with correct PI. Accept awrt 197 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, y=0 \Rightarrow 0 = A + B + \frac{2560}{13}\) | M1 | Uses initial conditions to set up equation in \(A\) and \(B\) |
| \(t=0, y=0, x=0 \Rightarrow \frac{dy}{dt} = 0.032\times0 - 0.025\times0 = 0\) | B1 | Uses initial conditions to find value of \(\frac{dy}{dt}\) when \(t=0\). Can be implied. |
| \(\frac{dy}{dt} = -\frac{A}{200}e^{\frac{-t}{200}} - \frac{13B}{200}e^{\frac{-13t}{200}} = 0 \Rightarrow -\frac{A}{200} - \frac{13B}{200} = 0 \Rightarrow A = -13B\) | M1 | Differentiates GS, substitutes \(t=0\) and sets equal to initial value of \(\frac{dy}{dt}\) to form another equation in \(A\) and \(B\) |
| \(y = -\frac{640}{3}e^{\frac{-t}{200}} + \frac{640}{39}e^{\frac{-13t}{200}} + \frac{2560}{13}\) | A1 | Correct particular solution, accepting awrt 197 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t \to \infty\), \(e^{-kt} \to 0\) for \(k > 0\) so \(y \to \ldots\) | M1 | Uses the limit of the exponential terms being zero to find the long term limit of the concentration |
| \(y \to \frac{2560}{13} \approx 196\) or \(197\), so the rate of administration is sufficient to reach the required level. | A1ft | Follow through on their constant term and draws a relevant conclusion |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dt^2} = 0.032\frac{dx}{dt} - 0.025\frac{dy}{dt}$ and $\frac{dx}{dt} = \frac{1}{0.032}\left(0.025\frac{dy}{dt} + \frac{d^2y}{dt^2}\right)$ | **B1** | Differentiates the second equation with respect to $t$ correctly. May have rearranged to make $x$ the subject first. Dot notation allowed. |
| Substitutes for $x$: $\frac{d^2y}{dt^2} = 0.0008y - \frac{0.00144}{0.032}\left(\frac{dy}{dt} + 0.025y\right) + 0.064 - 0.025\frac{dy}{dt}$ | **M1** | Uses the second equation to eliminate $x$ to achieve an equation in $y$, $\frac{dy}{dt}$, $\frac{d^2y}{dt^2}$ |
| $40000\frac{d^2y}{dt^2} + 2800\frac{dy}{dt} + 13y = 2560$ | **A1\*** | Achieves the printed answer with no errors. Allow dot notation. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $40000m^2 + 2800m + 13 = 0 \Rightarrow m = \ldots$ | **M1** | Uses the model to form and attempt to solve the auxiliary equation |
| CF: $y = Ae^{m_1 t} + Be^{m_2 t}$ | **M1** | Forms the complementary function correct for their roots. Must be in terms of $t$ only (not $x$) |
| CF: $y = Ae^{\frac{-t}{200}} + Be^{\frac{-13t}{200}}$ | **A1** | Correct CF |
| PI: Try $y = k \Rightarrow 13k = 2560 \Rightarrow k = \frac{2560}{13}$ | **M1** | Chooses correct form of PI and uses complete method to find PI |
| GS: $y = Ae^{-0.005t} + Be^{-0.065t} + \frac{2560}{13}$ | **A1ft** | Combines CF with correct PI. Accept awrt 197 |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, y=0 \Rightarrow 0 = A + B + \frac{2560}{13}$ | **M1** | Uses initial conditions to set up equation in $A$ and $B$ |
| $t=0, y=0, x=0 \Rightarrow \frac{dy}{dt} = 0.032\times0 - 0.025\times0 = 0$ | **B1** | Uses initial conditions to find value of $\frac{dy}{dt}$ when $t=0$. Can be implied. |
| $\frac{dy}{dt} = -\frac{A}{200}e^{\frac{-t}{200}} - \frac{13B}{200}e^{\frac{-13t}{200}} = 0 \Rightarrow -\frac{A}{200} - \frac{13B}{200} = 0 \Rightarrow A = -13B$ | **M1** | Differentiates GS, substitutes $t=0$ and sets equal to initial value of $\frac{dy}{dt}$ to form another equation in $A$ and $B$ |
| $y = -\frac{640}{3}e^{\frac{-t}{200}} + \frac{640}{39}e^{\frac{-13t}{200}} + \frac{2560}{13}$ | **A1** | Correct particular solution, accepting awrt 197 |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t \to \infty$, $e^{-kt} \to 0$ for $k > 0$ so $y \to \ldots$ | **M1** | Uses the limit of the exponential terms being zero to find the long term limit of the concentration |
| $y \to \frac{2560}{13} \approx 196$ or $197$, so the rate of administration is sufficient to reach the required level. | **A1ft** | Follow through on their constant term and draws a relevant conclusion |\begin{enumerate}
\item A patient is treated by administering an antibiotic intravenously at a constant rate for some time.
\end{enumerate}
Initially there is none of the antibiotic in the patient.\\
At time $t$ minutes after treatment began
\begin{itemize}
\item the concentration of the antibiotic in the blood of the patient is $x \mathrm { mg } / \mathrm { ml }$
\item the concentration of the antibiotic in the tissue of the patient is $y \mathrm { mg } / \mathrm { ml }$
\end{itemize}
The concentration of antibiotic in the patient is modelled by the equations
$$\begin{aligned}
& \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.025 y - 0.045 x + 2 \\
& \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.032 x - 0.025 y
\end{aligned}$$
(a) Show that
$$40000 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2800 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 2560$$
(b) Determine, according to the model, a general solution for the concentration of the antibiotic in the patient's tissue at time $t$ minutes after treatment began.\\
(c) Hence determine a particular solution for the concentration of the antibiotic in the tissue at time $t$ minutes after treatment began.
To be effective for the patient the concentration of antibiotic in the tissue must eventually reach a level between $185 \mathrm { mg } / \mathrm { ml }$ and $200 \mathrm { mg } / \mathrm { ml }$.\\
(d) Determine whether the rate of administration of the antibiotic is effective for the patient, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel CP2 2023 Q9 [14]}}