| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring knowledge of hyperbolic functions, their derivatives, and second-order differential equations. Part (a) demands careful algebraic manipulation using cosh²u - sinh²u = 1 and chain rule applications. Part (b) requires solving a non-homogeneous second-order ODE and applying initial conditions involving hyperbolic functions. While the techniques are standard for Further Maths, the multi-step nature and hyperbolic function manipulation elevate it above typical questions. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \sinh u \frac{du}{dx}\) | B1 | |
| \(\frac{d^2y}{dx^2} = \sinh u \frac{d^2u}{dx^2} + \cosh u \left(\frac{du}{dx}\right)^2\) | B1 | |
| \(\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = \sinh u \frac{d^2u}{dx^2} + \cosh u\left(\frac{du}{dx}\right)^2 + \sinh u \frac{du}{dx} - 2\cosh u\) | M1 | Uses substitution to find \(y\)-\(x\) equation, AG |
| \(\sqrt{\cosh^2 u - 1}\left(\frac{d^2u}{dx^2} + \frac{du}{dx}\right) + \cosh u \left(\frac{du}{dx}\right)^2 - 2\cosh u = 4e^{-x}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(m^2 + m - 2 = 0 \Rightarrow m = 1, -2\) | M1 | Auxiliary equation |
| \(y = Ae^x + Be^{-2x}\) | A1 | Complementary function. Allow \(y =\) missing |
| \(y = ke^{-x}\), \(y' = -ke^{-x}\), \(y'' = ke^{-x}\) | B1 | Particular integral and its derivatives |
| \(ke^{-x} - ke^{-x} - 2ke^{-x} = 4e^{-x}\) | M1 | Substitutes and equates coefficients |
| \(k = -2\) | A1 | WWW |
| \(y = \cosh u = Ae^x + Be^{-2x} - 2e^{-x}\) | A1 | Must have \(y =\) or \(\cosh u =\) |
| \(y' = \sinh u \frac{du}{dx} = Ae^x - 2Be^{-2x} + 2e^{-x}\) | B1 | |
| \(A + B - 2 = \frac{5}{3}\), \(A - 2B + 2 = 4\), \(A = \frac{28}{9}\), \(B = \frac{5}{9}\) | M1 A1 | Substitutes initial conditions; forms simultaneous equations. M1 allows substitution into their \(y\) and \(y'\) if two linear equations in two unknowns derived |
| \(u = \cosh^{-1}\!\left(\tfrac{28}{9}e^x + \tfrac{5}{9}e^{-2x} - 2e^{-x}\right)\) | A1 | Substitutes for \(y\) and finds \(u\) in terms of \(x\) |
## Question 8(a):
$\frac{dy}{dx} = \sinh u \frac{du}{dx}$ | B1 | |
$\frac{d^2y}{dx^2} = \sinh u \frac{d^2u}{dx^2} + \cosh u \left(\frac{du}{dx}\right)^2$ | B1 | |
$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = \sinh u \frac{d^2u}{dx^2} + \cosh u\left(\frac{du}{dx}\right)^2 + \sinh u \frac{du}{dx} - 2\cosh u$ | M1 | Uses substitution to find $y$-$x$ equation, AG |
$\sqrt{\cosh^2 u - 1}\left(\frac{d^2u}{dx^2} + \frac{du}{dx}\right) + \cosh u \left(\frac{du}{dx}\right)^2 - 2\cosh u = 4e^{-x}$ | A1 | AG |
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## Question 8(b):
$m^2 + m - 2 = 0 \Rightarrow m = 1, -2$ | M1 | Auxiliary equation |
$y = Ae^x + Be^{-2x}$ | A1 | Complementary function. Allow $y =$ missing |
$y = ke^{-x}$, $y' = -ke^{-x}$, $y'' = ke^{-x}$ | B1 | Particular integral and its derivatives |
$ke^{-x} - ke^{-x} - 2ke^{-x} = 4e^{-x}$ | M1 | Substitutes and equates coefficients |
$k = -2$ | A1 | WWW |
$y = \cosh u = Ae^x + Be^{-2x} - 2e^{-x}$ | A1 | Must have $y =$ or $\cosh u =$ |
$y' = \sinh u \frac{du}{dx} = Ae^x - 2Be^{-2x} + 2e^{-x}$ | B1 | |
$A + B - 2 = \frac{5}{3}$, $A - 2B + 2 = 4$, $A = \frac{28}{9}$, $B = \frac{5}{9}$ | M1 A1 | Substitutes initial conditions; forms simultaneous equations. M1 allows substitution into their $y$ and $y'$ if two linear equations in two unknowns derived |
$u = \cosh^{-1}\!\left(\tfrac{28}{9}e^x + \tfrac{5}{9}e^{-2x} - 2e^{-x}\right)$ | A1 | Substitutes for $y$ and finds $u$ in terms of $x$ |8 It is given that $\mathrm { y } = \operatorname { coshu }$, where $u > 0$, and
$$\sqrt { \cosh ^ { 2 } u - 1 } \left( \frac { d ^ { 2 } u } { d x ^ { 2 } } + \frac { d u } { d x } \right) + \cosh u \left( \frac { d u } { d x } \right) ^ { 2 } - 2 \cosh u = 4 e ^ { - x }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { d ^ { 2 } y } { d x ^ { 2 } } + \frac { d y } { d x } - 2 y = 4 e ^ { - x }$$
\item Find $u$ in terms of $x$, given that, when $x = 0 , u = \ln 3$ and $\frac { d u } { d x } = 3$.\\
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]}}