FP2 June 2012 Q5
5. \[x\frac{\mathrm{d}y}{\mathrm{d}x} = 3x + y^2\]
(a) Show that \[x\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (1 - 2y)\frac{\mathrm{d}y}{\mathrm{d}x} = 3\] (2)
Given that \(y = 1\) at \(x = 1\),
(b) find a series solution for \(y\) in ascending powers of \((x - 1)\), up to and including the term in \((x - 1)^3\). (8)
| Scheme | Marks |
|---|---|
| \(x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3 + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (Using differentiation of product or quotient and also differentiation of implicit function) | M1 |
| \(x\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + (1 - 2y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\) **ag** | A1 cso |
| (2) |
Notes
Finding second derivative and substituting into given answer acceptable
| Scheme | Marks |
|---|---|
| \(\left(x\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right) + \ldots\) | B1 |
| \(\ldots\left[(1 - 2y)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right] = 0\) | M1 A1 |
| At \(x = 1\): \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\) | B1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 7\) \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = 32\) | B1, B1 |
| \((y =)\mathrm{f}(1) + \mathrm{f}^{\prime}(1)(x - 1) + \dfrac{\mathrm{f}^{\prime\prime}(1)(x - 1)^2}{2} + \dfrac{\mathrm{f}^{\prime\prime\prime}(1)(x - 1)^3}{6}\ldots\) | M1 |
| \(y = 1 + 4(x - 1) + \dfrac{7}{2}(x - 1)^2 + \dfrac{16}{3}(x - 1)^3\) (or equiv.) | A1 ft |
| (8) | |
| (10 marks) |
Notes
1st M1 for differentiating second term to obtain an expression involving \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\)
B1B1B1 for 4, 7, 32 seen respectively
2nd M1 require \(\mathrm{f}(1)\) or 1, \(\mathrm{f}^{\prime}(1)\) etc and \(x - 1\) and at least first 3 terms
A1 for 4 terms following through their constants
Condone \(\mathrm{f}(x) =\) instead of \(y =\)