| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| 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 Core Pure 1 with straightforward steps: divide by cos x to get standard form, identify integrating factor as sec x, integrate e^(2x)cos x (a routine A-level integral), then apply initial condition and solve y=0. While it requires multiple techniques, each step follows a well-practiced algorithm with no novel insight needed, making it slightly easier than average. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| IF \(= e^{\int \tan x\, dx} = e^{\ln\sec x} = \sec x\); leading to \(y\sec x = \int e^{2x}\,dx\) | M1 | 3.1a — finds integrating factor and attempts solution |
| \(y\sec x = \dfrac{1}{2}e^{2x}(+c)\) | A1 | 1.1b — condone missing \(+c\) |
| \(y = \left(\dfrac{1}{2}e^{2x} + c\right)\cos x\) | A1 | 1.1b — correct general solution; accept \(y = \dfrac{e^{2x}}{2\sec x} + \dfrac{c}{\sec x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=0,\ y=3 \Rightarrow c = 2.5\) | M1 | 3.1a — uses initial condition; allow if done in part (a) |
| Sets \(y=0\): \(\left(\dfrac{1}{2}e^{2x}+\dfrac{5}{2}\right)\cos x = 0 \Rightarrow \cos x = 0\) | M1 | 1.1b — \(A\) is 1, 2 or \(\frac{1}{2}\); valid attempt to solve for \(x\) |
| \(x = \dfrac{\pi}{2}\) | A1 | 1.1b — depends on both M's; \(\frac{\pi}{2}\) only; must have attempted to find \(c\) |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| IF $= e^{\int \tan x\, dx} = e^{\ln\sec x} = \sec x$; leading to $y\sec x = \int e^{2x}\,dx$ | M1 | 3.1a — finds integrating factor and attempts solution |
| $y\sec x = \dfrac{1}{2}e^{2x}(+c)$ | A1 | 1.1b — condone missing $+c$ |
| $y = \left(\dfrac{1}{2}e^{2x} + c\right)\cos x$ | A1 | 1.1b — correct general solution; accept $y = \dfrac{e^{2x}}{2\sec x} + \dfrac{c}{\sec x}$ |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=0,\ y=3 \Rightarrow c = 2.5$ | M1 | 3.1a — uses initial condition; allow if done in part (a) |
| Sets $y=0$: $\left(\dfrac{1}{2}e^{2x}+\dfrac{5}{2}\right)\cos x = 0 \Rightarrow \cos x = 0$ | M1 | 1.1b — $A$ is 1, 2 or $\frac{1}{2}$; valid attempt to solve for $x$ |
| $x = \dfrac{\pi}{2}$ | A1 | 1.1b — depends on both M's; $\frac{\pi}{2}$ only; must have attempted to find $c$ |
---\begin{enumerate}
\item (a) Determine the general solution of the differential equation
\end{enumerate}
$$\cos x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y \sin x = \mathrm { e } ^ { 2 x } \cos ^ { 2 } x$$
giving your answer in the form $y = \mathrm { f } ( x )$
Given that $y = 3$ when $x = 0$\\
(b) determine the smallest positive value of $x$ for which $y = 0$
\hfill \mbox{\textit{Edexcel CP1 2022 Q3 [6]}}