| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | With preliminary integration |
| Difficulty | Challenging +1.2 Part (a) is a routine proof from definitions requiring straightforward algebraic manipulation of exponentials. Part (b) is a standard integrating factor problem, though it requires recognizing coth x, finding the integrating factor (1/sinh x), and performing integration involving hyperbolic functions. The preliminary identity helps with the integration. This is typical Further Maths fare but more mechanical than insightful, placing it slightly above average difficulty. |
| Spec | 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sinh x = \frac{1}{2}(e^x - e^{-x})\), \(\cosh x = \frac{1}{2}(e^x + e^{-x})\) | B1 | Writes in exponential form |
| \(\frac{1}{2}(e^x - e^{-x})^2 = \frac{1}{2}(e^{2x} - 2 + e^{-2x}) = \frac{1}{2}(e^{2x}+e^{-2x})-1\) | M1 | Expands |
| \(\cosh 2x - 1\) | A1 | AG |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{\int\coth x} = e^{\ln\sinh x} = \sinh x\) | M1 A1 | Finds integrating factor |
| \(\frac{d}{dx}(y\sinh x) = 4\sinh^2 x\) | M1 | Correct form on both sides |
| \(y\sinh x = 2\!\left(\frac{1}{2}\sinh 2x - x\right) + C = \sinh 2x - 2x + C\) | M1 A1 | Integrates \(\sinh^2 x\) using identity |
| \(\sinh\ln 3 = \sinh\ln 9 - 2\ln 3 + C\) leading to \(\frac{4}{3} = \frac{40}{9} - 2\ln 3 + C\) | M1 | Substitutes initial conditions into their solution |
| \(y\sinh x = \sinh 2x - 2x + 2\ln 3 - \frac{28}{9}\) | A1 | |
| Total: 7 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sinh x = \frac{1}{2}(e^x - e^{-x})$, $\cosh x = \frac{1}{2}(e^x + e^{-x})$ | B1 | Writes in exponential form |
| $\frac{1}{2}(e^x - e^{-x})^2 = \frac{1}{2}(e^{2x} - 2 + e^{-2x}) = \frac{1}{2}(e^{2x}+e^{-2x})-1$ | M1 | Expands |
| $\cosh 2x - 1$ | A1 | AG |
| **Total: 3** | | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{\int\coth x} = e^{\ln\sinh x} = \sinh x$ | M1 A1 | Finds integrating factor |
| $\frac{d}{dx}(y\sinh x) = 4\sinh^2 x$ | M1 | Correct form on both sides |
| $y\sinh x = 2\!\left(\frac{1}{2}\sinh 2x - x\right) + C = \sinh 2x - 2x + C$ | M1 A1 | Integrates $\sinh^2 x$ using identity |
| $\sinh\ln 3 = \sinh\ln 9 - 2\ln 3 + C$ leading to $\frac{4}{3} = \frac{40}{9} - 2\ln 3 + C$ | M1 | Substitutes initial conditions into their solution |
| $y\sinh x = \sinh 2x - 2x + 2\ln 3 - \frac{28}{9}$ | A1 | |
| **Total: 7** | | |6
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of sinh and cosh in terms of exponentials, prove that
$$2 \sinh ^ { 2 } x = \cosh 2 x - 1$$
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_67_1550_374_347}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_65_1569_468_328}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_67_1573_557_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_70_1573_646_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_72_1573_735_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_72_1570_826_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_74_1570_916_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e313d6f0-7615-4be5-b13e-2796fd6335e5-10_69_1570_1007_324}
\item Find the solution to the differential equation
$$\frac { d y } { d x } + y \operatorname { coth } x = 4 \sinh x$$
for which $y = 1$ when $x = \ln 3$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q6 [10]}}