A2 June 2023 Paper 1 Q11
11 Solve the differential equation \(\cosh x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\sinh x = \cosh x\), given that \(y = 1\) when \(x = 0\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\dfrac{\sinh x}{\cosh x} = 1\) | B1 | 3.1a |
| IF is \(\mathrm{e}^{\int -\frac{2\sinh x}{\cosh x}\,\mathrm{d}x}\) | M1 | 2.1 |
| \(= \mathrm{e}^{-2\ln\cosh x} = \left(\mathrm{e}^{\ln\operatorname{sech} x}\right)^2 = \operatorname{sech}^2 x\) | A1 | 2.2a |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y\operatorname{sech}^2 x\right) = \operatorname{sech}^2 x\) | M1 | 2.1 |
| \(y\operatorname{sech}^2 x = \int\operatorname{sech}^2 x\,\mathrm{d}x + c = \tanh x + c\) | A1 | 1.1 |
| substituting \(x = 0,\ y = 1\) gives \(c = 1\) | M1 | 1.1 |
| \(y = \cosh^2 x\,(\tanh x + 1)\) | A1 | 2.2a |
| [7] |
Notes
B1: \(\frac{\mathrm{d}y}{\mathrm{d}x} - 2y\tanh x = 1\)
M1: (1st) or \(\mathrm{e}^{\int -2\tanh x\,\mathrm{d}x}\). Integral must come from an attempt to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) on its own
M1: (2nd) or multiplying through by their IF
M1: (3rd) substituting \(x = 0\) and \(y = 1\) to lead to a value for \(c\) (corrected from the printed mark scheme, which says “substituting \(x = 1\) and \(y = 1\)”; the condition is \(y = 1\) when \(x = 0\))
A1: (final) oe, e.g. \(\cosh x\,(\sinh x + \cosh x)\) or \(\mathrm{e}^x\cosh x\)