FP2 June 2007 Q10
10. \[(1 - x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - x\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = 0.\]
At \(x = 0\), \(y = 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -1\).
(a) Find the value of \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) at \(x = 0\). (3)
(b) Express \(y\) as a series in ascending powers of \(x\), up to and including the term in \(x^3\). (4)
| Scheme | Marks |
|---|---|
| \((1 - x^2)\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} - 2x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - \dfrac{\mathrm{d}y}{\mathrm{d}x} + 2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 |
| At \(x = 0\), \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = -\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) | M1A1cso |
| (3) |
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = -4\) Allow anywhere | B1 |
| \(y = \mathrm{f}(0) + \mathrm{f}'(0)x + \dfrac{\mathrm{f}''(0)}{2}x^2 + \dfrac{\mathrm{f}'''(0)}{6}x^3 + \ldots\) | |
| \(= 2 - x - 2x^2,\ + \dfrac{1}{6}x^3 + \ldots\) | M1A1ft, A1 (dep) |
| (4) | |
| (7 marks) |