| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Challenging +1.2 This is a standard second-order linear differential equation with constant coefficients requiring complementary function (solving auxiliary equation with two real roots), particular integral (linear form), applying initial conditions, and finding/justifying a minimum. While multi-part with several techniques, each step follows routine procedures taught in Further Maths Core Pure 2. Part (d) requires interpretation but is straightforward. More demanding than basic A-level due to Further Maths content, but still a textbook-style question without novel insight required. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2m^2 + 5m + 2 = 0 \Rightarrow m = -\frac{1}{2}, -2\) | M1 | Attempts to solve auxiliary equation, usual rules for quadratic |
| \(x = Ae^{-0.5t} + Be^{-2t}\) | A1 | Correct CF, must be in terms of \(t\) |
| PI is of the form \(x = pt + q\) | B1 | Correct form for PI, accept \(x = at^2 + bt + c\) |
| \(\frac{dx}{dt} = p\), \(\frac{d^2x}{dt^2} = 0 \Rightarrow 5p + 2pt + 2q = 4t + 12 \Rightarrow p = ..., q = ...\) | M1 | Differentiates PI and substitutes into DE, \(p, q \neq 0\) |
| \(p = 2, q = 1\) | A1 | Correct PI |
| \(x = Ae^{-0.5t} + Be^{-2t} + 2t + 1\) | A1ft | Correct GS = CF + PI, must have \(x =\) in terms of \(t\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=0, x=3 \Rightarrow 3 = A + B + 1\) | M1 | Substitutes \(x=3\), \(t=0\) into GS |
| \(\frac{dx}{dt} = -0.5Ae^{-0.5t} - 2Be^{-2t} + 2\); \(t=0, \frac{dx}{dt} = -2 \Rightarrow -\frac{1}{2}A - 2B + 2 = -2\) | M1 | Differentiates GS, sets \(\frac{dx}{dt}=-2\) at \(t=0\), solves simultaneously |
| \(x = 2e^{-2t} + 2t + 1\) | A1 | Correct particular solution, need \(x =\) in terms of \(t\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = -4e^{-2t} + 2\) | M1 | Differentiates PS of form \(ae^{-kt} + bt + c\) to obtain \(Ce^{-kt} + D\) |
| \(e^{-2t} = \frac{1}{2} \Rightarrow -2t = \ln\frac{1}{2} \Rightarrow t = \frac{1}{2}\ln 2\) | dM1 | Solves \(Ce^{-kt} + D = 0\), \(C \times D < 0\), to obtain \(t = -\frac{1}{k}\ln\!\left(\frac{-D}{C}\right)\) |
| \(x = 2e^{-2t} + 2t + 1 = 1 + \ln 2 + 1 = 2 + \ln 2\) | A1* | Substitutes \(t = \frac{1}{2}\ln 2\) to obtain printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^2x}{dt^2} = 8e^{-2t} > 0\) for all values of \(t\), so distance is a minimum | B1ft | Obtains second derivative of form \(\lambda e^{-\mu t}\), \(\lambda, \mu > 0\), and makes conclusion; or substitutes \(t\) value and states \(\frac{d^2x}{dt^2} > 0\) hence minimum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For large values of \(t\): \(\left[e^{-2t} \to 0\right] \Rightarrow x \to 2t+1\) so constant speed; \(\frac{dx}{dt} \to 2\) constant speed; \(\frac{d^2x}{dt^2} \to 0\) constant speed. Conclusion: model is suitable | B1ft | PS must be of form \(f(t)+bt+c\) where \(f(t)\) has terms in \(ae^{-kt}\), \(k>0\), \(a,b \neq 0\); demonstrates \(e^{-kt} \to 0\) as \(t \to \infty\) giving constant speed, states model is suitable |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2m^2 + 5m + 2 = 0 \Rightarrow m = -\frac{1}{2}, -2$ | M1 | Attempts to solve auxiliary equation, usual rules for quadratic |
| $x = Ae^{-0.5t} + Be^{-2t}$ | A1 | Correct CF, must be in terms of $t$ |
| PI is of the form $x = pt + q$ | B1 | Correct form for PI, accept $x = at^2 + bt + c$ |
| $\frac{dx}{dt} = p$, $\frac{d^2x}{dt^2} = 0 \Rightarrow 5p + 2pt + 2q = 4t + 12 \Rightarrow p = ..., q = ...$ | M1 | Differentiates PI and substitutes into DE, $p, q \neq 0$ |
| $p = 2, q = 1$ | A1 | Correct PI |
| $x = Ae^{-0.5t} + Be^{-2t} + 2t + 1$ | A1ft | Correct GS = CF + PI, must have $x =$ in terms of $t$ only |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=0, x=3 \Rightarrow 3 = A + B + 1$ | M1 | Substitutes $x=3$, $t=0$ into GS |
| $\frac{dx}{dt} = -0.5Ae^{-0.5t} - 2Be^{-2t} + 2$; $t=0, \frac{dx}{dt} = -2 \Rightarrow -\frac{1}{2}A - 2B + 2 = -2$ | M1 | Differentiates GS, sets $\frac{dx}{dt}=-2$ at $t=0$, solves simultaneously |
| $x = 2e^{-2t} + 2t + 1$ | A1 | Correct particular solution, need $x =$ in terms of $t$ only |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = -4e^{-2t} + 2$ | M1 | Differentiates PS of form $ae^{-kt} + bt + c$ to obtain $Ce^{-kt} + D$ |
| $e^{-2t} = \frac{1}{2} \Rightarrow -2t = \ln\frac{1}{2} \Rightarrow t = \frac{1}{2}\ln 2$ | dM1 | Solves $Ce^{-kt} + D = 0$, $C \times D < 0$, to obtain $t = -\frac{1}{k}\ln\!\left(\frac{-D}{C}\right)$ |
| $x = 2e^{-2t} + 2t + 1 = 1 + \ln 2 + 1 = 2 + \ln 2$ | A1* | Substitutes $t = \frac{1}{2}\ln 2$ to obtain printed answer with no errors |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^2x}{dt^2} = 8e^{-2t} > 0$ for all values of $t$, so distance is a minimum | B1ft | Obtains second derivative of form $\lambda e^{-\mu t}$, $\lambda, \mu > 0$, and makes conclusion; or substitutes $t$ value and states $\frac{d^2x}{dt^2} > 0$ hence minimum |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For large values of $t$: $\left[e^{-2t} \to 0\right] \Rightarrow x \to 2t+1$ so constant speed; $\frac{dx}{dt} \to 2$ constant speed; $\frac{d^2x}{dt^2} \to 0$ constant speed. Conclusion: model is suitable | B1ft | PS must be of form $f(t)+bt+c$ where $f(t)$ has terms in $ae^{-kt}$, $k>0$, $a,b \neq 0$; demonstrates $e^{-kt} \to 0$ as $t \to \infty$ giving constant speed, states model is suitable |
---\begin{enumerate}
\item The motion of a particle $P$ along the $x$-axis is modelled by the differential equation
\end{enumerate}
$$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 4 t + 12$$
where $P$ is $x$ metres from the origin $O$ at time $t$ seconds, $t \geqslant 0$\\
(a) Determine the general solution of the differential equation.\\
(b) Hence determine the particular solution for which $x = 3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2$ when $t = 0$\\
(c) (i) Show that, according to the model, the minimum distance between $O$ and $P$ is $( 2 + \ln 2 )$ metres.\\
(ii) Justify that this distance is a minimum.
For large values of $t$ the particle is expected to move with constant speed.\\
(d) Comment on the suitability of the model in light of this information.
\hfill \mbox{\textit{Edexcel CP2 2024 Q6 [14]}}