| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | With preliminary integration |
| Difficulty | Challenging +1.2 This question combines a guided substitution integral with a standard integrating factor differential equation. Part (a) is straightforward with the substitution provided. Part (b) requires recognizing the integrating factor method (1/θ), applying it correctly, and using integration by parts with the result from (a). While it's a multi-step problem requiring several techniques, all steps are standard for Further Maths students and the structure is clearly signposted. The preliminary integration in part (a) directly feeds into part (b), making it easier than if students had to identify this connection themselves. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{du}{d\theta} = -2(\theta-1)\) leading to \((\theta-1)\,d\theta = -\frac{1}{2}\,du\) | B1 | |
| \(\int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta = -\frac{1}{2}\int \frac{1}{\sqrt{u}}\,du = -\sqrt{u}+C = -\sqrt{1-(\theta-1)^2}+C\) | M1 A1 | Applies substitution. For M1, integrand must be of the form \(\frac{k}{\sqrt{u}}\), where \(k \neq 0\) is constant. For A1, allow "\(+C\)" missing. |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{d\theta} - \frac{y}{\theta} = \theta\sin^{-1}(\theta-1)\) | B1 | Divides through by \(\theta\). |
| \(e^{-\int \theta^{-1}\,d\theta} = e^{-\ln\theta} = \theta^{-1}\) | M1 A1 | Finds integrating factor. |
| \(\frac{d}{d\theta}\!\left(\theta^{-1}y\right) = \sin^{-1}(\theta-1)\) | M1 | Correct form on LHS using their integrating factor. |
| \(\theta^{-1}y = \theta\sin^{-1}(\theta-1) - \int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta\) | M1 A1 | Integrates \(\sin^{-1}(\theta-1)\). |
| \(\int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta = \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta + \int \frac{1}{\sqrt{1-(\theta-1)^2}}\,d\theta\) | M1 | Applies part (a) and uses \(\int \frac{1}{1-x^2}\,dx = \sin^{-1}x + C\). |
| \(\theta^{-1}y = \theta\sin^{-1}(\theta-1) + \sqrt{1-(\theta-1)^2} - \sin^{-1}(\theta-1) + C\) | A1 | |
| \(1 = 1 + C\) | M1 | Substitutes initial conditions into their expression in \(y\) and \(\theta\). |
| \(y = \theta(\theta-1)\sin^{-1}(\theta-1) + \theta\sqrt{1-(\theta-1)^2}\) | M1 A1 | Divides through by their integrating factor. |
| 11 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{du}{d\theta} = -2(\theta-1)$ leading to $(\theta-1)\,d\theta = -\frac{1}{2}\,du$ | **B1** | |
| $\int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta = -\frac{1}{2}\int \frac{1}{\sqrt{u}}\,du = -\sqrt{u}+C = -\sqrt{1-(\theta-1)^2}+C$ | **M1 A1** | Applies substitution. For M1, integrand must be of the form $\frac{k}{\sqrt{u}}$, where $k \neq 0$ is constant. For A1, allow "$+C$" missing. |
| | **3** | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{d\theta} - \frac{y}{\theta} = \theta\sin^{-1}(\theta-1)$ | **B1** | Divides through by $\theta$. |
| $e^{-\int \theta^{-1}\,d\theta} = e^{-\ln\theta} = \theta^{-1}$ | **M1 A1** | Finds integrating factor. |
| $\frac{d}{d\theta}\!\left(\theta^{-1}y\right) = \sin^{-1}(\theta-1)$ | **M1** | Correct form on LHS using their integrating factor. |
| $\theta^{-1}y = \theta\sin^{-1}(\theta-1) - \int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta$ | **M1 A1** | Integrates $\sin^{-1}(\theta-1)$. |
| $\int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta = \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta + \int \frac{1}{\sqrt{1-(\theta-1)^2}}\,d\theta$ | **M1** | Applies part **(a)** and uses $\int \frac{1}{1-x^2}\,dx = \sin^{-1}x + C$. |
| $\theta^{-1}y = \theta\sin^{-1}(\theta-1) + \sqrt{1-(\theta-1)^2} - \sin^{-1}(\theta-1) + C$ | **A1** | |
| $1 = 1 + C$ | **M1** | Substitutes initial conditions into their expression in $y$ and $\theta$. |
| $y = \theta(\theta-1)\sin^{-1}(\theta-1) + \theta\sqrt{1-(\theta-1)^2}$ | **M1 A1** | Divides through by their integrating factor. |
| | **11** | |8
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u = 1 - ( \theta - 1 ) ^ { 2 }$ to find
$$\int \frac { \theta - 1 } { \sqrt { 1 - ( \theta - 1 ) ^ { 2 } } } \mathrm {~d} \theta$$
\item Find the solution of the differential equation
$$\theta \frac { d y } { d \theta } - y = \theta ^ { 2 } \sin ^ { - 1 } ( \theta - 1 ) ,$$
where $0 < \theta < 2$, given that $y = 1$ when $\theta = 1$. Give your answer in the form $y = \mathrm { f } ( \theta )$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q8 [14]}}