| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Bernoulli equation |
| Difficulty | Challenging +1.2 This is a structured Bernoulli equation problem with explicit guidance through substitution. Part (a) is routine verification of a given substitution, part (b) is standard integrating factor technique, and part (c) is direct back-substitution. While it requires multiple techniques and is Further Maths content, the heavy scaffolding and standard methods make it moderately above average difficulty rather than genuinely challenging. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y^2 = z^{-1} \Rightarrow 2y\frac{dy}{dx} = -\frac{1}{z^2}\frac{dz}{dx}\) | B1 | |
| \(-\frac{1}{z^2}\frac{dz}{dx} + \frac{4}{z} = \frac{6x}{z^2}\) | M1 | |
| \(\frac{dz}{dx} - 4z = -6x\) * | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| IF \(= e^{\int -4\,dx} = e^{-4x}\) | B1 | |
| \(ze^{-4x} = -6\int xe^{-4x}\,dx\) | M1 | |
| \(= -6\left[-\frac{1}{4}xe^{-4x} + \int \frac{1}{4}e^{-4x}\,dx\right]\) | M1 | |
| \(= -6\left[-\frac{1}{4}xe^{-4x} - \frac{1}{16}e^{-4x}\right](+c)\) | A1 | |
| \(z = \frac{3}{2}x + \frac{3}{8} + ce^{4x}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(m - 4 = 0 \Rightarrow m = 4\), CF: \(z = Ae^{4x}\) | B1 | |
| PI: \(z = \lambda + \mu x\) | M1 | |
| \(4\mu = 6,\ 4\lambda = \mu \Rightarrow \mu = \frac{3}{2},\ \lambda = \frac{3}{8}\) | M1, A1 | |
| \(z = \frac{3}{2}x + \frac{3}{8} + Ae^{4x}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y^2 = \frac{1}{\frac{3}{2}x + \frac{3}{8} + ce^{4x}} = \frac{8}{12x+3+Ae^{4x}}\) | B1ft | Follow through their \(z\) |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y^2 = z^{-1} \Rightarrow 2y\frac{dy}{dx} = -\frac{1}{z^2}\frac{dz}{dx}$ | B1 | |
| $-\frac{1}{z^2}\frac{dz}{dx} + \frac{4}{z} = \frac{6x}{z^2}$ | M1 | |
| $\frac{dz}{dx} - 4z = -6x$ * | A1* | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| IF $= e^{\int -4\,dx} = e^{-4x}$ | B1 | |
| $ze^{-4x} = -6\int xe^{-4x}\,dx$ | M1 | |
| $= -6\left[-\frac{1}{4}xe^{-4x} + \int \frac{1}{4}e^{-4x}\,dx\right]$ | M1 | |
| $= -6\left[-\frac{1}{4}xe^{-4x} - \frac{1}{16}e^{-4x}\right](+c)$ | A1 | |
| $z = \frac{3}{2}x + \frac{3}{8} + ce^{4x}$ | A1 | |
**ALT (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $m - 4 = 0 \Rightarrow m = 4$, CF: $z = Ae^{4x}$ | B1 | |
| PI: $z = \lambda + \mu x$ | M1 | |
| $4\mu = 6,\ 4\lambda = \mu \Rightarrow \mu = \frac{3}{2},\ \lambda = \frac{3}{8}$ | M1, A1 | |
| $z = \frac{3}{2}x + \frac{3}{8} + Ae^{4x}$ | A1 | |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y^2 = \frac{1}{\frac{3}{2}x + \frac{3}{8} + ce^{4x}} = \frac{8}{12x+3+Ae^{4x}}$ | B1ft | Follow through their $z$ |4. (a) Show that the substitution $y ^ { 2 } = \frac { 1 } { z }$ transforms the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y = 3 x y ^ { 3 } \quad y \neq 0$$
into the differential equation
$$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 z = - 6 x$$
(b) Obtain the general solution of differential equation (II).\\
(c) Hence obtain the general solution of differential equation (I), giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel F2 2021 Q4 [9]}}