| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Challenging +1.2 This is a standard integrating factor question from Further Maths with straightforward structure (P(x) = 2/x gives IF = x²). However, the right-hand side requires partial fractions decomposition with a quadratic factor, making the integration more involved than typical A-level examples. The particular solution adds routine work but no additional conceptual difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Selects integrating factor method: \(P = \frac{2}{x} \Rightarrow \int P\,dx = 2\ln x\) | M1 (3.1a) | |
| Integrating factor \(= e^{\int P\,dx} = x^2\) | A1 (1.1b) | |
| Multiplies DE by integrating factor: \(x^2\frac{dy}{dx} + 2xy = \frac{x(x+3)}{(x-1)(x^2+3)}\), giving \(\frac{d}{dx}(x^2 y) = \frac{x^2+3x}{(x-1)(x^2+3)}\) | M1 (1.1a) | |
| Correctly integrates LHS to obtain \(x^2 y\) | A1F (1.1b) | |
| Splits RHS: \(\frac{x^2+3x}{(x-1)(x^2+3)} \equiv \frac{A}{x-1} + \frac{Bx+C}{x^2+3}\), obtaining \(\frac{1}{x-1} + \frac{3}{x^2+3}\) | M1 (3.1a) | Numerators in correct form |
| Partial fractions of form \(\frac{A}{x-1} + \frac{Bx+C}{x^2+3}\) with correct values, no extra fractions | M1 (1.1a) | |
| Integrates \(\frac{1}{x^2+3}\) to obtain \(k\tan^{-1}\frac{x}{\sqrt{3}}\) | B1 (1.1b) | |
| \(x^2 y = \ln(x-1) + \frac{3}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} + c\), i.e. \(y = \frac{1}{x^2}\left(\ln(x-1) + \sqrt{3}\tan^{-1}\frac{x}{\sqrt{3}} + c\right)\) | A1 (1.1b) | Condone omission of \(c\) (ACF) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Substitutes \((3,0)\): \(0 = \frac{1}{9}(\ln 2 + \sqrt{3}\tan^{-1}\sqrt{3} + c)\), giving \(c = -\ln 2 - \frac{\pi\sqrt{3}}{3}\) | M1 (1.1a) | |
| \(y = \frac{1}{x^2}\left(\ln(x-1) + \sqrt{3}\tan^{-1}\frac{x}{\sqrt{3}} - \ln 2 - \frac{\pi\sqrt{3}}{3}\right)\) | A1F (1.1b) | FT their general solution; condone correct \(c\) as decimal |
## Question 10(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Selects integrating factor method: $P = \frac{2}{x} \Rightarrow \int P\,dx = 2\ln x$ | M1 (3.1a) | |
| Integrating factor $= e^{\int P\,dx} = x^2$ | A1 (1.1b) | |
| Multiplies DE by integrating factor: $x^2\frac{dy}{dx} + 2xy = \frac{x(x+3)}{(x-1)(x^2+3)}$, giving $\frac{d}{dx}(x^2 y) = \frac{x^2+3x}{(x-1)(x^2+3)}$ | M1 (1.1a) | |
| Correctly integrates LHS to obtain $x^2 y$ | A1F (1.1b) | |
| Splits RHS: $\frac{x^2+3x}{(x-1)(x^2+3)} \equiv \frac{A}{x-1} + \frac{Bx+C}{x^2+3}$, obtaining $\frac{1}{x-1} + \frac{3}{x^2+3}$ | M1 (3.1a) | Numerators in correct form |
| Partial fractions of form $\frac{A}{x-1} + \frac{Bx+C}{x^2+3}$ with correct values, no extra fractions | M1 (1.1a) | |
| Integrates $\frac{1}{x^2+3}$ to obtain $k\tan^{-1}\frac{x}{\sqrt{3}}$ | B1 (1.1b) | |
| $x^2 y = \ln(x-1) + \frac{3}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}} + c$, i.e. $y = \frac{1}{x^2}\left(\ln(x-1) + \sqrt{3}\tan^{-1}\frac{x}{\sqrt{3}} + c\right)$ | A1 (1.1b) | Condone omission of $c$ (ACF) |
## Question 10(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Substitutes $(3,0)$: $0 = \frac{1}{9}(\ln 2 + \sqrt{3}\tan^{-1}\sqrt{3} + c)$, giving $c = -\ln 2 - \frac{\pi\sqrt{3}}{3}$ | M1 (1.1a) | |
| $y = \frac{1}{x^2}\left(\ln(x-1) + \sqrt{3}\tan^{-1}\frac{x}{\sqrt{3}} - \ln 2 - \frac{\pi\sqrt{3}}{3}\right)$ | A1F (1.1b) | FT their general solution; condone correct $c$ as decimal |10
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 2 y } { x } = \frac { x + 3 } { x ( x - 1 ) \left( x ^ { 2 } + 3 \right) } \quad ( x > 1 )$$
10
\item Find the particular solution for which $y = 0$ when $x = 3$\\
Give your answer in the form $y = \mathrm { f } ( x )$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q10 [10]}}