A2 June 2022 Paper 2 Q9
9
(a) A curve passes through the point \((5, 12.3)\) and satisfies the differential equation\[\frac{\mathrm{d}y}{\mathrm{d}x} = (x^2 - 9)^{\frac{1}{2}} + \frac{2xy}{x^2 - 9} \qquad x \gt 3\]
Use Euler’s step by step method once, and then the midpoint formula
\[y_{r+1} = y_{r-1} + 2h\mathrm{f}(x_r, y_r), \quad x_{r+1} = x_r + h\]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. [4 marks]
(b)
(i) Find the general solution of the differential equation\[\frac{\mathrm{d}y}{\mathrm{d}x} = (x^2 - 9)^{\frac{1}{2}} + \frac{2xy}{x^2 - 9} \qquad (x \gt 3)\] [6 marks]
(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. [3 marks]
(c) Comment on the accuracy of your answer to part (a). [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Uses Euler method once (condone one slip) | M1 | 1.1a |
| Obtains correct value of \(y_1\) Allow AWRT 13.5 | A1 | 1.1b |
| Uses midpoint formula once (condone one slip) | M1 | 3.1a |
| Obtains the correct value to 6 sig. fig. | A1 | 1.1b |
| (4) |
Typical solution
\[x_0 = 5\]\[y_0 = 12.3\]\[h = 0.1\]\[y_1 = 12.3 + 0.1\left(4 + \frac{2 \times 5 \times 12.3}{16}\right) = \frac{431}{32}\]\[x_1 = 5.1\]\[y_2 = 12.3 + 0.2\left(\frac{9\sqrt{21}}{10} + \frac{2 \times 5.1 \times \frac{431}{32}}{17.01}\right) = 14.7402\]| Scheme | Marks | AO |
|---|---|---|
| (i) Applies the Integrating Factor Method | M1 | 3.1a |
| Finds correct integrating factor | A1 | 1.1b |
| Multiplies equation by their integrating factor | M1 | 1.1a |
| Integrates LHS to obtain \(\dfrac{y}{x^2 - 9}\) | A1 | 1.1b |
| Uses inverse cosh or logarithmic equivalent to integrate RHS | M1 | 1.1a |
| Finds correct solution including constant of integration. ACF Accept \(\dfrac{y}{x^2 - 9} = \cosh^{-1}\dfrac{x}{3} + c\) | A1 | 1.1b |
| (6) | ||
| (ii) Uses initial conditions to find constant of integration | M1 | 3.1a |
| Substitutes \(x = 5.2\) and their constant of integration into their solution of DE | M1 | 1.1a |
| Obtains correct answer | A1 | 1.1b |
| (3) |
Typical solution
(i)
\[\frac{\mathrm{d}y}{\mathrm{d}x} - \left(\frac{2x}{x^2 - 9}\right)y = (x^2 - 9)^{\frac{1}{2}}\]\[\begin{aligned}\int P\,\mathrm{d}x &= -\int \frac{2x}{x^2 - 9}\,\mathrm{d}x \\ &= -\ln(x^2 - 9)\end{aligned}\]Integrating factor \(= \mathrm{e}^{\int P\,\mathrm{d}x} = \dfrac{1}{x^2 - 9}\)
\[\left(\frac{1}{x^2 - 9}\right)\frac{\mathrm{d}y}{\mathrm{d}x} - \left(\frac{2x}{(x^2 - 9)^2}\right)y = (x^2 - 9)^{-\frac{1}{2}}\]\[\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{x^2 - 9}\right) = (x^2 - 9)^{-\frac{1}{2}}\]\[\frac{y}{x^2 - 9} = \int (x^2 - 9)^{-\frac{1}{2}}\,\mathrm{d}x\]\[y = (x^2 - 9)\left(\cosh^{-1}\frac{x}{3} + c\right)\](ii)
\[\frac{12.3}{16} = \cosh^{-1}\frac{5}{3} + c\]\[c = \frac{12.3}{16} - \cosh^{-1}\frac{5}{3} = -0.32986\ldots\]When \(x = 5.2\)
\[y = (5.2^2 - 9)\left(\cosh^{-1}\frac{5.2}{3} - 0.32986\ldots\right)\]\(y = 14.7434\) (6 sig. fig.)
| Scheme | Marks | AO |
|---|---|---|
| Compares the two correct values to make a correct evaluation | E1 | 3.2b |
| (1) | ||
| (14 marks) |
Typical solution
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.