Question 10 - A-Level Maths

CAIE FP1 2014 June — Question 10 12 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Asymptotic behavior for large values
Difficulty	Challenging +1.2 This is a standard second-order linear differential equation with constant coefficients and a polynomial forcing term. Students must find the complementary function (solving the auxiliary equation), particular integral (trying x = at + b), apply initial conditions, and analyze asymptotic behavior as exponential terms decay. While it requires multiple techniques and careful algebra, it follows a well-established procedure taught in FP1 with no novel insights required.
Spec	4.10e Second order non-homogeneous: complementary + particular integral

10 Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 0.16 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 0.0064 x = 8.64 + 0.32 t$$ given that when $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$. Show that, for large positive $t , \frac { \mathrm {~d} x } { \mathrm {~d} t } \approx 50$.

Show mark scheme Show mark scheme source

Question 10:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$m^2 + 0.16m + 0.0064 = 0 \Rightarrow (m+0.08)^2 = 0 \Rightarrow m = -0.08$	M1
CF: $x = (A+Bt)e^{-0.08t}$	A1
PI: $x = p + qt \Rightarrow \dot{x} = q, \ddot{x} = 0$	M1
$0.16q + 0.0064(p+qt) = 8.64 + 0.32t \Rightarrow p = 100$, $q = 50$	M1A1
$x = (A+Bt)e^{-0.08t} + 100 + 50t$	A1
$x = 0$ when $t = 0 \Rightarrow A = -100$	B1
$\dot{x} = -0.08(Bt-100)e^{-0.08t} + Be^{-0.08t} + 50$	M1	Correct form required for M mark
$\dot{x} = 0$ when $t = 0 \Rightarrow B = -58$	A1
$x = 100 + 50t - (100+58t)e^{-0.08t}$	A1
Total	[10]
From above: $e^{-0.08t} \to 0$ as $t \to \infty$	M1	Correct form required for M mark
$\therefore \dot{x} \to 50$	A1
Total	[2]

# Question 10:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $m^2 + 0.16m + 0.0064 = 0 \Rightarrow (m+0.08)^2 = 0 \Rightarrow m = -0.08$ | M1 | |
| CF: $x = (A+Bt)e^{-0.08t}$ | A1 | |
| PI: $x = p + qt \Rightarrow \dot{x} = q, \ddot{x} = 0$ | M1 | |
| $0.16q + 0.0064(p+qt) = 8.64 + 0.32t \Rightarrow p = 100$, $q = 50$ | M1A1 | |
| $x = (A+Bt)e^{-0.08t} + 100 + 50t$ | A1 | |
| $x = 0$ when $t = 0 \Rightarrow A = -100$ | B1 | |
| $\dot{x} = -0.08(Bt-100)e^{-0.08t} + Be^{-0.08t} + 50$ | M1 | Correct form required for M mark |
| $\dot{x} = 0$ when $t = 0 \Rightarrow B = -58$ | A1 | |
| $x = 100 + 50t - (100+58t)e^{-0.08t}$ | A1 | |
| Total | [10] | |
| From above: $e^{-0.08t} \to 0$ as $t \to \infty$ | M1 | Correct form required for M mark |
| $\therefore \dot{x} \to 50$ | A1 | |
| Total | [2] | |

Show LaTeX source

10 Find the particular solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 0.16 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 0.0064 x = 8.64 + 0.32 t$$

given that when $t = 0 , x = 0$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$.

Show that, for large positive $t , \frac { \mathrm {~d} x } { \mathrm {~d} t } \approx 50$.

\hfill \mbox{\textit{CAIE FP1 2014 Q10 [12]}}

This paper (12 questions)

View full paper

Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 10 Q7 10 Q8 11 Q9 10 Q10 12 Q11 EITHER Q11 OR