C4 January 2011 Q3
3.
(a) Express \(\dfrac{5}{(x - 1)(3x + 2)}\) in partial fractions. (3)
(b) Hence find \(\displaystyle\int \frac{5}{(x - 1)(3x + 2)}\,\mathrm{d}x\), where \(x > 1\). (3)
(c) Find the particular solution of the differential equation \[(x - 1)(3x + 2)\frac{\mathrm{d}y}{\mathrm{d}x} = 5y, \qquad x > 1,\] for which \(y = 8\) at \(x = 2\). Give your answer in the form \(y = \mathrm{f}(x)\). (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{(x - 1)(3x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{3x + 2}\) | |
| \(5 = A(3x + 2) + B(x - 1)\) | |
| \(x \to 1\) \(5 = 5A \Rightarrow A = 1\) | M1 A1 |
| \(x \to -\dfrac{2}{3}\) \(5 = -\dfrac{5}{3}B \Rightarrow B = -3\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{5}{(x - 1)(3x + 2)}\,\mathrm{d}x = \int \left(\frac{1}{x - 1} - \frac{3}{3x + 2}\right)\mathrm{d}x\) | |
| \(= \ln(x - 1) - \ln(3x + 2)\quad (+C)\) ft constants | M1 A1ft A1ft |
| (3) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{5}{(x - 1)(3x + 2)}\,\mathrm{d}x = \int \left(\frac{1}{y}\right)\mathrm{d}y\) | M1 |
| \(\ln(x - 1) - \ln(3x + 2) = \ln y\quad (+C)\) | M1 A1 |
| \(y = \dfrac{K(x - 1)}{3x + 2}\) depends on first two Ms in (c) | M1 dep |
| Using \((2, 8)\) \(8 = \dfrac{K}{8}\) depends on first two Ms in (c) | M1 dep |
| \(y = \dfrac{64(x - 1)}{3x + 2}\) | A1 |
| (6) | |
| (12 marks) |