Question 1 - A-Level Maths

CAIE Further Paper 2 2020 Specimen — Question 1 6 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2020
Session	Specimen
Marks	6
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 second-order linear differential equation with constant coefficients and polynomial RHS, requiring both complementary function (repeated root case) and particular integral (polynomial trial solution). While methodical, it's Further Maths content with multiple steps including handling the repeated root λ=-2 and finding a quadratic particular integral, placing it moderately above average difficulty.
Spec	4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

1 Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = 7 - 2 t ^ { 2 }$$

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Question 1:

Answer	Marks	Guidance
Answer	Marks	Guidance
$m^2 + 4m + 4 = 0$: $m = -2, -2$	M1	Auxiliary equation
$x = (At + B)e^{-2t}$	A1	Complementary function
$x = P + Qt + rt^2$: $\dot{x} = Q + 2Rt$; $\ddot{x} = 2R$	B1	Particular integral and its derivatives
Substitute and equate coefficients of powers of $t$	M1
$P = 1, Q = 1, R = -\frac{1}{2}$	A1
$x = (At + B)e^{-2t} + 1 + t - \frac{1}{2}t^2$	A1
6

# Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $m^2 + 4m + 4 = 0$: $m = -2, -2$ | M1 | Auxiliary equation |
| $x = (At + B)e^{-2t}$ | A1 | Complementary function |
| $x = P + Qt + rt^2$: $\dot{x} = Q + 2Rt$; $\ddot{x} = 2R$ | B1 | Particular integral and its derivatives |
| Substitute and equate coefficients of powers of $t$ | M1 | |
| $P = 1, Q = 1, R = -\frac{1}{2}$ | A1 | |
| $x = (At + B)e^{-2t} + 1 + t - \frac{1}{2}t^2$ | A1 | |
| | **6** | |

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1 Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 4 x = 7 - 2 t ^ { 2 }$$

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q1 [6]}}

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Q0 Q1 6 Q2 6 Q3 8 Q4 8 Q5 10 Q6 10 Q7 12 Q8 15