| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| 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 on second-order differential equations with substitution. Parts (a) and (b) are guided derivations using chain rule, part (c) is a standard constant-coefficient equation with polynomial particular integral, and part (d) requires simple back-substitution. While it involves multiple techniques and is Further Maths content (inherently harder), the question is highly scaffolded with clear steps, making it more accessible than typical FM questions requiring independent problem-solving. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10c Integrating factor: first order equations4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{1}{x}\frac{dy}{dt}\) or \(t = \ln x \Rightarrow \frac{dx}{dy} = e^t \frac{dt}{dy}\) | B1 | Any correct equation linking \(\frac{dy}{dx}\) and \(\frac{dy}{dt}\) (or their reciprocals). |
| \(\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\frac{d^2y}{dt^2}\frac{dt}{dx}\) or equivalent chain rule application | M1 | Differentiates again wrt \(x\) or \(t\) with attempts at both product rule and chain rule to obtain equation involving \(\frac{d^2y}{dx^2}\) and \(\frac{d^2y}{dt^2}\). |
| \(\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)\) | A1* | Eliminates the \(x\)'s and \(\frac{dx}{dt}\)'s and completes to printed answer with no errors. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2\frac{d^2y}{dx^2} - 2y = 1 + 4\ln x - 2(\ln x)^2 \Rightarrow e^{2t} \times e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) - 2y = 1 + 4t - 2t^2\) \(\Rightarrow \frac{d^2y}{dt^2} - \frac{dy}{dt} - 2y = 1 + 4t - 2t^2\) | B1* | Correct substitution leading to printed answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m^2 - m - 2 = 0 \Rightarrow m = 2, -1\) | M1 | Solves auxiliary equation. |
| \(y = Ae^{2t} + Be^{-t}\) | A1 | Correct complementary function. |
| \(y = at^2 + bt + c \Rightarrow \frac{dy}{dt} = 2at + b \Rightarrow \frac{d^2y}{dt^2} = 2a\) | M1 | Attempts correct form of particular integral and differentiates. |
| \(2a - 2at - b - 2at^2 - 2bt - 2c = 1 + 4t - 2t^2\) \(2a = 2 \Rightarrow a = 1\) \(-2a - 2b = 4 \Rightarrow b = -3\) \(2a - b - 2c = 1 \Rightarrow c = 2\) | dM1 | Substitutes and equates coefficients to find \(a\), \(b\), \(c\). |
| \(y = Ae^{2t} + Be^{-t} + t^2 - 3t + 2\) | A1 | Correct general solution. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = Ax^2 + Bx^{-1} + (\ln x)^2 - 3\ln x + 2\) | B1ft | Follow through from their answer to (c) converting back to \(x\). |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{1}{x}\frac{dy}{dt}$ or $t = \ln x \Rightarrow \frac{dx}{dy} = e^t \frac{dt}{dy}$ | B1 | Any correct equation linking $\frac{dy}{dx}$ and $\frac{dy}{dt}$ (or their reciprocals). |
| $\frac{d^2y}{dx^2} = -\frac{1}{x^2}\frac{dy}{dt} + \frac{1}{x}\frac{d^2y}{dt^2}\frac{dt}{dx}$ or equivalent chain rule application | M1 | Differentiates again wrt $x$ or $t$ with attempts at both product rule and chain rule to obtain equation involving $\frac{d^2y}{dx^2}$ and $\frac{d^2y}{dt^2}$. |
| $\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)$ | A1* | Eliminates the $x$'s and $\frac{dx}{dt}$'s and completes to printed answer with no errors. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2\frac{d^2y}{dx^2} - 2y = 1 + 4\ln x - 2(\ln x)^2 \Rightarrow e^{2t} \times e^{-2t}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) - 2y = 1 + 4t - 2t^2$ $\Rightarrow \frac{d^2y}{dt^2} - \frac{dy}{dt} - 2y = 1 + 4t - 2t^2$ | B1* | Correct substitution leading to printed answer. |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m^2 - m - 2 = 0 \Rightarrow m = 2, -1$ | M1 | Solves auxiliary equation. |
| $y = Ae^{2t} + Be^{-t}$ | A1 | Correct complementary function. |
| $y = at^2 + bt + c \Rightarrow \frac{dy}{dt} = 2at + b \Rightarrow \frac{d^2y}{dt^2} = 2a$ | M1 | Attempts correct form of particular integral and differentiates. |
| $2a - 2at - b - 2at^2 - 2bt - 2c = 1 + 4t - 2t^2$ $2a = 2 \Rightarrow a = 1$ $-2a - 2b = 4 \Rightarrow b = -3$ $2a - b - 2c = 1 \Rightarrow c = 2$ | dM1 | Substitutes and equates coefficients to find $a$, $b$, $c$. |
| $y = Ae^{2t} + Be^{-t} + t^2 - 3t + 2$ | A1 | Correct general solution. |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = Ax^2 + Bx^{-1} + (\ln x)^2 - 3\ln x + 2$ | B1ft | Follow through from their answer to (c) converting back to $x$. |\begin{enumerate}
\item (a) Given that $t = \ln x$, where $x > 0$, show that
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$
(b) Hence show that the transformation $t = \ln x$, where $x > 0$, transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 y = 1 + 4 \ln x - 2 ( \ln x ) ^ { 2 }$$
into the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } - 2 y = 1 + 4 t - 2 t ^ { 2 }$$
(c) Solve differential equation (II) to determine $y$ in terms of $t$.\\
(d) Hence determine the general solution of differential equation (I).
\hfill \mbox{\textit{Edexcel F2 2024 Q8 [10]}}