C4 June 2018 Q6
6. Given that \(y = 2\) when \(x = -\dfrac{\pi}{8}\), solve the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2}{3\cos^2 2x} \qquad -\frac{1}{2} < x < \frac{1}{2}\] giving your answer in the form \(y = \mathrm{f}(x)\).
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y^2}{3\cos^2 2x};\ -\dfrac{1}{2} < x < \dfrac{1}{2};\ y = 2\) at \(x = -\dfrac{\pi}{8}\) | |
| \(\displaystyle\int \frac{1}{y^2}\,\mathrm{d}y = \int \frac{1}{3\cos^2 2x}\,\mathrm{d}x\) Separates variables as shown. Can be implied by a correct attempt at integration. Ignore the integral signs | B1 |
| \(\displaystyle\int \frac{1}{y^2}\,\mathrm{d}y = \int \frac{1}{3}\sec^2 2x\,\mathrm{d}x\) | |
| \(-\dfrac{1}{y} = \dfrac{1}{3}\left(\dfrac{\tan 2x}{2}\right)\ \{+c\}\) \(\pm\dfrac{A}{y^2} \to \pm\dfrac{B}{y};\ A, B \neq 0\) \(\pm\lambda\tan 2x\) \(-\dfrac{1}{y} = \dfrac{1}{3}\left(\dfrac{\tan 2x}{2}\right)\) | M1 M1 A1 |
| \(-\dfrac{1}{2} = \dfrac{1}{6}\tan\left(2\left(-\dfrac{\pi}{8}\right)\right) + c\) Use of \(x = -\dfrac{\pi}{8}\) and \(y = 2\) in an integrated equation containing a constant of integration, e.g. \(c\) | M1 |
| \(-\dfrac{1}{2} = -\dfrac{1}{6} + c \Rightarrow c = -\dfrac{1}{3}\) | |
| \(-\dfrac{1}{y} = \dfrac{1}{6}\tan 2x - \dfrac{1}{3} = \dfrac{\tan(2x) - 2}{6}\) | |
| \(y = \dfrac{-1}{\frac{1}{6}\tan 2x - \frac{1}{3}}\) or \(y = \dfrac{6}{2 - \tan 2x}\) or \(y = \dfrac{6\cot 2x}{-1 + 2\cot 2x}\) \(\left\{-\dfrac{1}{2} < x < \dfrac{1}{2}\right\}\) | A1 o.e. |
| (6 marks) |
Notes
B1: Separates variables as shown. \(\mathrm{d}y\) and \(\mathrm{d}x\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. The number “3” may appear on either side.
E.g. \(\displaystyle\int \frac{1}{y^2}\,\mathrm{d}y = \int \frac{1}{3}\sec^2 2x\,\mathrm{d}x\) or \(\displaystyle\int \frac{3}{y^2}\,\mathrm{d}y = \int \frac{1}{\cos^2 2x}\,\mathrm{d}x\) are fine for B1
Note: Allow e.g. \(\displaystyle\int \frac{1}{y^2}\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x = \int \frac{1}{3}\sec^2 2x\,\mathrm{d}x\) for B1 or condone \(\displaystyle\int \frac{1}{y^2} = \int \frac{1}{3}\sec^2 2x\) for B1
Note: B1 can be implied by correct integration of both sides
M1: \(\pm\dfrac{A}{y^2} \to \pm\dfrac{B}{y};\ A, B \neq 0\)
M1: \(\dfrac{1}{\cos^2 2x}\) or \(\sec^2 2x \to \pm\lambda\tan 2x;\ \lambda \neq 0\)
A1: \(-\dfrac{1}{y} = \dfrac{1}{3}\left(\dfrac{\tan 2x}{2}\right)\) with or without '\(+ c\)'. E.g. \(-\dfrac{6}{y} = \tan 2x\)
M1: Evidence of using both \(x = -\dfrac{\pi}{8}\) and \(y = 2\) in an integrated or changed equation containing \(c\)
Note: This mark can be implied by the correct value of \(c\)
Note: You may need to use your calculator to check that they have satisfied the final M mark
Note: Condone using \(x = \dfrac{\pi}{8}\) instead of \(x = -\dfrac{\pi}{8}\)
A1: \(y = \dfrac{-1}{\frac{1}{6}\tan 2x - \frac{1}{3}}\) or \(y = \dfrac{6}{2 - \tan 2x}\) or any equivalent correct answer in the form \(y = \mathrm{f}(x)\)
Note: You can ignore subsequent working, which follows from a correct answer
Note: Writing \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y^2}{3\cos^2 2x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{3}y^2\sec^2 2x\) leading to e.g.
- \(y = \dfrac{1}{9}y^3\left(\dfrac{1}{2}\tan 2x\right)\) gets 2nd M0 for \(\pm\lambda\tan 2x\)
- \(u = \dfrac{1}{3}y^2,\ \dfrac{\mathrm{d}v}{\mathrm{d}x} = \sec^2 2x \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{2}{3}y,\ v = \dfrac{1}{2}\tan 2x\) gets 2nd M0 for \(\pm\lambda\tan 2x\)
because the variables have not been separated