| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Iterative/numerical methods |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring numerical methods (Euler and midpoint), solving a first-order linear ODE with integrating factor, and error analysis. While each technique is standard, the combination, the algebraic complexity of the integrating factor method with the given equation, and the need to compare numerical vs analytical solutions makes this moderately challenging for Further Maths students. |
| Spec | 1.09d Newton-Raphson method4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_0 = 5\), \(y_0 = 12.3\), \(h = 0.1\) — Uses Euler method once | M1 | Condone one slip |
| \(y_1 = 12.3 + 0.1\left(4 + \frac{2 \times 5 \times 12.3}{16}\right) = \frac{431}{32}\) | A1 | Allow AWRT 13.5 |
| \(x_1 = 5.1\) — Uses midpoint formula once | M1 | Condone one slip |
| \(y_2 = 12.3 + 0.2\left(\frac{9\sqrt{21}}{10} + \frac{2 \times 5.1 \times \frac{431}{32}}{17.01}\right) = 14.7402\) | A1 | Correct to 6 sig. fig. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} - \left(\frac{2x}{x^2-9}\right)y = (x^2-9)^{\frac{1}{2}}\) — Applies Integrating Factor Method | M1 | |
| \(\int P\,dx = -\int \frac{2x}{x^2-9}\,dx = -\ln(x^2-9)\) | A1 | Finds correct integrating factor |
| Integrating factor \(= e^{\int P\,dx} = \frac{1}{x^2-9}\) | M1 | Multiplies equation by their integrating factor |
| \(\frac{d}{dx}\left(\frac{y}{x^2-9}\right) = (x^2-9)^{-\frac{1}{2}}\) | A1 | Integrates LHS to obtain \(\frac{y}{x^2-9}\) |
| \(\frac{y}{x^2-9} = \int (x^2-9)^{-\frac{1}{2}}\,dx\) | M1 | Uses inverse cosh or logarithmic equivalent to integrate RHS |
| \(y = (x^2-9)\left(\cosh^{-1}\frac{x}{3} + c\right)\) | A1 | Correct solution including constant. ACF: \(\frac{y}{x^2-9} = \cosh^{-1}\frac{x}{3} + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{12.3}{16} = \cosh^{-1}\frac{5}{3} + c\) | M1 | Uses initial conditions to find constant of integration |
| \(c = \frac{12.3}{16} - \cosh^{-1}\frac{5}{3} = -0.32986\ldots\); when \(x = 5.2\) | M1 | Substitutes \(x = 5.2\) and their constant into their solution of DE |
| \(y = (5.2^2-9)\left(\cosh^{-1}\frac{5.2}{3} - 0.32986\ldots\right) = 14.7434\) (6 sig. fig.) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The value from part (a) is equal to the answer to part (b)(ii) to four significant figures, meaning that the estimate in part (a) is very accurate. | E1 | Compares the two correct values to make a correct evaluation |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_0 = 5$, $y_0 = 12.3$, $h = 0.1$ — Uses Euler method once | M1 | Condone one slip |
| $y_1 = 12.3 + 0.1\left(4 + \frac{2 \times 5 \times 12.3}{16}\right) = \frac{431}{32}$ | A1 | Allow AWRT 13.5 |
| $x_1 = 5.1$ — Uses midpoint formula once | M1 | Condone one slip |
| $y_2 = 12.3 + 0.2\left(\frac{9\sqrt{21}}{10} + \frac{2 \times 5.1 \times \frac{431}{32}}{17.01}\right) = 14.7402$ | A1 | Correct to 6 sig. fig. |
---
## Question 9(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} - \left(\frac{2x}{x^2-9}\right)y = (x^2-9)^{\frac{1}{2}}$ — Applies Integrating Factor Method | M1 | |
| $\int P\,dx = -\int \frac{2x}{x^2-9}\,dx = -\ln(x^2-9)$ | A1 | Finds correct integrating factor |
| Integrating factor $= e^{\int P\,dx} = \frac{1}{x^2-9}$ | M1 | Multiplies equation by their integrating factor |
| $\frac{d}{dx}\left(\frac{y}{x^2-9}\right) = (x^2-9)^{-\frac{1}{2}}$ | A1 | Integrates LHS to obtain $\frac{y}{x^2-9}$ |
| $\frac{y}{x^2-9} = \int (x^2-9)^{-\frac{1}{2}}\,dx$ | M1 | Uses inverse cosh or logarithmic equivalent to integrate RHS |
| $y = (x^2-9)\left(\cosh^{-1}\frac{x}{3} + c\right)$ | A1 | Correct solution including constant. ACF: $\frac{y}{x^2-9} = \cosh^{-1}\frac{x}{3} + c$ |
---
## Question 9(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{12.3}{16} = \cosh^{-1}\frac{5}{3} + c$ | M1 | Uses initial conditions to find constant of integration |
| $c = \frac{12.3}{16} - \cosh^{-1}\frac{5}{3} = -0.32986\ldots$; when $x = 5.2$ | M1 | Substitutes $x = 5.2$ and their constant into their solution of DE |
| $y = (5.2^2-9)\left(\cosh^{-1}\frac{5.2}{3} - 0.32986\ldots\right) = 14.7434$ (6 sig. fig.) | A1 | Correct answer |
---
## Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The value from part (a) is equal to the answer to part (b)(ii) to four significant figures, meaning that the estimate in part (a) is very accurate. | E1 | Compares the two correct values to make a correct evaluation |
---9
\begin{enumerate}[label=(\alph*)]
\item A curve passes through the point (5, 12.3) and satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( x ^ { 2 } - 9 \right) ^ { \frac { 1 } { 2 } } + \frac { 2 x y } { x ^ { 2 } - 9 } \quad x > 3$$
Use Euler's step by step method once, and then the midpoint formula
$$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right) , \quad x _ { r + 1 } = x _ { r } + h$$
once, each with a step length of 0.1 , to estimate the value of $y$ when $x = 5.2$\\
Give your answer to six significant figures.\\
9
\item (i) Find the general solution of the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \left( x ^ { 2 } - 9 \right) ^ { \frac { 1 } { 2 } } + \frac { 2 x y } { x ^ { 2 } - 9 } \quad ( x > 3 )$$
9 (b) (ii) Given that $y$ satisfies the differential equation in part (b)(i) and that $y = 12.3$ when $x = 5$, find the value of $y$ when $x = 5.2$
Give your answer to six significant figures.\\[0pt]
[3 marks]\\
9
\item Comment on the accuracy of your answer to part (a).
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q9 [14]}}