| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| 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 Further Maths second-order DE requiring substitution to simplify, then solving a constant-coefficient equation with complex roots and particular integral. Part (a) involves careful differentiation using product rule twice and algebraic manipulation to verify the simplified form. Part (b) requires standard auxiliary equation methods (yielding complex roots -2±3i), finding a particular integral (trial solution Ae^(t/2)), then back-substituting x=t³y. While systematic, it demands multiple techniques, careful algebra across several steps, and is beyond standard A-level—typical of Further Maths Paper 2 difficulty. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = t^3\frac{dy}{dt} + 3t^2 y\) | B1 | |
| \(\frac{d^2x}{dt^2} = t^3\frac{d^2y}{dt^2} + 6t^2\frac{dy}{dt} + 6ty\) | B1 | |
| \(\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 13x = t^3\frac{d^2y}{dt^2} + 6t^2\frac{dy}{dt} + 6ty + 4t^3\frac{dy}{dt} + 12t^2y + 13t^3y = 6le^{\frac{1}{2}t}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m^2 + 4m + 13 = 0 \Rightarrow m = -2 \pm 3i\) | M1 | |
| \(x = e^{-2t}(A\cos 3t + B\sin 3t)\) | A1 | |
| \(x = ke^{\frac{1}{2}t} \Rightarrow \dot{x} = \frac{1}{2}ke^{\frac{1}{2}t} \Rightarrow \ddot{x} = \frac{1}{4}ke^{\frac{1}{2}t}\) | B1 | |
| \(\frac{1}{4}k + 2k + 13k = 61 \Rightarrow k = 4\) | M1 A1 | |
| \(t^3 y = e^{-2t}(A\cos 3t + B\sin 3t) + 4e^{\frac{1}{2}t} \Rightarrow y = t^{-3}e^{-2t}(A\cos 3t + B\sin 3t) + 4t^{-3}e^{\frac{1}{2}t}\) | M1 A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = t^3\frac{dy}{dt} + 3t^2 y$ | B1 | |
| $\frac{d^2x}{dt^2} = t^3\frac{d^2y}{dt^2} + 6t^2\frac{dy}{dt} + 6ty$ | B1 | |
| $\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 13x = t^3\frac{d^2y}{dt^2} + 6t^2\frac{dy}{dt} + 6ty + 4t^3\frac{dy}{dt} + 12t^2y + 13t^3y = 6le^{\frac{1}{2}t}$ | M1 A1 | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + 4m + 13 = 0 \Rightarrow m = -2 \pm 3i$ | M1 | |
| $x = e^{-2t}(A\cos 3t + B\sin 3t)$ | A1 | |
| $x = ke^{\frac{1}{2}t} \Rightarrow \dot{x} = \frac{1}{2}ke^{\frac{1}{2}t} \Rightarrow \ddot{x} = \frac{1}{4}ke^{\frac{1}{2}t}$ | B1 | |
| $\frac{1}{4}k + 2k + 13k = 61 \Rightarrow k = 4$ | M1 A1 | |
| $t^3 y = e^{-2t}(A\cos 3t + B\sin 3t) + 4e^{\frac{1}{2}t} \Rightarrow y = t^{-3}e^{-2t}(A\cos 3t + B\sin 3t) + 4t^{-3}e^{\frac{1}{2}t}$ | M1 A1 | |7 It is given that $x = t ^ { 3 } y$ and
$$t ^ { 3 } \frac { d ^ { 2 } y } { d t ^ { 2 } } + \left( 4 t ^ { 3 } + 6 t ^ { 2 } \right) \frac { d y } { d t } + \left( 13 t ^ { 3 } + 12 t ^ { 2 } + 6 t \right) y = 61 e ^ { \frac { 1 } { 2 } t }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 13 x = 61 e ^ { \frac { 1 } { 2 } t }$$
\item Find the general solution for $y$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q7 [11]}}