| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (requires integration by parts) |
| Difficulty | Standard +0.3 This is a straightforward separable variables question with standard integration techniques. Part (a) is routine integration by parts (twice), and part (b) requires recognizing the separation, integrating both sides, and using the result from (a) with substitution. The 'hence' signpost makes the connection explicit, reducing problem-solving demand to below-average difficulty. |
| Spec | 1.08i Integration by parts1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int x^2 \cos 2x\, dx = x^2 \cdot \frac{1}{2}\sin 2x - \int 2x \cdot \frac{1}{2}\sin 2x\, dx\) | M1 | Attempts parts in correct direction; look for \(Ax^2\sin 2x \pm \int Bx\sin 2x\, dx\) |
| First application correct as shown | A1 | Correct first application of parts |
| \(= x^2 \cdot \frac{1}{2}\sin 2x - \left(-\frac{1}{2}x\cos 2x - \int -\frac{1}{2}\cos 2x\, dx\right)\) | M1 | Attempts parts again in same direction; \(\int x\sin 2x\, dx \to Kx\cos 2x \pm L\int \cos 2x\, dx\) |
| \(= \frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x\ (+c)\) | A1 | Correct answer with or without constant of integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sets up correct table with \(x^2\) and \(\cos x\) as leads | M1 | Correct table of derivatives and integrals |
| Correct tables of derivatives and integrals | A1 | |
| Extracts answer from table in correct form | M1 | |
| \(= \frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x\ (+c)\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dt} = \frac{t^2\cos^2 t}{y^2} \Rightarrow \int y^2\frac{dy}{dt}\,dt\) or \(\int y^2\,dy = \int t^2\cos^2 t\,dt\) | B1 | Correct separation of variables with indication to integrate both sides |
| \(\frac{y^3}{3} = \ldots\) | M1 | For \(y^n \to \ldots y^{n+1}\) on left hand side |
| \(\ldots = \frac{1}{2}\int t^2 + t^2\cos 2t\, dt\) | M1 | Applies double angle formula; accept \(\cos^2 t \to \frac{1}{2}(\pm 1 \pm \cos 2t)\) |
| \(\ldots = \frac{1}{6}t^3 + \frac{1}{2}\left(\frac{1}{2}t^2\sin 2t + \frac{1}{2}t\cos 2t - \frac{1}{4}\sin 2t\right)\ (+c)\) | M1 | Applies result from (a) and integrates \(t^2\) term |
| \(y^3 = \frac{1}{2}t^3 + \frac{3}{4}t^2\sin 2t + \frac{3}{4}t\cos 2t - \frac{3}{8}\sin 2t + c\) | A1ft | Correct result; must include \(+c\); follow through on answer to (a) providing isw was not applied |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int x^2 \cos 2x\, dx = x^2 \cdot \frac{1}{2}\sin 2x - \int 2x \cdot \frac{1}{2}\sin 2x\, dx$ | M1 | Attempts parts in correct direction; look for $Ax^2\sin 2x \pm \int Bx\sin 2x\, dx$ |
| First application correct as shown | A1 | Correct first application of parts |
| $= x^2 \cdot \frac{1}{2}\sin 2x - \left(-\frac{1}{2}x\cos 2x - \int -\frac{1}{2}\cos 2x\, dx\right)$ | M1 | Attempts parts again in same direction; $\int x\sin 2x\, dx \to Kx\cos 2x \pm L\int \cos 2x\, dx$ |
| $= \frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x\ (+c)$ | A1 | Correct answer with or without constant of integration |
**DI Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Sets up correct table with $x^2$ and $\cos x$ as leads | M1 | Correct table of derivatives and integrals |
| Correct tables of derivatives and integrals | A1 | |
| Extracts answer from table in correct form | M1 | |
| $= \frac{1}{2}x^2\sin 2x + \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x\ (+c)$ | A1 | Correct answer |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dt} = \frac{t^2\cos^2 t}{y^2} \Rightarrow \int y^2\frac{dy}{dt}\,dt$ or $\int y^2\,dy = \int t^2\cos^2 t\,dt$ | B1 | Correct separation of variables with indication to integrate both sides |
| $\frac{y^3}{3} = \ldots$ | M1 | For $y^n \to \ldots y^{n+1}$ on left hand side |
| $\ldots = \frac{1}{2}\int t^2 + t^2\cos 2t\, dt$ | M1 | Applies double angle formula; accept $\cos^2 t \to \frac{1}{2}(\pm 1 \pm \cos 2t)$ |
| $\ldots = \frac{1}{6}t^3 + \frac{1}{2}\left(\frac{1}{2}t^2\sin 2t + \frac{1}{2}t\cos 2t - \frac{1}{4}\sin 2t\right)\ (+c)$ | M1 | Applies result from (a) and integrates $t^2$ term |
| $y^3 = \frac{1}{2}t^3 + \frac{3}{4}t^2\sin 2t + \frac{3}{4}t\cos 2t - \frac{3}{8}\sin 2t + c$ | A1ft | Correct result; must include $+c$; follow through on answer to (a) providing isw was not applied |
---\begin{enumerate}
\item (a) Find $\int x ^ { 2 } \cos 2 x d x$\\
(b) Hence solve the differential equation
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( \frac { t \cos t } { y } \right) ^ { 2 }$$
giving your answer in the form $y ^ { n } = \mathrm { f } ( t )$ where $n$ is an integer.
\hfill \mbox{\textit{Edexcel P4 2024 Q5 [9]}}