| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Standard +0.3 This is a standard integrating factor problem requiring students to divide by x to get standard form, identify the integrating factor as x², integrate 4x ln(x), and apply the initial condition. While it involves multiple steps and integration by parts, it follows a completely routine procedure taught explicitly in the syllabus with no novel insight required. The presence of ln(x) adds minor computational complexity but this remains a textbook exercise, placing it slightly above average difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
$\frac{\mathrm{d}y}{\mathrm{d}x} + \frac{2}{x}y = \frac{4\ln x}{x}$ I.F. is $\exp\left\{\int\frac{2}{x}\ \mathrm{d}x\right\} = x^2$ **M1 A1**
$x^2\frac{\mathrm{d}y}{\mathrm{d}x} + 2xy = 4x\ln x \Rightarrow x^2 y =$ LHS correct **B1**
RHS $= 4\ln x \cdot \frac{1}{2}x^2 - \int\frac{1}{2}x^2 \cdot \frac{4}{x}\ \mathrm{d}x$ M1 Use of parts, right way round **M1**
$= 2x^2\ln x - x^2 + C$ Ignore the "$+C$" here **A1**
$y = 2\ln x - 1 + \frac{C}{x^2}$ **A1**
Use of $x = 1$, $y = 1$ to find $C$: $C = 2$ i.e. $y = 2\ln x - 1 + \frac{2}{x^2}$ (any correct form) **M1 A1** [8]6 Solve the first-order differential equation $x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 4 \ln x$ given that $y = 1$ when $x = 1$. Give your answer in the form $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q6 [8]}}