| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Standard non-homogeneous with polynomial RHS |
| Difficulty | Standard +0.8 This is a standard Further Maths second-order linear differential equation requiring both complementary function (solving auxiliary equation with complex roots) and particular integral (polynomial trial solution with coefficient matching). Part (b) adds a conceptual element about asymptotic behavior. Solidly above average difficulty due to being Further Maths content with multiple technical steps, but follows standard procedures without requiring novel insight. |
| 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 |
| \(z^2 + z + 1 = 0\) leading to \(m = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\) | M1 | Auxiliary equation |
| \(x = e^{-\frac{1}{2}t}\left(A\sin\frac{\sqrt{3}}{2}t + B\cos\frac{\sqrt{3}}{2}t\right)\) | A1 | Complimentary function |
| \(x = pt^2 + qt + r\) leading to \(\dot{x} = 2pt + q\) leading to \(\ddot{x} = 2p\) | B1 | Particular integral and its derivatives |
| \(2p + 2pt + q + pt^2 + qt + r = t^2 + 1\) | M1 | Substitutes and equates coefficients. PI must have the correct form |
| \(r = 1,\quad q = -2,\quad p = 1\) | A1 | |
| \(x = e^{-\frac{1}{2}t}\left(A\sin\frac{\sqrt{3}}{2}t + B\cos\frac{\sqrt{3}}{2}t\right) + t^2 - 2t + 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{-\frac{1}{2}t} \to 0\) as \(t \to \infty\) | M1 | |
| \(\ddot{x} \to 2\) | A1 FT |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z^2 + z + 1 = 0$ leading to $m = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$ | M1 | Auxiliary equation |
| $x = e^{-\frac{1}{2}t}\left(A\sin\frac{\sqrt{3}}{2}t + B\cos\frac{\sqrt{3}}{2}t\right)$ | A1 | Complimentary function |
| $x = pt^2 + qt + r$ leading to $\dot{x} = 2pt + q$ leading to $\ddot{x} = 2p$ | B1 | Particular integral and its derivatives |
| $2p + 2pt + q + pt^2 + qt + r = t^2 + 1$ | M1 | Substitutes and equates coefficients. PI must have the correct form |
| $r = 1,\quad q = -2,\quad p = 1$ | A1 | |
| $x = e^{-\frac{1}{2}t}\left(A\sin\frac{\sqrt{3}}{2}t + B\cos\frac{\sqrt{3}}{2}t\right) + t^2 - 2t + 1$ | A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{-\frac{1}{2}t} \to 0$ as $t \to \infty$ | M1 | |
| $\ddot{x} \to 2$ | A1 FT | |3 The variables $t$ and $x$ are related by the differential equation
$$\frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x } { d t } + x = t ^ { 2 } + 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the general solution for $x$ in terms of $t$.
\item Deduce an approximate value of $\frac { \mathrm { d } ^ { 2 } \mathrm { x } } { \mathrm { dt } ^ { 2 } }$ for large positive values of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q3 [8]}}