A2 June 2023 Paper 1 Q5
5
(a) Find the general solution of the differential equation \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 5y = 0\). [2]
(b) Hence find the general solution of the differential equation \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} - 2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 5y = x(4 - 5x)\). [4]
| Scheme | Marks |
|---|---|
| Auxiliary equation: \(n^2 - 2n + 5 = 0\) \(\Rightarrow n = 1 \pm 2\mathrm{i}\) | B1 |
| \(\Rightarrow y = \mathrm{e}^x(A\cos 2x + B\sin 2x)\) oe | B1 |
| [2] |
Notes
B1: Correct roots of auxiliary equation.
B1: ft complex \(k\) only.
Or \(y = R\mathrm{e}^x\cos(2x + \varphi)\) or \(y = R\mathrm{e}^x\sin(2x + \varphi)\)
Or \(y = A\mathrm{e}^{(2\mathrm{i} + 1)x} + B\mathrm{e}^{(-2\mathrm{i} + 1)x}\)
Final equation must be \(y = \mathrm{f}(x)\)
| Scheme | Marks |
|---|---|
| Trial function: \(y = ax^2 + bx + c\) | B1 |
| \(\Rightarrow y^{\prime} = 2ax + b, \quad y^{\prime\prime} = 2a\) \(\Rightarrow 2a - 2(2ax + b) + 5(ax^2 + bx + c) \equiv 4x - 5x^2\) \(\Rightarrow 5ax^2 + x(-4a + 5b) + 5c - 2b + 2a \equiv 4x - 5x^2\) | M1 |
| \(\Rightarrow a = -1, b = 0, c = \dfrac{2}{5}\) | A1 |
| \(\Rightarrow \text{GS: } y = \mathrm{e}^x(A\cos 2x + B\sin 2x) - x^2 + \dfrac{2}{5}\) | A1ft |
| [4] |
Notes
B1: Allow any extraneous terms in the trial function (eg. \(dx^3\)) as long as \(d\) shown to be zero, \(a, b, c \neq 0\)
M1: Differentiates their trial function to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and substitutes
A1ft: ft their particular integral, and their CF from (a) (dependent on CF containing exactly two arbitrary constants)