| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| 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 FP2 second-order differential equation with particular integral, requiring auxiliary equation solution (distinct real roots), finding constants from initial conditions, then calculus to find and verify a maximum. While methodical, it involves multiple techniques (complementary function, particular integral, applying conditions, optimization) and the verification step adds rigor beyond routine exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| AE: \(m^2+5m+6=0 \Rightarrow (m+3)(m+2)=0 \Rightarrow m=-3,-2\) | ||
| \(x_{CF} = Ae^{-3t} + Be^{-2t}\) | M1, A1 | \(Ae^{m_1 t}+Be^{m_2 t}\) where \(m_1 \neq m_2\); \(Ae^{-3t}+Be^{-2t}\) |
| Try \(x = ke^{-t} \Rightarrow \frac{dx}{dt}=-ke^{-t}, \frac{d^2x}{dt^2}=ke^{-t}\); substituting gives \(2ke^{-t}=2e^{-t} \Rightarrow k=1\) | M1, A1 | Substitutes \(ke^{-t}\) into DE; finds \(k=1\) |
| \(x = Ae^{-3t} + Be^{-2t} + e^{-t}\) | M1* | Their \(x_{CF}\) + their \(x_{PI}\) |
| \(\frac{dx}{dt} = -3Ae^{-3t} - 2Be^{-2t} - e^{-t}\) | dM1* | Finds \(\frac{dx}{dt}\) by differentiating |
| \(t=0, x=0 \Rightarrow 0=A+B+1\); \(t=0, \frac{dx}{dt}=2 \Rightarrow 2=-3A-2B-1\) | ddM1* | Applies both initial conditions to form simultaneous equations |
| \(A=-1, B=0\) | ||
| \(x = -e^{-3t} + e^{-t}\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dt} = 3e^{-3t} - e^{-t} = 0\) | M1 | Differentiates \(x\), sets \(\frac{dx}{dt}=0\) |
| \(3 - e^{2t} = 0 \Rightarrow t = \frac{1}{2}\ln 3\) | dM1*, A1 | Credible attempt to solve; \(t=\frac{1}{2}\ln 3\) or \(\ln\sqrt{3}\) or awrt \(0.55\) |
| \(x = -3^{-\frac{3}{2}} + 3^{-\frac{1}{2}} = -\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}\) | ddM1, A1 AG | Substitutes \(t\) back, eliminates ln's; uses exact values to give \(\frac{2\sqrt{3}}{9}\) |
| \(\frac{d^2x}{dt^2} = -9e^{-3t}+e^{-t}\); at \(t=\frac{1}{2}\ln 3\): \(= -\frac{9}{3\sqrt{3}}+\frac{1}{\sqrt{3}} = -\frac{2}{\sqrt{3}} < 0\), therefore maximum | dM1*, A1 | Finds \(\frac{d^2x}{dt^2}\) and substitutes \(t\); \(-\frac{9}{3\sqrt{3}}+\frac{1}{\sqrt{3}}<0\) and maximum conclusion |
## Question 8:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| AE: $m^2+5m+6=0 \Rightarrow (m+3)(m+2)=0 \Rightarrow m=-3,-2$ | | |
| $x_{CF} = Ae^{-3t} + Be^{-2t}$ | M1, A1 | $Ae^{m_1 t}+Be^{m_2 t}$ where $m_1 \neq m_2$; $Ae^{-3t}+Be^{-2t}$ |
| Try $x = ke^{-t} \Rightarrow \frac{dx}{dt}=-ke^{-t}, \frac{d^2x}{dt^2}=ke^{-t}$; substituting gives $2ke^{-t}=2e^{-t} \Rightarrow k=1$ | M1, A1 | Substitutes $ke^{-t}$ into DE; finds $k=1$ |
| $x = Ae^{-3t} + Be^{-2t} + e^{-t}$ | M1* | Their $x_{CF}$ + their $x_{PI}$ |
| $\frac{dx}{dt} = -3Ae^{-3t} - 2Be^{-2t} - e^{-t}$ | dM1* | Finds $\frac{dx}{dt}$ by differentiating |
| $t=0, x=0 \Rightarrow 0=A+B+1$; $t=0, \frac{dx}{dt}=2 \Rightarrow 2=-3A-2B-1$ | ddM1* | Applies both initial conditions to form simultaneous equations |
| $A=-1, B=0$ | | |
| $x = -e^{-3t} + e^{-t}$ | A1 **cao** | |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dt} = 3e^{-3t} - e^{-t} = 0$ | M1 | Differentiates $x$, sets $\frac{dx}{dt}=0$ |
| $3 - e^{2t} = 0 \Rightarrow t = \frac{1}{2}\ln 3$ | dM1*, A1 | Credible attempt to solve; $t=\frac{1}{2}\ln 3$ or $\ln\sqrt{3}$ or awrt $0.55$ |
| $x = -3^{-\frac{3}{2}} + 3^{-\frac{1}{2}} = -\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$ | ddM1, A1 **AG** | Substitutes $t$ back, eliminates ln's; uses exact values to give $\frac{2\sqrt{3}}{9}$ |
| $\frac{d^2x}{dt^2} = -9e^{-3t}+e^{-t}$; at $t=\frac{1}{2}\ln 3$: $= -\frac{9}{3\sqrt{3}}+\frac{1}{\sqrt{3}} = -\frac{2}{\sqrt{3}} < 0$, therefore maximum | dM1*, A1 | Finds $\frac{d^2x}{dt^2}$ and substitutes $t$; $-\frac{9}{3\sqrt{3}}+\frac{1}{\sqrt{3}}<0$ and maximum conclusion |8.
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 6 x = 2 \mathrm { e } ^ { - t }$$
Given that $x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 2$ at $t = 0$,
\begin{enumerate}[label=(\alph*)]
\item find $x$ in terms of $t$.
The solution to part (a) is used to represent the motion of a particle $P$ on the $x$-axis. At time $t$ seconds, where $t > 0 , P$ is $x$ metres from the origin $O$.
\item Show that the maximum distance between $O$ and $P$ is $\frac { 2 \sqrt { } 3 } { 9 } \mathrm {~m}$ and justify that this\\
distance is a maximum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2009 Q8 [15]}}