Edexcel C4 2014 January — Question 5 6 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2014
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (polynomial/exponential x-side)
DifficultyModerate -0.3 This is a straightforward separable variables question requiring standard techniques: separate variables, integrate both sides (using substitution u=2x for the right side), apply initial conditions, and rearrange. While it involves sin²(2x) which requires a substitution, this is a routine C4 exercise with no conceptual challenges—slightly easier than the average A-level question due to its mechanical nature.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)
  1. Given that \(y = 2\) at \(x = \frac { \pi } { 8 }\), solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y ^ { 2 } } { 2 \sin ^ { 2 } 2 x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-17_81_102_2649_1779}
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx}=\frac{3y^2}{2\sin^2 2x}\), \(y=2\) at \(x=\frac{\pi}{8}\)
\(\int\frac{1}{y^2}dy=\int\frac{3}{2\sin^2 2x}dx\)B1 Separates variables as shown. Can be implied. Ignore the integral signs.
\(\int\frac{1}{y^2}dy=\int\frac{3}{2}\cosec^2 2x\,dx\)B1 \(\frac{1}{y^2}\to\frac{1}{y}\) (See notes).
\(-\frac{1}{y}=\frac{3}{2}\left(-\frac{\cot 2x}{2}\right)\{+c\}\)M1 \(\pm\lambda\cot 2x, \lambda \neq 0\)
\(-\frac{1}{y}=-\frac{3}{4}\cot 2x+\frac{1}{4}\)A1 oe
\(\left\{y=2, x=\frac{\pi}{8}\Rightarrow-\frac{1}{2}=-\frac{3}{4}\cot\left(2\left(\frac{\pi}{8}\right)\right)+c\right\}\)M1 Use of \(x=\frac{\pi}{8}\) and \(y=2\) in an integrated equation containing \(c\)
\(y=\frac{-1}{-\frac{3}{4}\cot 2x+\frac{1}{4}}\) or \(y=\frac{4}{3\cot 2x-1}\) or \(y=\frac{4\tan 2x}{3-\tan 2x}\)A1 oe
[6]
Notes
B1: Separates variables as shown. \(dy\) and \(dx\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. The numbers "\(3\)" and "\(2\)" may appear on either side.
Eg: \(\int\frac{2}{y^2}dy=\int\frac{3}{\sin^2 2x}dx\), \(\int\frac{3y^2}dy=\int\frac{2}{\sin^2 2x}dx\), \(\int\frac{1}{3y^2}dy=\int\frac{1}{2\sin^2 2x}dx\) are all fine for B1.
B1: \(\frac{1}{y^2}\to\frac{1}{y}\) or \(\frac{2}{y^2}\to\frac{2}{y}\) or \(\frac{2}{3y^2}\to\frac{2}{3y}\) or \(\frac{3y^2}\to\frac{3}{y}\)
M1: \(\frac{1}{\sin^2 2x}\) or \(\cosec^2 2x\to\pm\lambda\cot 2x, \lambda \neq 0\)
A1: \(-\frac{1}{y}=\frac{3}{2}\left(-\frac{\cot 2x}{2}\right)\) with/without \(+c\) or equivalent. Eg: \(\frac{4}{3y}=-\cot 2x\) or \(\frac{4}{3y}=\cot 2x\)
M1: Some evidence of using both \(x=\frac{\pi}{8}\) and \(y=2\) in an integrated or changed equation containing \(c\).
Note that is mark can be implied by the correct value of \(c\).
A1: \(y=\frac{-1}{-\frac{3}{4}\cot 2x+\frac{1}{4}}\) or \(y=\frac{4}{3\cot 2x-1}\) or \(y=\frac{4\tan 2x}{3-\tan 2x}\) or any equivalent correct answer.
Note: You can ignore subsequent working which follows from a correct answer.
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=\frac{3y^2}{2\sin^2 2x}$, $y=2$ at $x=\frac{\pi}{8}$ | | |
| $\int\frac{1}{y^2}dy=\int\frac{3}{2\sin^2 2x}dx$ | B1 | Separates variables as shown. Can be implied. Ignore the integral signs. |
| $\int\frac{1}{y^2}dy=\int\frac{3}{2}\cosec^2 2x\,dx$ | B1 | $\frac{1}{y^2}\to\frac{1}{y}$ (See notes). |
| $-\frac{1}{y}=\frac{3}{2}\left(-\frac{\cot 2x}{2}\right)\{+c\}$ | M1 | $\pm\lambda\cot 2x, \lambda \neq 0$ |
| $-\frac{1}{y}=-\frac{3}{4}\cot 2x+\frac{1}{4}$ | A1 oe | |
| $\left\{y=2, x=\frac{\pi}{8}\Rightarrow-\frac{1}{2}=-\frac{3}{4}\cot\left(2\left(\frac{\pi}{8}\right)\right)+c\right\}$ | M1 | Use of $x=\frac{\pi}{8}$ and $y=2$ in an integrated equation containing $c$ |
| $y=\frac{-1}{-\frac{3}{4}\cot 2x+\frac{1}{4}}$ or $y=\frac{4}{3\cot 2x-1}$ or $y=\frac{4\tan 2x}{3-\tan 2x}$ | A1 oe | |
| | | [6] |

**Notes**

B1: Separates variables as shown. $dy$ and $dx$ should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs. The numbers "$3$" and "$2$" may appear on either side.

Eg: $\int\frac{2}{y^2}dy=\int\frac{3}{\sin^2 2x}dx$, $\int\frac{3y^2}dy=\int\frac{2}{\sin^2 2x}dx$, $\int\frac{1}{3y^2}dy=\int\frac{1}{2\sin^2 2x}dx$ are all fine for B1.

B1: $\frac{1}{y^2}\to\frac{1}{y}$ or $\frac{2}{y^2}\to\frac{2}{y}$ or $\frac{2}{3y^2}\to\frac{2}{3y}$ or $\frac{3y^2}\to\frac{3}{y}$

M1: $\frac{1}{\sin^2 2x}$ or $\cosec^2 2x\to\pm\lambda\cot 2x, \lambda \neq 0$

A1: $-\frac{1}{y}=\frac{3}{2}\left(-\frac{\cot 2x}{2}\right)$ with/without $+c$ or equivalent. Eg: $\frac{4}{3y}=-\cot 2x$ or $\frac{4}{3y}=\cot 2x$

M1: Some evidence of using both $x=\frac{\pi}{8}$ and $y=2$ in an integrated or changed equation containing $c$.

Note that is mark can be implied by the correct value of $c$.

A1: $y=\frac{-1}{-\frac{3}{4}\cot 2x+\frac{1}{4}}$ or $y=\frac{4}{3\cot 2x-1}$ or $y=\frac{4\tan 2x}{3-\tan 2x}$ or any equivalent correct answer.

Note: You can ignore subsequent working which follows from a correct answer.

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\begin{enumerate}
  \item Given that $y = 2$ at $x = \frac { \pi } { 8 }$, solve the differential equation
\end{enumerate}

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y ^ { 2 } } { 2 \sin ^ { 2 } 2 x }$$

giving your answer in the form $y = \mathrm { f } ( x )$.\\

\includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-17_81_102_2649_1779}\\

\hfill \mbox{\textit{Edexcel C4 2014 Q5 [6]}}
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