C4 June 2011 Q8
8.
(a) Find \(\displaystyle\int (4y + 3)^{-\frac{1}{2}}\,\mathrm{d}y\) (2)
(b) Given that \(y = 1.5\) at \(x = -2\), solve the differential equation\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\sqrt{(4y + 3)}}{x^2}\]giving your answer in the form \(y = \mathrm{f}(x)\). (6)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int (4y + 3)^{-\frac{1}{2}}\,\mathrm{d}y = \frac{(4y + 3)^{\frac{1}{2}}}{(4)\left(\frac{1}{2}\right)} \quad (+C)\) \(\left(= \tfrac{1}{2}(4y + 3)^{\frac{1}{2}} + C\right)\) | M1 A1 |
| (2) |
Notes
(corrected from the printed mark scheme: the integral is printed with \(\mathrm{d}x\) for \(\mathrm{d}y\))
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{1}{\sqrt{(4y + 3)}}\,\mathrm{d}y = \int \frac{1}{x^2}\,\mathrm{d}x\) | B1 |
| \(\displaystyle\int (4y + 3)^{-\frac{1}{2}}\,\mathrm{d}y = \int x^{-2}\,\mathrm{d}x\) \(\dfrac{1}{2}(4y + 3)^{\frac{1}{2}} = -\dfrac{1}{x} \quad (+C)\) | M1 |
| Using \((-2, 1.5) \qquad \dfrac{1}{2}(4 \times 1.5 + 3)^{\frac{1}{2}} = -\dfrac{1}{-2} + C\) | M1 |
| leading to \(\qquad C = 1\) | A1 |
| \(\dfrac{1}{2}(4y + 3)^{\frac{1}{2}} = -\dfrac{1}{x} + 1\) \((4y + 3)^{\frac{1}{2}} = 2 - \dfrac{2}{x}\) | M1 |
| \(y = \dfrac{1}{4}\left(2 - \dfrac{2}{x}\right)^2 - \dfrac{3}{4}\) or equivalent | A1 |
| (6) | |
| (8 marks) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.