FP2 June 2005 Q5
5. Find the general solution of the differential equation \[(x + 1)\frac{\mathrm{d}y}{\mathrm{d}x} + 2y = \frac{1}{x}, \qquad x \gt 0.\] giving your answer in the form \(y = \mathrm{f}(x)\). (7)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{2}{1 + x}y = \dfrac{1}{x(x + 1)}\) Attempt \(y' + Py = Q\) form | M1 |
| I.F. \(= \mathrm{e}^{\int \frac{2}{1 + x}\mathrm{d}x} = \mathrm{e}^{2\ln(1 + x)},\ = (1 + x)^2\) | M1, A1 |
| \(\therefore y(1 + x)^2 = \displaystyle\int\left(\frac{x + 1}{x}\right)\mathrm{d}x\) OR \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(y(1 + x)^2\right) = \dfrac{x + 1}{x}\) | M1 (ft I.F.) |
| i.e. \((y(1 + x)^2 =)\ x + \ln x + (C)\) | M1 A1 |
| \(\underline{y =}\ \dfrac{x + \ln x + C}{(1 + x)^2}\) | A1 c.a.o. |
| (7 marks) |
Notes
(corrected from the printed mark scheme: the guidance reads “Attempt \(y' = Py = Q\) form”)