FP2 June 2008 Q9
9. \[(x^2 + 1)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2y^2 + (1 - 2x)\frac{\mathrm{d}y}{\mathrm{d}x} \qquad \text{(I)}\]
(a) By differentiating equation (I) with respect to \(x\), show that \[(x^2 + 1)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = (1 - 4x)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (4y - 2)\frac{\mathrm{d}y}{\mathrm{d}x}.\] (3)
Given that \(y = 1\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) at \(x = 0\),
(b) find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x^3\). (4)
(c) Use your series to estimate the value of \(y\) at \(x = -0.5\), giving your answer to two decimal places. (1)
| Scheme | Marks |
|---|---|
| \((x^2 + 1)\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 2x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4y\dfrac{\mathrm{d}y}{\mathrm{d}x} + (1 - 2x)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1A1 |
| \((x^2 + 1)\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = (1 - 4x)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + (4y - 2)\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (*) | A1 |
| (3) |
Notes
M: Use of product rule (at least once) and implicit differentiation (at least once).
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = 3\) | B1 |
| \(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = 5\) Follow through: \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\) | B1ft |
| \(y = 1 + x + \dfrac{3}{2}x^2 + \dfrac{5}{6}x^3 \ldots\) | M1A1 |
| (4) |
Notes
M: Use of series expansion with values for the derivatives (can be allowed without the first term 1, and can also be allowed if final term uses 3 rather than 3!)
| Scheme | Marks |
|---|---|
| \(x = -0.5,\ y \approx 1 - 0.5 + 0.375 - 0.104166\ldots\) \(= 0.77\) (2 d.p.) [awrt 0.77] | B1 |
| (1) | |
| (8 marks) |