FP2 June 2007 Q6
6. \[\frac{\mathrm{d}y}{\mathrm{d}x} - y\tan x = 2\sec^3 x.\]
Given that \(y = 3\) at \(x = 0\), find \(y\) in terms of \(x\)
| Scheme | Marks |
|---|---|
| Integrating factor \(\mathrm{e}^{\int -\tan x\,\mathrm{d}x} = \mathrm{e}^{\ln(\cos x)}\) (or \(\mathrm{e}^{-\ln(\sec x)}\)), \(= \cos x\left(\text{or } \dfrac{1}{\sec x}\right)\) \(\left(\cos x\dfrac{\mathrm{d}y}{\mathrm{d}x} - y\sin x = 2\sec^2 x\right)\) | M1, A1 |
| \(y\cos x = \displaystyle\int 2\sec^2 x\,\mathrm{d}x\) (or equiv.) \(\left(\text{Or: } \dfrac{\mathrm{d}}{\mathrm{d}x}(y\cos x) = 2\sec^2 x\right)\) | M1A1(ft) |
| \(y\cos x = 2\tan x\ (+C)\) (or equiv.) | A1 |
| \(y = 3\) at \(x = 0\): \(C = 3\) | M1 |
| \(y = \dfrac{2\tan x + 3}{\cos x}\) (Or equiv. in the form \(y = \mathrm{f}(x)\)) | A1 |
| (7 marks) |
Notes
1st M: Also scored for \(\mathrm{e}^{\int \tan x\,\mathrm{d}x} = \mathrm{e}^{-\ln(\cos x)}\) (or \(\mathrm{e}^{\ln(\sec x)}\)), then A0 for \(\sec x\).
2nd M: Attempt to use their integrating factor (requires one side of the equation ‘correct’ for their integrating factor).
2nd A: The follow-through is allowed only in the case where the integrating factor used is \(\sec x\) or \(-\sec x\). \(\left(y\sec x = \int 2\sec^4 x\,\mathrm{d}x\right)\)
3rd M: Using \(y = 3\) at \(x = 0\) to find a value for \(C\) (dependent on an integration attempt, however poor, on the RHS).
Alternative1st M: Multiply through the given equation by \(\cos x\).
1st A: Achieving \(\cos x\dfrac{\mathrm{d}y}{\mathrm{d}x} - y\sin x = 2\sec^2 x\). (Allowing the possibility of integrating by inspection).