FP2 June 2011 Q3
3. Find the general solution of the differential equation \[x\frac{\mathrm{d}y}{\mathrm{d}x} + 5y = \frac{\ln x}{x}, \qquad x > 0\] giving your answer in the form \(y = \mathrm{f}(x)\). (8)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} + 5\dfrac{y}{x} = \dfrac{\ln x}{x^2}\) Integrating factor \(\mathrm{e}^{\int \frac{5}{x}}\) | M1 |
| \(\mathrm{e}^{\int \frac{5}{x}} = \mathrm{e}^{5\ln x} = x^5\) | A1 |
| \(\displaystyle\int x^3\ln x\,\mathrm{d}x = \frac{x^4\ln x}{4} - \int \frac{x^3}{4}\,\mathrm{d}x\) | M1 M1 A1 |
| \(= \dfrac{x^4\ln x}{4} - \dfrac{x^4}{16}\ (+C)\) | A1 |
| \(x^5y = \dfrac{x^4\ln x}{4} - \dfrac{x^4}{16} + C \qquad y = \dfrac{\ln x}{4x} - \dfrac{1}{16x} + \dfrac{C}{x^5}\) | M1 A1 |
| (8 marks) |
Notes
1st M1 for attempt at correct Integrating Factor
1st A1 for simplified IF
2nd M1 for \(\dfrac{\ln x}{x^2}\) times their IF to give their ‘\(x^3\ln x\)’
3rd M1 for attempt at correct Integration by Parts
2nd A1 for both terms correct
3rd A1 constant not required
4th M1 \(x^5y =\) their answer \(+ C\)