| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a standard Further Maths second-order differential equation with complex roots and particular integral, requiring auxiliary equation solution, applying initial conditions, and finding a turning point. While methodical, it involves multiple techniques (complementary function with complex roots giving damped oscillation, particular integral by inspection, applying two initial conditions, then differentiation and solving transcendental equation for the turning point). The 'no calculator' requirement for part (c) adds modest difficulty. Typical for F2 but more involved than average A-level questions overall. |
| 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 |
| \(m^2 + 6m + 13 = 0 \Rightarrow m = \frac{-6 \pm \sqrt{36-52}}{2} = \{-3 \pm 2\text{i}\}\) | M1 | Forms correct auxiliary equation and obtains correct numerical expression for at least one root by formula or CTS |
| CF: \(x = e^{-3t}(A\cos 2t + B\sin 2t)\) | A1 | Correct complementary function in any form |
| PI: \(\{x =\}\ \lambda e^{-3t}\) | B1 | Correct form for PI; must include \(\lambda e^{-3t}\); if \(\lambda e^{pt}\) used, \(p=-3\) must be seen later |
| \(\frac{dx}{dt} = -3\lambda e^{-3t}\); \(\frac{d^2x}{dt^2} = 9\lambda e^{-3t}\) \(\Rightarrow 9\lambda e^{-3t} + 6(-3\lambda e^{-3t}) + 13\lambda e^{-3t} = 8e^{-3t}\) | M1 | Differentiates PI of any form twice (provided it has at least one constant and is a function of \(t\)) and substitutes; allow sign/coefficient errors only |
| \(\Rightarrow 9\lambda - 18\lambda + 13\lambda = 8 \Rightarrow \lambda = 2\) | dM1 | Finds value of constant following use of PI of correct form |
| \(x = e^{-3t}(A\cos 2t + B\sin 2t) + 2e^{-3t}\) | A1ft | Correct general solution ft on their CF only; must have \(x = \ldots\) in terms of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \frac{1}{2}\) at \(t=0 \Rightarrow \frac{1}{2} = A + 2 \Rightarrow A = -\frac{3}{2}\) | M1 | Uses initial condition for \(x\) in GS to find linear equation in one or two constants |
| \(\frac{dx}{dt} = e^{-3t}(-2A\sin 2t + 2B\cos 2t) - 3e^{-3t}(A\cos 2t + B\sin 2t) - 6e^{-3t}\) | M1 | Uses product rule to differentiate real GS; sign and coefficient errors only |
| \(t=0, \frac{dx}{dt} = \frac{1}{2} \Rightarrow \frac{1}{2} = 2B - 3A - 6 \Rightarrow B = 1\) | ddM1 | Uses both initial conditions to find values for 2 constants in GS = (CF with 2 constants) + PI; requires both previous M marks |
| \(x = e^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) + 2e^{-3t}\) | A1 | Correct particular solution in terms of \(t\); must be \(x = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = e^{-3t}(3\sin 2t + 2\cos 2t) - 3e^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) - 6e^{-3t} = 0\) | M1 | Sets expression for \(\frac{dx}{dt} = 0\); accept with any unfound constants provided \(\frac{dx}{dt} = f(t)\) |
| \((3\sin 2t + 2\cos 2t) - 3\left(-\frac{3}{2}\cos 2t + \sin 2t\right) - 6 = 0\) | dM1 | Achieves equation of form \(a\sin bt + c\cos bt + d = 0\) with terms uncollected; must follow GS = CF + PI where two constants found for CF and one for PI |
| \(\cos 2t = \frac{12}{13} \Rightarrow t = 0.1973955598\ldots\) | ddM1 | Finds value of \(t\) from \(\cos kt = c\) \((k\neq 1, -1 < c < 1)\) and uses positive value of \(t\) to find \(x\); requires both previous M marks |
| \(x\) or \(a = 0.553(1164729\ldots)\) | A1 | awrt 0.553 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 + 6m + 13 = 0 \Rightarrow m = \frac{-6 \pm \sqrt{36-52}}{2} = \{-3 \pm 2\text{i}\}$ | M1 | Forms correct auxiliary equation and obtains correct numerical expression for at least one root by formula or CTS |
| CF: $x = e^{-3t}(A\cos 2t + B\sin 2t)$ | A1 | Correct complementary function in any form |
| PI: $\{x =\}\ \lambda e^{-3t}$ | B1 | Correct form for PI; must include $\lambda e^{-3t}$; if $\lambda e^{pt}$ used, $p=-3$ must be seen later |
| $\frac{dx}{dt} = -3\lambda e^{-3t}$; $\frac{d^2x}{dt^2} = 9\lambda e^{-3t}$ $\Rightarrow 9\lambda e^{-3t} + 6(-3\lambda e^{-3t}) + 13\lambda e^{-3t} = 8e^{-3t}$ | M1 | Differentiates PI of **any** form twice (provided it has at least one constant and is a function of $t$) and substitutes; allow sign/coefficient errors only |
| $\Rightarrow 9\lambda - 18\lambda + 13\lambda = 8 \Rightarrow \lambda = 2$ | dM1 | Finds value of constant following use of PI of correct form |
| $x = e^{-3t}(A\cos 2t + B\sin 2t) + 2e^{-3t}$ | A1ft | Correct general solution ft on their CF only; must have $x = \ldots$ in terms of $t$ |
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# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{1}{2}$ at $t=0 \Rightarrow \frac{1}{2} = A + 2 \Rightarrow A = -\frac{3}{2}$ | M1 | Uses initial condition for $x$ in GS to find linear equation in one or two constants |
| $\frac{dx}{dt} = e^{-3t}(-2A\sin 2t + 2B\cos 2t) - 3e^{-3t}(A\cos 2t + B\sin 2t) - 6e^{-3t}$ | M1 | Uses product rule to differentiate real GS; sign and coefficient errors only |
| $t=0, \frac{dx}{dt} = \frac{1}{2} \Rightarrow \frac{1}{2} = 2B - 3A - 6 \Rightarrow B = 1$ | ddM1 | Uses both initial conditions to find values for 2 constants in GS = (CF with 2 constants) + PI; requires both previous M marks |
| $x = e^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) + 2e^{-3t}$ | A1 | Correct particular solution in terms of $t$; must be $x = \ldots$ |
---
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = e^{-3t}(3\sin 2t + 2\cos 2t) - 3e^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) - 6e^{-3t} = 0$ | M1 | Sets expression for $\frac{dx}{dt} = 0$; accept with any unfound constants provided $\frac{dx}{dt} = f(t)$ |
| $(3\sin 2t + 2\cos 2t) - 3\left(-\frac{3}{2}\cos 2t + \sin 2t\right) - 6 = 0$ | dM1 | Achieves equation of form $a\sin bt + c\cos bt + d = 0$ with terms uncollected; must follow GS = CF + PI where two constants found for CF and one for PI |
| $\cos 2t = \frac{12}{13} \Rightarrow t = 0.1973955598\ldots$ | ddM1 | Finds value of $t$ from $\cos kt = c$ $(k\neq 1, -1 < c < 1)$ and uses positive value of $t$ to find $x$; requires both previous M marks |
| $x$ or $a = 0.553(1164729\ldots)$ | A1 | awrt 0.553 |
---\begin{enumerate}
\item The differential equation
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 6 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 13 x = 8 \mathrm { e } ^ { - 3 t } \quad t \geqslant 0$$
describes the motion of a particle along the $x$-axis.\\
(a) Determine the general solution of this differential equation.
Given that the motion of the particle satisfies $x = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 2 }$ when $t = 0$\\
(b) determine the particular solution for the motion of the particle.
On the graph of the particular solution found in part (b), the first turning point for $t > 0$ occurs at $x = a$.\\
(c) Determine, to 3 significant figures, the value of $a$.\\[0pt]
[Solutions relying entirely on calculator technology are not acceptable.]
\hfill \mbox{\textit{Edexcel F2 2024 Q6 [14]}}