FP2 June 2014 (R) Q5
5. \[y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 2y = 0\]
Given that \(y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0.5\) at \(x = 0\),
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{2}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 - 2\) | |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) seen | B1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -\dfrac{4}{y}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + \dfrac{2}{y^2}\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^3\) | M1 (\(\div\) and diff) A1A1 |
| Alt: \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) + y\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 4\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{y}\left(-5\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | A1 A1 |
| (4) |
Notes
B1 \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) seen in the differentiation
M1 divide equation by \(y\) and differentiate wrt \(x\) chain and product/quotient rules needed
A1A1 -1 for each error. Ignore any simplification following the differentiation and obtaining \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \ldots\)
ALT:
B1 as above
M1 differentiating before dividing
A1A1 rearrange to a correct expression for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\), -1 each error
| Scheme | Marks |
|---|---|
| At \(x = 0\) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{2}\left(-2\times\left(\dfrac{1}{2}\right)^2 - 4\right) = -\dfrac{9}{4}\) (or \(-2.25\)) | M1A1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{1}{2}\left(-5\times\dfrac{1}{2}\times-\dfrac{9}{4} - 2\times\dfrac{1}{2}\right) = \dfrac{37}{16}\) (or 2.3125) | A1 |
| \(y = 2 + \dfrac{1}{2}x + \left(-\dfrac{9}{4}\right)\dfrac{x^2}{2!} + \left(\dfrac{37}{16}\right)\dfrac{x^3}{3!} + \ldots\) | M1(2! or 2, 3! or 6) |
| \(y = 2 + \dfrac{1}{2}x - \dfrac{9}{8}x^2 + \dfrac{37}{96}x^3 + \ldots\) 0.5 1.125 0.3854 3 sf or better | A1 |
| (5) | |
| (9 marks) |
Notes
M1 using values for \(x\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain a value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1 correct value for \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)
A1 correct value for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\)
M1 Taylor’s series formed using their values for the differentials, accept 2! or 2 and 3! or 6.
A1 correct series, must start \(y =\) (or end \(= y\))
(Corrected from the printed mark scheme: the decimal for \(\dfrac{37}{16}\) is printed as 2.325; \(\dfrac{37}{16} = 2.3125\).)