| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.3 This is a structured Further Maths question with clear guidance (substitution given, transformed equation provided). Part (a) requires careful but routine differentiation using the product rule twice and algebraic manipulation. Part (b) involves solving a standard constant-coefficient second-order DE with particular integral, then applying boundary conditions. While more demanding than typical A-level questions due to the algebraic complexity and being Further Maths content, the scaffolding and standard techniques place it moderately above average difficulty. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = tu \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}u}{\mathrm{d}t}t + u\) or \(\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{t\frac{\mathrm{d}x}{\mathrm{d}t} - x}{t^2}\) (oe) | B1 | Deduces correct first derivative of \(x=tu\) connecting \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}u}{\mathrm{d}t}\) |
| \(\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}u}{\mathrm{d}t}t + u \Rightarrow \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = \frac{\mathrm{d}^2u}{\mathrm{d}t^2}t + 2\frac{\mathrm{d}u}{\mathrm{d}t}\) or e.g. \(\frac{\mathrm{d}^2u}{\mathrm{d}t^2} = -\frac{1}{t^2}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{1}{t}\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - \frac{1}{t^4}\left(t^2\frac{\mathrm{d}x}{\mathrm{d}t} - 2tx\right)\) | B1 | Correct second derivative of \(x=tu\) connecting \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\) and \(\frac{\mathrm{d}^2u}{\mathrm{d}t^2}\) |
| \(t^2\left[\frac{\mathrm{d}^2u}{\mathrm{d}t^2}t + 2\frac{\mathrm{d}u}{\mathrm{d}t}\right] - 2t(t+1)\left[\frac{\mathrm{d}u}{\mathrm{d}t}t + u\right] + 2(t+1)tu = 8t^3e^t\) leading to \(t^3\frac{\mathrm{d}^2u}{\mathrm{d}t^2} + 2t^2\frac{\mathrm{d}u}{\mathrm{d}t} - 2t(t+1)\frac{\mathrm{d}u}{\mathrm{d}t}t - 2t(t+1)u + 2t(t+1)u = 8t^3e^t\) | M1 | Substitutes first and second derivatives into equation and makes some attempt to expand brackets |
| \(\frac{\mathrm{d}^2u}{\mathrm{d}t^2} - 2\frac{\mathrm{d}u}{\mathrm{d}t} = 8e^t\) * | A1\* | Fully correct proof with no errors or omissions; clear line where relevant terms cancel |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(m^2 - 2m = 0 \Rightarrow m = 0, 2\) | M1 | Forms and solves quadratic auxiliary equation \(m^2 - 2m = 0\) |
| \(u = A + Be^{2t}\) | A1 | Correct CF |
| \(\text{PS} \Rightarrow u = \lambda e^t\) | B1 | Deduces correct form of PI |
| \(\frac{\mathrm{d}u}{\mathrm{d}t} = \lambda e^t,\ \frac{\mathrm{d}^2u}{\mathrm{d}t^2} = \lambda e^t \Rightarrow \lambda e^t - 2\lambda e^t = 8e^t \Rightarrow \lambda = \ldots\{-8\}\), \(\Rightarrow u = \text{PS} + \text{CF}\) | M1 | Differentiates PI twice, substitutes into DE to find \(\lambda\) |
| \(u = \text{"}A + Be^{2t}\text{"} - 8e^t\) or \(x = \left(\text{"}A + Be^{2t}\text{"} - 8e^t\right)t\) | A1ft | General solution for \(u\) or \(x\); follow through on their CF |
| \(x = \left(A + Be^{2t} - 8e^t\right)t\); using \(x=0,\ t=\ln 3\) and \(t=\ln 5\): \(0 = A + 9B - 24\); \(0 = A + 25B - 40\) | M1 | Uses \(x=0,\ t=\ln 3\) and \(t=\ln 5\) to form two simultaneous equations |
| \(A + 9B = 24 \Big\} \Rightarrow A = \ldots\{15\},\ B = \ldots\{1\}\) \(A + 25B = 40\) | dM1 | Solves to find values of constants (dependent on previous M) |
| \(x = \left(15 + e^{2t} - 8e^t\right)t\) (oe) | A1 | Correct particular solution |
# Question 10:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = tu \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}u}{\mathrm{d}t}t + u$ **or** $\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{t\frac{\mathrm{d}x}{\mathrm{d}t} - x}{t^2}$ (oe) | **B1** | Deduces correct first derivative of $x=tu$ connecting $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}u}{\mathrm{d}t}$ |
| $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}u}{\mathrm{d}t}t + u \Rightarrow \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = \frac{\mathrm{d}^2u}{\mathrm{d}t^2}t + 2\frac{\mathrm{d}u}{\mathrm{d}t}$ or e.g. $\frac{\mathrm{d}^2u}{\mathrm{d}t^2} = -\frac{1}{t^2}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{1}{t}\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - \frac{1}{t^4}\left(t^2\frac{\mathrm{d}x}{\mathrm{d}t} - 2tx\right)$ | **B1** | Correct second derivative of $x=tu$ connecting $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}$ and $\frac{\mathrm{d}^2u}{\mathrm{d}t^2}$ |
| $t^2\left[\frac{\mathrm{d}^2u}{\mathrm{d}t^2}t + 2\frac{\mathrm{d}u}{\mathrm{d}t}\right] - 2t(t+1)\left[\frac{\mathrm{d}u}{\mathrm{d}t}t + u\right] + 2(t+1)tu = 8t^3e^t$ leading to $t^3\frac{\mathrm{d}^2u}{\mathrm{d}t^2} + 2t^2\frac{\mathrm{d}u}{\mathrm{d}t} - 2t(t+1)\frac{\mathrm{d}u}{\mathrm{d}t}t - 2t(t+1)u + 2t(t+1)u = 8t^3e^t$ | **M1** | Substitutes first and second derivatives into equation and makes some attempt to expand brackets |
| $\frac{\mathrm{d}^2u}{\mathrm{d}t^2} - 2\frac{\mathrm{d}u}{\mathrm{d}t} = 8e^t$ * | **A1\*** | Fully correct proof with no errors or omissions; clear line where relevant terms cancel |
**(4 marks)**
---
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $m^2 - 2m = 0 \Rightarrow m = 0, 2$ | **M1** | Forms and solves quadratic auxiliary equation $m^2 - 2m = 0$ |
| $u = A + Be^{2t}$ | **A1** | Correct CF |
| $\text{PS} \Rightarrow u = \lambda e^t$ | **B1** | Deduces correct form of PI |
| $\frac{\mathrm{d}u}{\mathrm{d}t} = \lambda e^t,\ \frac{\mathrm{d}^2u}{\mathrm{d}t^2} = \lambda e^t \Rightarrow \lambda e^t - 2\lambda e^t = 8e^t \Rightarrow \lambda = \ldots\{-8\}$, $\Rightarrow u = \text{PS} + \text{CF}$ | **M1** | Differentiates PI twice, substitutes into DE to find $\lambda$ |
| $u = \text{"}A + Be^{2t}\text{"} - 8e^t$ or $x = \left(\text{"}A + Be^{2t}\text{"} - 8e^t\right)t$ | **A1ft** | General solution for $u$ or $x$; follow through on their CF |
| $x = \left(A + Be^{2t} - 8e^t\right)t$; using $x=0,\ t=\ln 3$ and $t=\ln 5$: $0 = A + 9B - 24$; $0 = A + 25B - 40$ | **M1** | Uses $x=0,\ t=\ln 3$ and $t=\ln 5$ to form two simultaneous equations |
| $A + 9B = 24 \Big\} \Rightarrow A = \ldots\{15\},\ B = \ldots\{1\}$ $A + 25B = 40$ | **dM1** | Solves to find values of constants (dependent on previous M) |
| $x = \left(15 + e^{2t} - 8e^t\right)t$ (oe) | **A1** | Correct particular solution |
**(8 marks)**
**Total: 12 marks**\begin{enumerate}
\item The motion of a particle $P$ along the $x$-axis is modelled by the differential equation
\end{enumerate}
$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t ( t + 1 ) \frac { \mathrm { d } x } { \mathrm {~d} t } + 2 ( t + 1 ) x = 8 t ^ { 3 } \mathrm { e } ^ { t }$$
where $P$ has displacement $x$ metres from the origin $O$ at time $t$ minutes, $t > 0$\\
(a) Show that the transformation $x = t u$ transforms the differential equation (I) into the differential equation
$$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} u } { \mathrm {~d} t } = 8 \mathrm { e } ^ { t }$$
Given that $P$ is at $O$ when $t = \ln 3$ and when $t = \ln 5$\\
(b) determine the particular solution of the differential equation (I)
\hfill \mbox{\textit{Edexcel FP1 2024 Q10 [12]}}