| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 standard separable variables question with a helpful part (a) that guides students to the integration needed in part (b). The separation is straightforward, and the integration follows directly from the given result. While it requires chain rule proficiency and double angle knowledge, it's a routine application with clear scaffolding, making it slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07l Derivative of ln(x): and related functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dy}(\ln\tan 2y) = \frac{1}{\tan 2y} \times 2\sec^2 2y\) | M1A1 | Chain rule to obtain \(\frac{1}{\tan 2y} \times A\sec^2 2y\), A can be 1 |
| \(= \frac{\cos 2y}{\sin 2y} \times \frac{2}{\cos^2 2y} = \frac{k}{\sin 2y \cos 2y}\) | M1 | Uses identity \(\tan 2y = \frac{\sin 2y}{\cos 2y}\) and \(\sec^2 2y = \frac{1}{\cos^2 2y}\) reaching \(\frac{k}{\sin 2y \cos 2y}\) |
| \(= \frac{4}{2\sin 2y \cos 2y} = \frac{4}{\sin 4y}\) | A1* cso | Must multiply top and bottom by 2, using \(\sin 4y = 2\sin 2y\cos 2y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 2\cos x \sin 4y \Rightarrow \int\frac{dy}{\sin 4y} = \int 2\cos x\, dx\) | B1 | Separate terms; accept without integral sign if integration implied |
| \(\Rightarrow \frac{1}{4}\ln\tan 2y = 2\sin x (+c)\) | M1A1 | Use inverse of part (a); know \(\int\cos x\,dx = \pm\sin x\); correct answer |
| Put \(x=0, y=\frac{\pi}{6} \Rightarrow \frac{1}{4}\ln\tan\frac{2\pi}{6} = 2\sin 0 + c \Rightarrow c = \frac{1}{4}\ln\sqrt{3}\) or \(\frac{1}{8}\ln 3\) | M1 | Substitute \(x=0, y=\frac{\pi}{6}\) to find \(c\) |
| Takes exponentials so \(\tan 2y = e^{8\sin x + c}\) | M1 | Uses correct ln work for unsimplified \(\tan 2y\) (must have \(c\)) |
| \(\tan 2y = e^{8\sin x + c}\) so \(\tan 2y = Ae^{8\sin x}\) | M1 | |
| Put \(x=0, y=\frac{\pi}{6}\), so \(A = \sqrt{3}\) or \(e^c =\) | M1 | |
| \(\tan 2y = \sqrt{3}\, e^{8\sin x}\) | A1 | Correct answer; do not accept \(A = \tan\left(\frac{\pi}{3}\right)\), need \(A=\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 2\cos x \sin 4y \Rightarrow \int\frac{dy}{\sin 4y} = \int 2\cos x\, dx\) | B1 | Separate terms |
| \(\Rightarrow -\frac{1}{4}\ln(\csc 4y + \cot 4y) = 2\sin x (+c)\) | M1A1 | Use standard integral; correct answer |
| Sub \(x=0, y=\frac{\pi}{6}\): \(c = -\frac{1}{4}\ln\frac{1}{\sqrt{3}}\) or \(\frac{1}{4}\ln\sqrt{3}\) | M1 | |
| \(-\frac{1}{4}\ln\left(\frac{1+\cos 4y}{\sin 4y}\right) = \frac{1}{4}\ln(\tan 2y) = 2\sin x + \frac{1}{4}\ln\sqrt{3}\) so \(\tan 2y = \sqrt{3}\,e^{8\sin x}\) | M1A1 | Uses cosec/cot definitions with double angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan 2y = Ae^{B\sin x} \rightarrow (2)\sec^2 2y\frac{dy}{dx} = AB\cos x\, e^{B\sin x}\) | M1 | Implicit differentiation |
| \(\frac{dy}{dx} = \frac{B\cos x\tan 2y\cos^2 2y}{(2)}\) | M1 | |
| \(\frac{dy}{dx} = \frac{B\cos x \cdot 2\sin 2y\cos 2y}{(4)} = \frac{B\cos x\sin 4y}{(4)}\) | M1 | Uses double angle formulae |
| \(B=8\) and \(A=\sqrt{3}\) | A1 | Max 4/6 as B0, A0 not available |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dy}(\ln\tan 2y) = \frac{1}{\tan 2y} \times 2\sec^2 2y$ | M1A1 | Chain rule to obtain $\frac{1}{\tan 2y} \times A\sec^2 2y$, A can be 1 |
| $= \frac{\cos 2y}{\sin 2y} \times \frac{2}{\cos^2 2y} = \frac{k}{\sin 2y \cos 2y}$ | M1 | Uses identity $\tan 2y = \frac{\sin 2y}{\cos 2y}$ and $\sec^2 2y = \frac{1}{\cos^2 2y}$ reaching $\frac{k}{\sin 2y \cos 2y}$ |
| $= \frac{4}{2\sin 2y \cos 2y} = \frac{4}{\sin 4y}$ | A1* cso | Must multiply top and bottom by 2, using $\sin 4y = 2\sin 2y\cos 2y$ |
## Part (b) Way 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2\cos x \sin 4y \Rightarrow \int\frac{dy}{\sin 4y} = \int 2\cos x\, dx$ | B1 | Separate terms; accept without integral sign if integration implied |
| $\Rightarrow \frac{1}{4}\ln\tan 2y = 2\sin x (+c)$ | M1A1 | Use inverse of part (a); know $\int\cos x\,dx = \pm\sin x$; correct answer |
| Put $x=0, y=\frac{\pi}{6} \Rightarrow \frac{1}{4}\ln\tan\frac{2\pi}{6} = 2\sin 0 + c \Rightarrow c = \frac{1}{4}\ln\sqrt{3}$ or $\frac{1}{8}\ln 3$ | M1 | Substitute $x=0, y=\frac{\pi}{6}$ to find $c$ |
| Takes exponentials so $\tan 2y = e^{8\sin x + c}$ | M1 | Uses correct ln work for unsimplified $\tan 2y$ (must have $c$) |
| $\tan 2y = e^{8\sin x + c}$ so $\tan 2y = Ae^{8\sin x}$ | M1 | |
| Put $x=0, y=\frac{\pi}{6}$, so $A = \sqrt{3}$ or $e^c =$ | M1 | |
| $\tan 2y = \sqrt{3}\, e^{8\sin x}$ | A1 | Correct answer; do not accept $A = \tan\left(\frac{\pi}{3}\right)$, need $A=\sqrt{3}$ |
## Part (b) Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2\cos x \sin 4y \Rightarrow \int\frac{dy}{\sin 4y} = \int 2\cos x\, dx$ | B1 | Separate terms |
| $\Rightarrow -\frac{1}{4}\ln(\csc 4y + \cot 4y) = 2\sin x (+c)$ | M1A1 | Use standard integral; correct answer |
| Sub $x=0, y=\frac{\pi}{6}$: $c = -\frac{1}{4}\ln\frac{1}{\sqrt{3}}$ or $\frac{1}{4}\ln\sqrt{3}$ | M1 | |
| $-\frac{1}{4}\ln\left(\frac{1+\cos 4y}{\sin 4y}\right) = \frac{1}{4}\ln(\tan 2y) = 2\sin x + \frac{1}{4}\ln\sqrt{3}$ so $\tan 2y = \sqrt{3}\,e^{8\sin x}$ | M1A1 | Uses cosec/cot definitions with double angle formulae |
## Part (b) Way 3 (Special Case):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 2y = Ae^{B\sin x} \rightarrow (2)\sec^2 2y\frac{dy}{dx} = AB\cos x\, e^{B\sin x}$ | M1 | Implicit differentiation |
| $\frac{dy}{dx} = \frac{B\cos x\tan 2y\cos^2 2y}{(2)}$ | M1 | |
| $\frac{dy}{dx} = \frac{B\cos x \cdot 2\sin 2y\cos 2y}{(4)} = \frac{B\cos x\sin 4y}{(4)}$ | M1 | Uses double angle formulae |
| $B=8$ and $A=\sqrt{3}$ | A1 | Max 4/6 as B0, A0 not available |
---\begin{enumerate}
\item (a) Prove by differentiation that
\end{enumerate}
$$\frac { \mathrm { d } } { \mathrm {~d} y } ( \ln \tan 2 y ) = \frac { 4 } { \sin 4 y } , \quad 0 < y < \frac { \pi } { 4 }$$
(b) Given that $y = \frac { \pi } { 6 }$ when $x = 0$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos x \sin 4 y , \quad 0 < y < \frac { \pi } { 4 }$$
Give your answer in the form $\tan 2 y = A \mathrm { e } ^ { B \sin x }$, where $A$ and $B$ are constants to be determined.
\hfill \mbox{\textit{Edexcel C34 2015 Q8 [10]}}