FP2 June 2011 Q2
2. \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \mathrm{e}^x\left(2y\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1\right)\]
(a) Show that \[\frac{\mathrm{d}^3y}{\mathrm{d}x^3} = \mathrm{e}^x\left[2y\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + ky\frac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1\right],\] where \(k\) is a constant to be found. (3)
Given that, at \(x = 0\), \(y = 1\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\),
(b) find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x^3\). (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \mathrm{e}^x\left(2y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) + \mathrm{e}^x\left(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1\right)\) | M1 A1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \mathrm{e}^x\left(2y\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 2\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 4y\dfrac{\mathrm{d}y}{\mathrm{d}x} + y^2 + 1\right)\) \((k = 4)\) | A1 |
| (3) |
Notes
1st M1 for evidence of Product Rule
1st A1 for completely correct expression or equivalent
2nd A1 for correct expression or \(k = 4\) stated
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = \mathrm{e}^0(4 + 1 + 1) = 6\) | B1 |
| \(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = \mathrm{e}^0(12 + 8 + 8 + 1 + 1) = 30\) | B1 |
| \(y = 1 + 2x + \dfrac{6x^2}{2} + \dfrac{30x^3}{6} = 1 + 2x + 3x^2 + 5x^3\) | M1 A1ft |
| (4) | |
| (7 marks) |
Notes
2nd M1 require four terms and denominators of 2 and 6 (might be implied)
A1 follow through from their values in the final answer.