| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| 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) requires differentiation of a composite hyperbolic function using quotient/chain rule and hyperbolic identities—moderately technical but standard Further Maths material. Part (b) is a first-order linear ODE requiring recognition of integrating factor form, integration involving hyperbolic functions, and application of initial conditions. While the hyperbolic context adds complexity beyond standard A-level, the structure is routine for Further Maths students who have practiced integrating factor methods. The 10-mark allocation and multi-step nature place it above average difficulty, but it doesn't require novel insight or particularly sophisticated problem-solving. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.10c Integrating factor: first order equations |
| Answer | Marks |
|---|---|
| 7(a) | d sech2x |
| Answer | Marks | Guidance |
|---|---|---|
| dx tanhx | M1A1 | Applies chain rule. |
| Answer | Marks | Guidance |
|---|---|---|
| sinhxcoshx sinh2x | A1 | AG. |
| Answer | Marks |
|---|---|
| 7(b) | dy 2y |
| Answer | Marks | Guidance |
|---|---|---|
| dx sinh(2x) | B1 | Divides through by sinh(2x). |
| Answer | Marks | Guidance |
|---|---|---|
| dx | M1A1 | Correct form on LHS and RHS. |
| ytanhx=ln(coshx)+C | M1A1 | Integrates RHS. |
| Answer | Marks | Guidance |
|---|---|---|
| 5 4 | M1 | Substitutes initial conditions. |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | A1 | Accept equivalent exact form. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(a) ---
7(a) | d sech2x
( ln(tanhx))=
dx tanhx | M1A1 | Applies chain rule.
1 2
= = .
sinhxcoshx sinh2x | A1 | AG.
3
--- 7(b) ---
7(b) | dy 2y
+ =1
dx sinh(2x) | B1 | Divides through by sinh(2x).
d
(ytanhx)=tanhx
dx | M1A1 | Correct form on LHS and RHS.
ytanhx=ln(coshx)+C | M1A1 | Integrates RHS.
5 (3) =ln5+C
5 4 | M1 | Substitutes initial conditions.
ytanhx=ln(coshx)+3−ln5
4 | A1 | Accept equivalent exact form.
7
Question | Answer | Marks | Guidance\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{d}{dx}(\ln(\tanh x)) = 2\cosh 2x$. [3]
\item Find the solution of the differential equation
$$\sinh 2x \frac{dy}{dx} + 2y = \sinh 2x$$
for which $y = 5$ when $x = \ln 2$. Give your answer in an exact form. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q7 [10]}}