A2 June 2019 Q3
3.
\[\dfrac{\mathrm{d}y}{\mathrm{d}x} = x - y^2 \qquad \text{(I)}\]| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} \Rightarrow \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -2y\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\) | M1 A1 | 1.1b 1.1b |
| \(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} = -2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2y\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 4\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = -6\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} - 2y\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) | dM1 | 2.1 |
| \(\begin{aligned}&\dfrac{\mathrm{d}^{5}y}{\mathrm{d}x^{5}} = -6\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 6\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2 - 2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\\[6pt] &= -2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 8\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 6\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2\end{aligned}\) | A1 | 2.1 |
| (4) |
Notes
M1: Attempts to find the second and third derivatives:
This requires \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) followed by \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \pm 2y\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} \pm \ldots\) or \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = \pm \ldots \pm 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)
A1: Correct second and third derivatives.
dM1: Continues to differentiate to reach the 5th derivative. This is dependent on the first method mark but there is no need to check the detail and the mark can be awarded as long as the 5th derivative is reached.
A1: Completes the process, collecting terms if necessary, to obtain the correct expression
(NB \(a = -2\), \(b = -8\), \(c = -6\))
Allow dash/dot notation for the derivatives but the final answer must be in the correct form.
Note that if \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is obtained initially, allow a full recovery in (a).
Note that (a) can be found using Leibnitz’s theorem and the following scheme should be applied:
M1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) followed by an attempt to differentiate \(y\) 3 times and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) 3 times.
A1: All correct
dM1: \(\dfrac{\mathrm{d}^{5}y}{\mathrm{d}x^{5}} = -2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 3 \times 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 3 \times 2\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2 - 2\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (correct application of Leibnitz)
A1: \(= -2y\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}} - 8\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} - 6\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)^2\)
(Corrected from the printed mark scheme: the Leibnitz line is printed with the first term as \(-2\dfrac{\mathrm{d}y}{\mathrm{d}x}\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\) and the last as \(-2y\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\dfrac{\mathrm{d}y}{\mathrm{d}x}\); the factors \(y\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) are the wrong way round.)
As in the main scheme, if \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = \pm 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is obtained initially, allow a full recovery in (a).
Alternative for (a)
M1: \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 1 - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - 2y\left(x - y^2\right) = 1 - 2xy + 2y^3 \Rightarrow \dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}} = -2y - 2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 6y^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Score for the second derivative form as in the main scheme and then an attempt at the third derivative with at least 2 terms correct.
A1: Fully correct
Then as main scheme.
| Scheme | Marks | AO |
|---|---|---|
| \(x = 0,\ y = 1 \Rightarrow \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 = 0 - 1^2 = -1\) | B1 | 2.2a |
| \(\begin{aligned}&\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)_0 = 1 - 2(1)(-1) = 3,\ \left(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\right)_0 = -2(1)(3) - 2(-1)^2 = -8\\[6pt] &\left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\right)_0 = -6(-1)(3) - 2(1)(-8) = 34,\\[6pt] &\left(\dfrac{\mathrm{d}^{5}y}{\mathrm{d}x^{5}}\right)_0 = -2(1)(34) - 8(-1)(-8) - 6(3)^2 = -186\end{aligned}\) | M1 A1 | 1.1b 1.1b |
| \(y = y(0) + x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 + \dfrac{x^2}{2!}\left(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\right)_0 + \dfrac{x^3}{3!}\left(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\right)_0 + \dfrac{x^4}{4!}\left(\dfrac{\mathrm{d}^{4}y}{\mathrm{d}x^{4}}\right)_0 + \dfrac{x^5}{5!}\left(\dfrac{\mathrm{d}^{5}y}{\mathrm{d}x^{5}}\right)_0 + \ldots\) With their values | M1 | 2.5 |
| \(\begin{aligned}&(y =)\,1 - x + \dfrac{3}{2}x^2 - \dfrac{8}{6}x^3 + \dfrac{34}{24}x^4 - \dfrac{186}{120}x^5 + \ldots\\[6pt] &(y =)\,1 - x + \dfrac{3}{2}x^2 - \dfrac{4}{3}x^3 + \dfrac{17}{12}x^4 - \dfrac{31}{20}x^5 + \ldots\end{aligned}\) | A1ft | 1.1b |
| (5) | ||
| (9 marks) |
Notes
B1: Deduces the correct value for \(y^{\prime}(0)\)
M1: Finds the values of all the other derivatives at \(x = 0\) up to 5th. There is no need to check their values as long as there is no obvious incorrect work, but values for all the derivatives up to the 5th must be found
A1: All values correct (as single values – e.g. do not allow unsimplified)
M1: Applies the correct Maclaurin series for their values including the factorials up to the term in \(x^5\)
A1ft: Correct expansion, follow through their values for the derivatives. This does not have to be simplified but the factorials need to be evaluated. Once a correct, or correct follow through expression is seen apply isw.