| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 17 |
| 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 coupled differential equations problem requiring systematic elimination to form a second-order DE, solving the auxiliary equation (complex roots), finding particular integrals, and applying initial conditions. While it involves multiple steps and techniques (differentiation, substitution, complex exponentials/trigonometric form), each part is clearly signposted and follows standard procedures taught in Core Pure 1. The context is straightforward and part (e) only requires interpreting the solution behavior. More challenging than average due to length and coordination of techniques, but no novel insight required. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral4.10h Coupled systems: simultaneous first order DEs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2x}{dt^2} = -5\frac{dx}{dt} + 10\frac{dy}{dt}\) | B1 | Differentiates first equation w.r.t. \(t\) correctly |
| Substitutes second equation to eliminate \(y\): \(\frac{d^2x}{dt^2} = -5\frac{dx}{dt} + 10(-2x+3y-4)\) | M1 | Uses second equation to eliminate \(y\), achieving equation in \(x\), \(\frac{dx}{dt}\), \(\frac{d^2x}{dt^2}\) |
| \(\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 50\) | A1* | Printed answer, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m^2 + 2m + 5 = 0 \Rightarrow m = \ldots\) | M1 | Attempts auxiliary equation |
| \(m = -1 \pm 2i\) | A1 | Correct roots |
| \(m = \alpha \pm \beta i \Rightarrow x = e^{\alpha t}(A\cos\beta t + B\sin\beta t)\) | M1 | Forms complementary function, must be in terms of \(t\) only |
| \(x = e^{-t}(A\cos 2t + B\sin 2t)\) | A1 | Correct CF |
| PI: Try \(x = k \Rightarrow 5k = 50 \Rightarrow k = 10\) | M1 | Correct form of PI, complete method |
| \(GS: x = e^{-t}(A\cos 2t + B\sin 2t) + 10\) | A1ft | CF + correct PI |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = e^{-t}(2B\cos 2t - 2A\sin 2t) - e^{-t}(A\cos 2t + B\sin 2t)\) | B1ft | Correct differentiation of their \(x\) |
| \(y = \frac{1}{10}\left(\frac{dx}{dt} + 5x + 30\right)\) | M1 | Uses model and part (b) to find \(y\) in terms of \(t\) |
| \(y = \frac{1}{10}e^{-t}\big((4A+2B)\cos 2t + (4B-2A)\sin 2t\big) + 8\) | A1 | Correct equation, must have \(y = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, x=2 \Rightarrow 2 = A + 10 \Rightarrow A = -8\) | M1 | Uses initial conditions in equation for \(x\) |
| \(t=0, y=5 \Rightarrow 5 = \frac{1}{10}(2B-32)+8 \Rightarrow B=1\) | M1 | Uses initial conditions in equation for \(y\) to find both constants |
| \(x = e^{-t}(\sin 2t - 8\cos 2t) + 10\) | A1 | Correct equation for \(x\) |
| \(y = e^{-t}(2\sin 2t - 3\cos 2t) + 8\) | A1 | Correct equation for \(y\), constants must be gathered |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. When \(t > 8\), compounds \(X\) and \(Y\) remain approximately constant at 10 and 8 respectively, suggesting reaction has stopped. This supports the scientist's claim. | B1 | Valid comment with supporting reason; must have positive limits for both \(x\) and \(y\); both \(x\) and \(y\) must be considered |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2x}{dt^2} = -5\frac{dx}{dt} + 10\frac{dy}{dt}$ | B1 | Differentiates first equation w.r.t. $t$ correctly |
| Substitutes second equation to eliminate $y$: $\frac{d^2x}{dt^2} = -5\frac{dx}{dt} + 10(-2x+3y-4)$ | M1 | Uses second equation to eliminate $y$, achieving equation in $x$, $\frac{dx}{dt}$, $\frac{d^2x}{dt^2}$ |
| $\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 50$ | A1* | Printed answer, no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 2m + 5 = 0 \Rightarrow m = \ldots$ | M1 | Attempts auxiliary equation |
| $m = -1 \pm 2i$ | A1 | Correct roots |
| $m = \alpha \pm \beta i \Rightarrow x = e^{\alpha t}(A\cos\beta t + B\sin\beta t)$ | M1 | Forms complementary function, must be in terms of $t$ only |
| $x = e^{-t}(A\cos 2t + B\sin 2t)$ | A1 | Correct CF |
| PI: Try $x = k \Rightarrow 5k = 50 \Rightarrow k = 10$ | M1 | Correct form of PI, complete method |
| $GS: x = e^{-t}(A\cos 2t + B\sin 2t) + 10$ | A1ft | CF + correct PI |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = e^{-t}(2B\cos 2t - 2A\sin 2t) - e^{-t}(A\cos 2t + B\sin 2t)$ | B1ft | Correct differentiation of their $x$ |
| $y = \frac{1}{10}\left(\frac{dx}{dt} + 5x + 30\right)$ | M1 | Uses model and part (b) to find $y$ in terms of $t$ |
| $y = \frac{1}{10}e^{-t}\big((4A+2B)\cos 2t + (4B-2A)\sin 2t\big) + 8$ | A1 | Correct equation, must have $y = \ldots$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, x=2 \Rightarrow 2 = A + 10 \Rightarrow A = -8$ | M1 | Uses initial conditions in equation for $x$ |
| $t=0, y=5 \Rightarrow 5 = \frac{1}{10}(2B-32)+8 \Rightarrow B=1$ | M1 | Uses initial conditions in equation for $y$ to find both constants |
| $x = e^{-t}(\sin 2t - 8\cos 2t) + 10$ | A1 | Correct equation for $x$ |
| $y = e^{-t}(2\sin 2t - 3\cos 2t) + 8$ | A1 | Correct equation for $y$, constants must be gathered |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. When $t > 8$, compounds $X$ and $Y$ remain approximately constant at 10 and 8 respectively, suggesting reaction has stopped. This supports the scientist's claim. | B1 | Valid comment with supporting reason; must have positive limits for both $x$ and $y$; both $x$ and $y$ must be considered |
---\begin{enumerate}
\item Two compounds, $X$ and $Y$, are involved in a chemical reaction. The amounts in grams of these compounds, $t$ minutes after the reaction starts, are $x$ and $y$ respectively and are modelled by the differential equations
\end{enumerate}
$$\begin{aligned}
& \frac { \mathrm { d } x } { \mathrm {~d} t } = - 5 x + 10 y - 30 \\
& \frac { \mathrm {~d} y } { \mathrm {~d} t } = - 2 x + 3 y - 4
\end{aligned}$$
(a) Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 50$$
(b) Find, according to the model, a general solution for the amount in grams of compound $X$ present at time $t$ minutes.\\
(c) Find, according to the model, a general solution for the amount in grams of compound $Y$ present at time $t$ minutes.
Given that $x = 2$ and $y = 5$ when $t = 0$\\
(d) find\\
(i) the particular solution for $x$,\\
(ii) the particular solution for $y$.
A scientist thinks that the chemical reaction will have stopped after 8 minutes.\\
(e) Explain whether this is supported by the model.
\hfill \mbox{\textit{Edexcel CP1 2020 Q5 [17]}}