| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| 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 substitution integral that's straightforward once the substitution is given. Part (b) requires recognizing the integrating factor method (standard for first-order linear ODEs), applying it correctly, then using the result from part (a) to evaluate the integral involving sinh^(-1)x. While this is a multi-step Further Maths question requiring careful algebra and knowledge of hyperbolic functions, it follows a standard template with the preliminary work already done in part (a), making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.08h Integration by substitution4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{u}}\,du = \sqrt{u}\;[+C] = \sqrt{1+x^2}\;[+C]\) | M1 A1 | Applies substitution. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} - \frac{y}{x} = x\sinh^{-1}x\) | B1 | Divides through by \(x\). |
| \(e^{-\int x^{-1}dx} = e^{-\ln x} = x^{-1}\) | M1 A1 | Finds integrating factor. |
| \(\frac{d}{dx}(x^{-1}y) = \sinh^{-1}x\) | M1 | Correct form on LHS: \(\frac{d}{dx}(Iy)\) for their integrating factor \(I\). |
| \(x^{-1}y = x\sinh^{-1}x - \int\frac{x}{\sqrt{1+x^2}}\,dx\) | M1 A1 | Integrates RHS. RHS must be of the form \(c\sinh^{-1}x\). |
| \(x^{-1}y = x\sinh^{-1}x - \sqrt{1+x^2} + C\) | A1 | \(\sinh^{-1}x = \ln\!\left(x+\sqrt{x^2+1}\right)\) |
| \(1 = \ln(1+\sqrt{2}) - \sqrt{2} + C\) | \*M1 | Substitutes initial conditions. |
| \(y = x^2\sinh^{-1}x - x\sqrt{1+x^2} - x\ln(1+\sqrt{2}) + x(1+\sqrt{2})\) | DM1 A1 | Divides through by integrating factor. Accept \(y = x^2\sinh^{-1}x - x\sqrt{1+x^2} - x\sinh^{-1}1 + x(1+\sqrt{2})\). |
## Question 7(a):
$\int \frac{x}{\sqrt{1+x^2}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{u}}\,du = \sqrt{u}\;[+C] = \sqrt{1+x^2}\;[+C]$ | **M1 A1** | Applies substitution.
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## Question 7(b):
$\frac{dy}{dx} - \frac{y}{x} = x\sinh^{-1}x$ | **B1** | Divides through by $x$.
$e^{-\int x^{-1}dx} = e^{-\ln x} = x^{-1}$ | **M1 A1** | Finds integrating factor.
$\frac{d}{dx}(x^{-1}y) = \sinh^{-1}x$ | **M1** | Correct form on LHS: $\frac{d}{dx}(Iy)$ for their integrating factor $I$.
$x^{-1}y = x\sinh^{-1}x - \int\frac{x}{\sqrt{1+x^2}}\,dx$ | **M1 A1** | Integrates RHS. RHS must be of the form $c\sinh^{-1}x$.
$x^{-1}y = x\sinh^{-1}x - \sqrt{1+x^2} + C$ | **A1** | $\sinh^{-1}x = \ln\!\left(x+\sqrt{x^2+1}\right)$
$1 = \ln(1+\sqrt{2}) - \sqrt{2} + C$ | **\*M1** | Substitutes initial conditions.
$y = x^2\sinh^{-1}x - x\sqrt{1+x^2} - x\ln(1+\sqrt{2}) + x(1+\sqrt{2})$ | **DM1 A1** | Divides through by integrating factor. Accept $y = x^2\sinh^{-1}x - x\sqrt{1+x^2} - x\sinh^{-1}1 + x(1+\sqrt{2})$.
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\begin{enumerate}[label=(\alph*)]
\item Use the substitution $\mathrm { u } = 1 + \mathrm { x } ^ { 2 }$ to find
$$\int \frac { x } { \sqrt { 1 + x ^ { 2 } } } d x$$
\item Find the solution of the differential equation
$$x \frac { d y } { d x } - y = x ^ { 2 } \sinh ^ { - 1 } x$$
given that $y = 1$ when $x = 1$. Give your answer in the form $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q7 [12]}}