A2 June 2024 Paper 1 Q14
14 Solve the differential equation
\[\frac{\mathrm{d}y}{\mathrm{d}x} + y\tanh x = \sinh^3 x\]given that \(y = 3\) when \(x = \ln 2\)
Give your answer in an exact form. [7 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes \(\mathrm{e}^{\int \tanh x\,\mathrm{d}x}\) | M1 | 3.1a |
| Obtains \(\cosh x\) | A1 | 1.1b |
| Obtains \(y\cosh x\) | B1 | 1.1b |
| Writes down \(\displaystyle\int \sinh^3 x\cosh x\,\mathrm{d}x\) (PI) | M1 | 1.1a |
| Obtains \(k\sinh^4 x\) or \(k_1\cosh 4x + k_2\cosh 2x\) (Accept equivalent exponential form) | A1 | 1.1b |
| Substitutes \(x = \ln 2\) and \(y = 3\) into \(y\cosh x = k\sinh^4 x + c\) or \(y\cosh x = k_1\cosh 4x + k_2\cosh 2x + c\) (Accept equivalent exponential form) and uses to evaluate \(c\) | M1 | 1.1a |
| Deduces \(y\cosh x = \dfrac{1}{4}\sinh^4 x + \dfrac{3759}{1024}\) Or \(y\cosh x = \dfrac{1}{32}\cosh 4x - \dfrac{1}{8}\cosh 2x + \dfrac{3855}{1024}\) ACF. ISW | A1 | 2.2a |
| (7 marks) |
Typical solution
\[\text{Integrating factor} = \mathrm{e}^{\int \tanh x\,\mathrm{d}x}\]\[\int \tanh x\,\mathrm{d}x = \ln(\cosh x)\]Integrating factor \(= \cosh x\)
\[\begin{aligned}y\cosh x &= \int \sinh^3 x\cosh x\,\mathrm{d}x \\ &= \frac{1}{4}\sinh^4 x + c\end{aligned}\]When \(x = \ln 2\), \(\sinh x = \dfrac{3}{4}\) and \(\cosh x = \dfrac{5}{4}\)
\[3 \times \frac{5}{4} = \frac{1}{4} \times \frac{81}{256} + c\]\[c = \frac{3759}{1024}\]\[y\cosh x = \frac{1}{4}\sinh^4 x + \frac{3759}{1024}\]