| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - variable coefficients |
| Difficulty | Standard +0.3 This is a standard integrating factor question from FP2 with straightforward application of the method (dividing by x, finding integrating factor x², integrating). Parts (b) and (c) add routine calculus (applying initial condition, finding turning points), but the overall structure is textbook-standard with no novel insights required. Slightly above average difficulty due to being Further Maths content and multi-part nature. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} + 2\dfrac{y}{x} = 4x\) | M1 | Dividing given equation by \(x\); may be implied |
| I.F.: \(e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2\) | M1 | \(\int \frac{2}{x}dx\) must be seen; \(\ln x\) must appear |
| \(x^2\dfrac{dy}{dx} + 2xy = 4x^3\) | M1dep | Multiplying equation by I.F.; dep on 2nd M |
| \(yx^2 = \int 4x^3 dx = x^4 (+c)\) | M1dep | Attempting integration; dep on 2nd and 3rd M |
| \(y = x^2 + \dfrac{c}{x^2}\) | A1cso | oe e.g. \(yx^2 = x^4 + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=1,\ y=5 \Rightarrow c=4\) | M1 | Using \(x=1, y=5\) in their expression |
| \(y = x^2 + \dfrac{4}{x^2}\) | A1ft | Follow through from (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = 2x - \dfrac{8}{x^3}\) | ||
| \(\dfrac{dy}{dx} = 0 \Rightarrow x^4 = 4,\ x = \pm\sqrt{2}\) or \(\pm\sqrt[4]{4}\) | M1, A1 | No follow through; ignore imaginary roots |
| \(y = 2 + \dfrac{4}{2} = 4\) | A1cao | Using particular solution; no extra values |
| Alt: \(\dfrac{dy}{dx}=0\) in original DE gives \(y=2x^2\); or complete the square to get \(y=\left(x+\dfrac{2}{x}\right)^2 - 4\) | M1 | |
| \(x = \pm\sqrt{2},\quad y=4\) | A1, A1 | |
| Correct shape (two minima, two branches asymptotic to \(y\)-axis) | B1 | |
| Turning points \((-\sqrt{2}, 4)\) and \((\sqrt{2}, 4)\) shown on sketch | B1 | Coordinates shown somewhere on/beside sketch |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} + 2\dfrac{y}{x} = 4x$ | M1 | Dividing given equation by $x$; may be implied |
| I.F.: $e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2$ | M1 | $\int \frac{2}{x}dx$ must be seen; $\ln x$ must appear |
| $x^2\dfrac{dy}{dx} + 2xy = 4x^3$ | M1dep | Multiplying equation by I.F.; dep on 2nd M |
| $yx^2 = \int 4x^3 dx = x^4 (+c)$ | M1dep | Attempting integration; dep on 2nd and 3rd M |
| $y = x^2 + \dfrac{c}{x^2}$ | A1cso | oe e.g. $yx^2 = x^4 + c$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=1,\ y=5 \Rightarrow c=4$ | M1 | Using $x=1, y=5$ in their expression |
| $y = x^2 + \dfrac{4}{x^2}$ | A1ft | Follow through from (a) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = 2x - \dfrac{8}{x^3}$ | | |
| $\dfrac{dy}{dx} = 0 \Rightarrow x^4 = 4,\ x = \pm\sqrt{2}$ or $\pm\sqrt[4]{4}$ | M1, A1 | No follow through; ignore imaginary roots |
| $y = 2 + \dfrac{4}{2} = 4$ | A1cao | Using particular solution; no extra values |
| **Alt:** $\dfrac{dy}{dx}=0$ in original DE gives $y=2x^2$; or complete the square to get $y=\left(x+\dfrac{2}{x}\right)^2 - 4$ | M1 | |
| $x = \pm\sqrt{2},\quad y=4$ | A1, A1 | |
| Correct shape (two minima, two branches asymptotic to $y$-axis) | B1 | |
| Turning points $(-\sqrt{2}, 4)$ and $(\sqrt{2}, 4)$ shown on sketch | B1 | Coordinates shown somewhere on/beside sketch |
---\begin{enumerate}
\item (a) Find the general solution of the differential equation\\
(b) Find the particular solution for which $y = 5$ at $x = 1$, giving your answer in the form $y = \mathrm { f } ( x )$.
\end{enumerate}
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 4 x ^ { 2 }$$
(c) (i) Find the exact values of the coordinates of the turning points of the curve with equation $y = \mathrm { f } ( x )$, making your method clear.\\
(ii) Sketch the curve with equation $y = \mathrm { f } ( x )$, showing the coordinates of the turning points.
\hfill \mbox{\textit{Edexcel FP2 2013 Q5 [12]}}