| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| 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 +1.2 This is a structured FP3 differential equations question with clear guidance through substitution. Part (a) is verification algebra (4 marks), part (b) is a standard separable DE requiring integration of 2x/(x²-1), and part (c) requires integrating back. While it involves second-order DEs and requires careful algebraic manipulation across multiple steps, the substitution is given and each part follows logically from the previous one with standard techniques throughout. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = \frac{dy}{dx} + x \Rightarrow \frac{du}{dx} = \frac{d^2y}{dx^2} + 1\) | M1A1 | |
| \((x^2 - 1)\left(\frac{du}{dx} - 1\right) - 2x(u - x) = x^2 + 1\) | M1 | Substitution into LHS of DE as far as no \(y\)s |
| \(DE \Rightarrow (x^2 - 1)\frac{du}{dx} - 2xu = 0\) | ||
| \(\Rightarrow \frac{du}{dx} = \frac{2xu}{x^2 - 1}\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{1}{u}\, du = \int \frac{2x}{x^2 - 1}\, dx\) | M1, A1 | Separate variables |
| \(\ln u = \ln\lvert x^2 - 1\rvert + \ln A\) | A1A1 | |
| \(u = A(x^2 - 1)\) | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} + x = A(x^2 - 1)\) | M1 | Use (b) \((\neq 0)\) to form DE in \(y\) and \(x\) |
| \(\frac{dy}{dx} = A(x^2 - 1) - x\) | ||
| \(y = A\left(\frac{x^3}{3} - x\right) - \frac{x^2}{2} + B\) | M1, A1ft | 3 marks |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = \frac{dy}{dx} + x \Rightarrow \frac{du}{dx} = \frac{d^2y}{dx^2} + 1$ | M1A1 | |
| $(x^2 - 1)\left(\frac{du}{dx} - 1\right) - 2x(u - x) = x^2 + 1$ | M1 | Substitution into LHS of DE as far as no $y$s |
| $DE \Rightarrow (x^2 - 1)\frac{du}{dx} - 2xu = 0$ | | |
| $\Rightarrow \frac{du}{dx} = \frac{2xu}{x^2 - 1}$ | A1 | **4 marks** | CSO; AG |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{1}{u}\, du = \int \frac{2x}{x^2 - 1}\, dx$ | M1, A1 | Separate variables |
| $\ln u = \ln\lvert x^2 - 1\rvert + \ln A$ | A1A1 | |
| $u = A(x^2 - 1)$ | A1 | **5 marks** |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} + x = A(x^2 - 1)$ | M1 | Use (b) $(\neq 0)$ to form DE in $y$ and $x$ |
| $\frac{dy}{dx} = A(x^2 - 1) - x$ | | |
| $y = A\left(\frac{x^3}{3} - x\right) - \frac{x^2}{2} + B$ | M1, A1ft | **3 marks** | Solution must have two different constants and correct method used to solve the DE |5
\begin{enumerate}[label=(\alph*)]
\item A differential equation is given by
$$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$
Show that the substitution
$$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + x$$
transforms this differential equation into
$$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$
(4 marks)
\item Find the general solution of
$$\frac { \mathrm { d } u } { \mathrm {~d} x } = \frac { 2 x u } { x ^ { 2 } - 1 }$$
giving your answer in the form $u = \mathrm { f } ( x )$.
\item Hence find the general solution of the differential equation
$$\left( x ^ { 2 } - 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = x ^ { 2 } + 1$$
giving your answer in the form $y = \mathrm { g } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q5 [12]}}