| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Substitution reducing to first order linear ODE |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear guidance through each step. Part (a) is a standard integrating factor problem, part (b) is a 'show that' verification requiring product rule application, and part (c) requires integrating the result from (a). While it involves second-order DEs and substitution (FP3 content), the scaffolding makes it more accessible than typical unguided Further Maths problems. The integrating factor and subsequent integration require careful algebra but follow established procedures. |
| Spec | 4.10c Integrating factor: first order equations4.10d Second order homogeneous: auxiliary equation method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrating factor: \(e^{\int -\frac{2x}{x^2+4}\,dx}\) | M1 | Attempt at integrating factor |
| \(= e^{-\ln(x^2+4)} = \frac{1}{x^2+4}\) | A1 | Correct IF |
| \(\frac{d}{dx}\left(\frac{u}{x^2+4}\right) = 3\) | M1 | Multiply through and recognise derivative |
| \(\frac{u}{x^2+4} = 3x + C\) | A1 | Correct integration |
| \(u = (x^2+4)(3x+C)\) | A1 | Correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = x^2\frac{dy}{dx}\) | B1 | Statement of substitution |
| \(\frac{du}{dx} = 2x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}\) | M1 | Differentiating \(u\) |
| Substituting into LHS: \(x^2(x^2+4)\frac{d^2y}{dx^2} + 8x\frac{dy}{dx}\) | M1 | Substitution attempted |
| \(= (x^2+4)\left(x^2\frac{d^2y}{dx^2}+2x\frac{dy}{dx}\right) + (8x - 2x(x^2+4)/(x^2+4))\cdot x^2\frac{dy}{dx}\) | ||
| \(= (x^2+4)\frac{du}{dx} - 2x\cdot u\) giving required equation | A1 | Completion with all steps shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| From (a): \(u = (x^2+4)(3x+C)\) and \(u = x^2\frac{dy}{dx}\) | M1 | Use result from (a) with substitution |
| \(\frac{dy}{dx} = \frac{(x^2+4)(3x+C)}{x^2}\) | ||
| \(y = \int \frac{(x^2+4)(3x+C)}{x^2}\,dx\) | A1 | Correct general solution (integration implied) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrating factor: $e^{\int -\frac{2x}{x^2+4}\,dx}$ | M1 | Attempt at integrating factor |
| $= e^{-\ln(x^2+4)} = \frac{1}{x^2+4}$ | A1 | Correct IF |
| $\frac{d}{dx}\left(\frac{u}{x^2+4}\right) = 3$ | M1 | Multiply through and recognise derivative |
| $\frac{u}{x^2+4} = 3x + C$ | A1 | Correct integration |
| $u = (x^2+4)(3x+C)$ | A1 | Correct solution |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = x^2\frac{dy}{dx}$ | B1 | Statement of substitution |
| $\frac{du}{dx} = 2x\frac{dy}{dx} + x^2\frac{d^2y}{dx^2}$ | M1 | Differentiating $u$ |
| Substituting into LHS: $x^2(x^2+4)\frac{d^2y}{dx^2} + 8x\frac{dy}{dx}$ | M1 | Substitution attempted |
| $= (x^2+4)\left(x^2\frac{d^2y}{dx^2}+2x\frac{dy}{dx}\right) + (8x - 2x(x^2+4)/(x^2+4))\cdot x^2\frac{dy}{dx}$ | | |
| $= (x^2+4)\frac{du}{dx} - 2x\cdot u$ giving required equation | A1 | Completion with all steps shown |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| From (a): $u = (x^2+4)(3x+C)$ and $u = x^2\frac{dy}{dx}$ | M1 | Use result from (a) with substitution |
| $\frac{dy}{dx} = \frac{(x^2+4)(3x+C)}{x^2}$ | | |
| $y = \int \frac{(x^2+4)(3x+C)}{x^2}\,dx$ | A1 | Correct general solution (integration implied) |
---6
\begin{enumerate}[label=(\alph*)]
\item By using an integrating factor, find the general solution of the differential equation
$$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
giving your answer in the form $u = \mathrm { f } ( x )$.\\[0pt]
[6 marks]
\item Show that the substitution $u = x ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x }$ transforms the differential equation
$$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$
into
$$\frac { \mathrm { d } u } { \mathrm {~d} x } - \frac { 2 x } { x ^ { 2 } + 4 } u = 3 \left( x ^ { 2 } + 4 \right)$$
\item Hence, given that $x > 0$, find the general solution of the differential equation
$$x ^ { 2 } \left( x ^ { 2 } + 4 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 8 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 3 \left( x ^ { 2 } + 4 \right) ^ { 2 }$$
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2014 Q6 [8]}}