Standard +0.3 This is a standard integrating factor question from FP2 requiring routine application of the method: divide by x to get standard form, identify integrating factor as x^5, multiply through, integrate, and solve for y. While it's a Further Maths topic (making it inherently harder than Core), it's a textbook application with no tricks or novel insights required, placing it slightly above average difficulty overall.
3. Find the general solution of the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = \frac { \ln x } { x } , \quad x > 0$$
giving your answer in the form \(y = \mathrm { f } ( x )\).
1st M1 for attempt at correct Integrating Factor. 1st A1 for simplified IF. 2nd M1 for \(\frac{\ln x}{x^2}\) 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 \(+ C\).
| $\frac{dy}{dx} + 5\frac{y}{x} = \frac{\ln x}{x^2}$ | M1 | Integrating factor $e^{\int 5}$ |
| $e^{\int 5} = e^{5\ln x} = x^5$ | A1 | |
| $\int x^3 \ln xdx = \frac{x^4\ln x}{4} - \int \frac{x^4}{4}dx$ | M1 M1 A1 | |
| $= \frac{x^4\ln x}{4} - \frac{x^4}{16} + C$ | A1 | |
| $x^5y = \frac{x^4\ln x}{4} - \frac{x^4}{16} + C$ | M1 A1 | |
| $y = \frac{\ln x}{4x} - \frac{1}{16x} + \frac{C}{x^5}$ | | |
| | (8) 8 | 1st M1 for attempt at correct Integrating Factor. 1st A1 for simplified IF. 2nd M1 for $\frac{\ln x}{x^2}$ 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 $+ C$. |
3. Find the general solution of the differential equation
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = \frac { \ln x } { x } , \quad x > 0$$
giving your answer in the form $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{Edexcel FP2 2011 Q3 [8]}}