| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.3 This is a standard second-order linear differential equation with constant coefficients requiring finding the complementary function (solving 4m²-1=0), particular integral (constant -3), applying initial conditions to find constants, then solving y=0. While it involves multiple steps and is Further Maths content, it follows a completely routine procedure with no conceptual challenges or novel insights required. |
| Spec | 4.10a General/particular solutions: of differential equations4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4m^2 - 1 = 0\) leading to \(m = \pm\frac{1}{2}\) | M1 | Auxiliary equation |
| \(y = Ae^{-\frac{1}{2}x} + Be^{\frac{1}{2}x}\) | A1 | Complementary function. Allow "\(y=\)" missing |
| \(-k = 3\) leading to \(k = -3\) | M1 A1 | Substitutes \(y = k\) |
| \(y = Ae^{-\frac{1}{2}x} + Be^{\frac{1}{2}x} - 3\) | A1 FT | General solution. Must have \(y=\) or sensible equivalent, such as \(y_{GS}=\). Don't allow \(GS=\) |
| \(y' = -\frac{1}{2}Ae^{-\frac{1}{2}x} + \frac{1}{2}Be^{\frac{1}{2}x}\) | B1 | Derivative of general solution |
| \(A + B = 0\), \(-\frac{1}{2}A + \frac{1}{2}B = 2\) leading to \(A = -2, B = 2\) | M1 | Substitutes initial conditions into general solution, solves simultaneous equations |
| \(y = 2e^{\frac{1}{2}x} - 2e^{-\frac{1}{2}x} - 3\) | A1 | Particular solution. Must have \(y=\) or sensible equivalent, such as \(y_{PS}=\). Don't allow \(PS=\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 0\) leading to \(4\sinh\frac{1}{2}x = 3\) leading to \(\frac{1}{2}x = \sinh^{-1}\frac{3}{4}\) | M1 | Makes \(x\) the subject. \(y\) must be of the form \(a_1e^{bx} \pm a_2e^{-bx} + c\) |
| \(\sinh^{-1}\frac{3}{4} = \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16}+1}\right) = \ln 2\) | M1 | Uses logarithmic form. \(y\) must be of the form \(a_1e^{bx} \pm a_2e^{-bx} + c\) |
| \(x = 2\ln 2 = \ln 4\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4m^2 - 1 = 0$ leading to $m = \pm\frac{1}{2}$ | M1 | Auxiliary equation |
| $y = Ae^{-\frac{1}{2}x} + Be^{\frac{1}{2}x}$ | A1 | Complementary function. Allow "$y=$" missing |
| $-k = 3$ leading to $k = -3$ | M1 A1 | Substitutes $y = k$ |
| $y = Ae^{-\frac{1}{2}x} + Be^{\frac{1}{2}x} - 3$ | A1 FT | General solution. Must have $y=$ or sensible equivalent, such as $y_{GS}=$. Don't allow $GS=$ |
| $y' = -\frac{1}{2}Ae^{-\frac{1}{2}x} + \frac{1}{2}Be^{\frac{1}{2}x}$ | B1 | Derivative of general solution |
| $A + B = 0$, $-\frac{1}{2}A + \frac{1}{2}B = 2$ leading to $A = -2, B = 2$ | M1 | Substitutes initial conditions into general solution, solves simultaneous equations |
| $y = 2e^{\frac{1}{2}x} - 2e^{-\frac{1}{2}x} - 3$ | A1 | Particular solution. Must have $y=$ or sensible equivalent, such as $y_{PS}=$. Don't allow $PS=$ |
**Total: 8 marks**
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 0$ leading to $4\sinh\frac{1}{2}x = 3$ leading to $\frac{1}{2}x = \sinh^{-1}\frac{3}{4}$ | M1 | Makes $x$ the subject. $y$ must be of the form $a_1e^{bx} \pm a_2e^{-bx} + c$ |
| $\sinh^{-1}\frac{3}{4} = \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16}+1}\right) = \ln 2$ | M1 | Uses logarithmic form. $y$ must be of the form $a_1e^{bx} \pm a_2e^{-bx} + c$ |
| $x = 2\ln 2 = \ln 4$ | A1 | |
**Total: 3 marks**
---7 The variables $x$ and $y$ are related by the differential equation
$$4 \frac { d ^ { 2 } y } { d x ^ { 2 } } - y = 3$$
It is given that, when $x = 0 , y = - 3$ and $\frac { \mathrm { dy } } { \mathrm { dx } } = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find $y$ in terms of $x$.
\item Deduce the exact value of $x$ for which $y = 0$. Give your answer in logarithmic form.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q7 [11]}}