| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Hyperbolic function manipulation then solve |
| Difficulty | Challenging +1.2 Part (a) is a straightforward hyperbolic identity verification using definitions. Part (b) is a standard second-order linear ODE with constant coefficients requiring auxiliary equation, particular integral (simplified by part (a) to e^(x/2)), and applying initial conditions. While this is Further Maths content, it follows a well-rehearsed procedure with no novel insight required, making it moderately above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cosh x + \sinh x)^{\frac{1}{2}} = \left(\frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})\right)^{\frac{1}{2}}\) | M1 | Substitutes sinh and cosh in terms of exponentials |
| \((e^x)^{\frac{1}{2}} = e^{\frac{1}{2}x}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m^2+m+3=0 \Rightarrow m=-\frac{1}{2}\pm\frac{1}{2}i\sqrt{11}\) | M1 | Auxiliary equation |
| \(y=e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)\) | A1 | Complementary function; allow \(y=\) missing |
| \(y=ke^{\frac{1}{2}x} \Rightarrow y'=\frac{1}{2}ke^{\frac{1}{2}x} \Rightarrow y''=\frac{1}{4}ke^{\frac{1}{2}x}\) | B1 | Particular integral and its derivatives |
| \(\frac{1}{4}ke^{\frac{1}{2}x}+\frac{1}{2}ke^{\frac{1}{2}x}+3ke^{\frac{1}{2}x}=5e^{\frac{1}{2}x} \Rightarrow \frac{1}{4}k+\frac{1}{2}k+3k=5\) | M1 | Substitutes and equates coefficients; PI must be correct form (\(ae^{bx}\)) |
| \(k=\frac{4}{3}\) | A1 | |
| \(y=e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)+\frac{4}{3}e^{\frac{1}{2}x}\) | A1 FT | Must have \(y=\); FT on their CF |
| \(\frac{dy}{dx}=e^{-\frac{1}{2}x}\!\left(-\frac{\sqrt{11}}{2}A\sin\frac{\sqrt{11}}{2}x+\frac{\sqrt{11}}{2}B\cos\frac{\sqrt{11}}{2}x\right)-\frac{1}{2}e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)+\frac{2}{3}e^{\frac{1}{2}x}\) | B1 | |
| \(1=A+\frac{4}{3},\quad \frac{4}{3}=\frac{\sqrt{11}}{2}B-\frac{1}{2}A+\frac{2}{3} \Rightarrow A=-\frac{1}{3},\; B=\frac{1}{\sqrt{11}}\) | M1 A1 | Substitutes initial conditions and forms simultaneous equations; CF must be correct form |
| \(y=e^{-\frac{1}{2}x}\!\left(-\frac{1}{3}\cos\frac{\sqrt{11}}{2}x+\frac{1}{\sqrt{11}}\sin\frac{\sqrt{11}}{2}x\right)+\frac{4}{3}e^{\frac{1}{2}x}\) | A1 | Must have \(y=\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cosh x + \sinh x)^{\frac{1}{2}} = \left(\frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})\right)^{\frac{1}{2}}$ | M1 | Substitutes sinh and cosh in terms of exponentials |
| $(e^x)^{\frac{1}{2}} = e^{\frac{1}{2}x}$ | A1 | AG |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2+m+3=0 \Rightarrow m=-\frac{1}{2}\pm\frac{1}{2}i\sqrt{11}$ | M1 | Auxiliary equation |
| $y=e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)$ | A1 | Complementary function; allow $y=$ missing |
| $y=ke^{\frac{1}{2}x} \Rightarrow y'=\frac{1}{2}ke^{\frac{1}{2}x} \Rightarrow y''=\frac{1}{4}ke^{\frac{1}{2}x}$ | B1 | Particular integral and its derivatives |
| $\frac{1}{4}ke^{\frac{1}{2}x}+\frac{1}{2}ke^{\frac{1}{2}x}+3ke^{\frac{1}{2}x}=5e^{\frac{1}{2}x} \Rightarrow \frac{1}{4}k+\frac{1}{2}k+3k=5$ | M1 | Substitutes and equates coefficients; PI must be correct form ($ae^{bx}$) |
| $k=\frac{4}{3}$ | A1 | |
| $y=e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)+\frac{4}{3}e^{\frac{1}{2}x}$ | A1 FT | Must have $y=$; FT on their CF |
| $\frac{dy}{dx}=e^{-\frac{1}{2}x}\!\left(-\frac{\sqrt{11}}{2}A\sin\frac{\sqrt{11}}{2}x+\frac{\sqrt{11}}{2}B\cos\frac{\sqrt{11}}{2}x\right)-\frac{1}{2}e^{-\frac{1}{2}x}\!\left(A\cos\frac{\sqrt{11}}{2}x+B\sin\frac{\sqrt{11}}{2}x\right)+\frac{2}{3}e^{\frac{1}{2}x}$ | B1 | |
| $1=A+\frac{4}{3},\quad \frac{4}{3}=\frac{\sqrt{11}}{2}B-\frac{1}{2}A+\frac{2}{3} \Rightarrow A=-\frac{1}{3},\; B=\frac{1}{\sqrt{11}}$ | M1 A1 | Substitutes initial conditions and forms simultaneous equations; CF must be correct form |
| $y=e^{-\frac{1}{2}x}\!\left(-\frac{1}{3}\cos\frac{\sqrt{11}}{2}x+\frac{1}{\sqrt{11}}\sin\frac{\sqrt{11}}{2}x\right)+\frac{4}{3}e^{\frac{1}{2}x}$ | A1 | Must have $y=$ |6
\begin{enumerate}[label=(\alph*)]
\item Show that $( \cosh x + \sinh x ) ^ { \frac { 1 } { 2 } } = \mathrm { e } ^ { \frac { 1 } { 2 } x }$.
\item Find the particular solution of the differential equation
$$\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } + 3 y = 5 ( \cosh x + \sinh x ) ^ { \frac { 1 } { 2 } }$$
given that, when $x = 0 , y = 1$ and $\frac { d y } { d x } = \frac { 4 } { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q6 [12]}}