| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Standard +0.3 This is a straightforward separable variables question from Further Maths requiring standard algebraic manipulation to separate variables, followed by routine integration (including partial fractions or substitution for the x/(x²+1) term) and application of initial conditions. While it's Further Maths content, the techniques are mechanical and well-practiced, making it slightly above average difficulty but not requiring novel insight. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(\frac{dy}{dx} + \frac{xy}{(1+x^2)} = \frac{x}{(1+x^2)}\) | B1 | Correct form |
| \(I = e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2}\ln(1+x^2)} = (1+x^2)^{\frac{1}{2}}\) | M1A1 | M1: \(I = e^{\int \frac{x}{1+x^2}dx} = e^{k\ln(1+x^2)}\); A1: correct integrating factor \((1+x^2)^{\frac{1}{2}}\) |
| \(y(1+x^2)^{\frac{1}{2}} = \int \frac{x}{(1+x^2)^{\frac{1}{2}}}dx\) | M1 | Uses integrating factor to reach form \(yI = \int Q I\, dx\) |
| \(= (1+x^2)^{\frac{1}{2}} (+c)\) | A1 | Correct integration (\(+c\) not needed here) |
| \(y = 1 + c(1+x^2)^{-\frac{1}{2}}\) oe | A1 | cao with constant correctly placed; "\(y=\)" must appear at some point |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(\int \frac{dy}{1-y} = \int \frac{x}{x^2+1}dx\) | B1 | Separates variables correctly |
| \(\int \frac{x}{x^2+1}dx = \frac{1}{2}\ln(x^2+1)\) | M1A1 | M1: result \(= k\ln(x^2+1)\); A1: correct integration \(\frac{1}{2}\ln(x^2+1)\) |
| \(\int \frac{dy}{1-y} = -\ln(1-y)\) | M1 | \(= k\ln(1-y)\) or \(k\ln(y-1)\) |
| \(-\ln(1-y) = \frac{1}{2}\ln(x^2+1)(+c)\) | A1 | Fully correct integration |
| \(y = 1 + c(1+x^2)^{-\frac{1}{2}}\) oe | A1 | cao, isw if necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance Notes |
| \(2 = 1 + c(1+3^2)^{-\frac{1}{2}} \Rightarrow c = \ldots\) | M1 | Substitutes \(x=3\) and \(y=2\), attempts to find \(c\) |
| \(y = 1 + \sqrt{10}(1+x^2)^{-\frac{1}{2}}\) oe | A1 | cao ("\(y=\)" not needed); apply isw if necessary |
# Question 2(a):
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $\frac{dy}{dx} + \frac{xy}{(1+x^2)} = \frac{x}{(1+x^2)}$ | B1 | Correct form |
| $I = e^{\int \frac{x}{1+x^2}dx} = e^{\frac{1}{2}\ln(1+x^2)} = (1+x^2)^{\frac{1}{2}}$ | M1A1 | M1: $I = e^{\int \frac{x}{1+x^2}dx} = e^{k\ln(1+x^2)}$; A1: correct integrating factor $(1+x^2)^{\frac{1}{2}}$ |
| $y(1+x^2)^{\frac{1}{2}} = \int \frac{x}{(1+x^2)^{\frac{1}{2}}}dx$ | M1 | Uses integrating factor to reach form $yI = \int Q I\, dx$ |
| $= (1+x^2)^{\frac{1}{2}} (+c)$ | A1 | Correct integration ($+c$ not needed here) |
| $y = 1 + c(1+x^2)^{-\frac{1}{2}}$ oe | A1 | cao with constant correctly placed; "$y=$" must appear at some point |
**Alternative (separation of variables):**
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $\int \frac{dy}{1-y} = \int \frac{x}{x^2+1}dx$ | B1 | Separates variables correctly |
| $\int \frac{x}{x^2+1}dx = \frac{1}{2}\ln(x^2+1)$ | M1A1 | M1: result $= k\ln(x^2+1)$; A1: correct integration $\frac{1}{2}\ln(x^2+1)$ |
| $\int \frac{dy}{1-y} = -\ln(1-y)$ | M1 | $= k\ln(1-y)$ or $k\ln(y-1)$ |
| $-\ln(1-y) = \frac{1}{2}\ln(x^2+1)(+c)$ | A1 | Fully correct integration |
| $y = 1 + c(1+x^2)^{-\frac{1}{2}}$ oe | A1 | cao, isw if necessary |
# Question 2(b):
| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $2 = 1 + c(1+3^2)^{-\frac{1}{2}} \Rightarrow c = \ldots$ | M1 | Substitutes $x=3$ and $y=2$, attempts to find $c$ |
| $y = 1 + \sqrt{10}(1+x^2)^{-\frac{1}{2}}$ oe | A1 | cao ("$y=$" not needed); apply isw if necessary |
---\begin{enumerate}
\item (a) Find the general solution of the differential equation
\end{enumerate}
$$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + x y - x = 0$$
giving your answer in the form $y = \mathrm { f } ( x )$.\\
(b) Find the particular solution for which $y = 2$ when $x = 3$\\
\hfill \mbox{\textit{Edexcel F2 2018 Q2 [8]}}