FP2 June 2013 Q3
3. \[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 4y - \sin x = 0\]
Given that \(y = \dfrac{1}{2}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{8}\) at \(x = 0\),
find a series expansion for \(y\) in terms of \(x\), up to and including the term in \(x^3\). (5)
| Scheme | Marks |
|---|---|
| \((x = 0)\quad \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \sin 0 - 4 \times \dfrac{1}{2} = -2\) | B1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} + 4\dfrac{\mathrm{d}y}{\mathrm{d}x} - \cos x\ (= 0)\) | M1 |
| \((x = 0)\quad \dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = \cos 0 - 4 \times \dfrac{1}{8} = \dfrac{1}{2}\) | A1 |
| \((y =)\ y_0 + x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 + \dfrac{x^2}{2!}\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 + \dfrac{x^3}{3!}\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 + \ldots\) | M1 (2! or 2 and 3! or 6) |
| \((y =)\ \dfrac{1}{2} + x \times \dfrac{1}{8} + \dfrac{x^2}{2} \times (-2) + \dfrac{x^3}{6} \times \dfrac{1}{2}\) | |
| \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) | A1 cao |
| (5 marks) |
Notes
B1 for \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_0 = -2\) wherever seen
M1 for attempting the differentiation of the given equation. To obtain \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} \pm k\dfrac{\mathrm{d}y}{\mathrm{d}x} \pm \cos x\ (= 0)\) oe
A1 for substituting \(x = 0\) to obtain \(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_0 = \dfrac{1}{2}\)
M1 for using the expansion \(\left[y = \mathrm{f}(x)\right] = \mathrm{f}(0) + x\mathrm{f}^{\prime}(0) + \dfrac{x^2}{2(!)}\mathrm{f}^{\prime\prime}(0) + \dfrac{x^3}{3!}\mathrm{f}^{\prime\prime\prime}(0)\) with their values for \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\). Factorial can be omitted in the \(x^2\) term but must be shown explicitly in the \(x^3\) term or implied by further working eg using 6.
A1cao for \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) (Ignore any higher powers included) Exact decimals allowed. Must include \(y = \ldots\)
Alternative
| Scheme | Marks |
|---|---|
| \(y = \dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3 + \ldots\) | B1 |
| \(y'' = 2a + 6bx + \ldots\) | M1 Diff twice |
| \(2a + 6bx + \ldots = \sin x - 4\left(\dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3\ldots\right)\) | A1 Correct differentiation and equation used |
| \(2a + 2 = 0\quad a = -1\) | M1 |
| \(6b + \dfrac{1}{2} = 1\quad b = \dfrac{1}{12}\) | |
| \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) | A1cao |
(corrected from the printed mark scheme: the printed alternative omits the factor 4 in front of the bracket, which is shown in the notes below.)
B1 for \(y = \dfrac{1}{2} + \dfrac{x}{8} + ax^2 + bx^3 + \ldots\)
M1 for differentiating this twice to get \(y'' = 2a + 6bx + \ldots\) (may not be completely correct)
A1 for correct differentiation and using the given equation and the expansion of \(\sin x\) to get \(2a + 6bx + \ldots = \left(x - \dfrac{x^3}{6} + \ldots\right) - 4\left(\dfrac{1}{2} + \dfrac{x}{8} + \ldots\right)\) (corrected from the printed mark scheme: \(\dfrac{x^3}{3}\) is printed for \(\dfrac{x^3}{6}\))
M1 for equating coefficients to obtain a value for \(a\) or \(b\)
A1 cao for \(y = \dfrac{1}{2} + \dfrac{x}{8} - x^2 + \dfrac{x^3}{12}\) (Ignore any higher powers included)