Question 1 - A-Level Maths

CAIE Further Paper 2 2020 June — Question 1 6 marks

Exam Board	CAIE
Module	Further Paper 2 (Further Paper 2)
Year	2020
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Standard linear first order - constant coefficients
Difficulty	Standard +0.3 This is a straightforward integrating factor question with a simple exponential right-hand side. The integrating factor is e^(5x), leading to a routine integration of e^(-2x). While it's a Further Maths topic (making it slightly above average), the method is completely standard with no complications, requiring only direct application of the integrating factor technique and one initial condition.
Spec	4.10c Integrating factor: first order equations

1 Find the solution of the differential equation $$\frac { d y } { d x } + 5 y = e ^ { - 7 x }$$ for which $y = 0$ when $x = 0$. Give your answer in the form $y = f ( x )$.

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
Answer	Marks	Guidance
$e^{\int 5\,dx} = e^{5x}$	M1 A1	Integrating factor
$\frac{d}{dx}(ye^{5x}) = e^{-2x}$	M1
$ye^{5x} = -\frac{1}{2}e^{-2x} + C$	A1
$0 = -\frac{1}{2} + C$	M1	Using initial conditions
$y = \frac{1}{2}e^{-5x} - \frac{1}{2}e^{-7x}$	A1
6

**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{\int 5\,dx} = e^{5x}$ | M1 A1 | Integrating factor |
| $\frac{d}{dx}(ye^{5x}) = e^{-2x}$ | M1 | |
| $ye^{5x} = -\frac{1}{2}e^{-2x} + C$ | A1 | |
| $0 = -\frac{1}{2} + C$ | M1 | Using initial conditions |
| $y = \frac{1}{2}e^{-5x} - \frac{1}{2}e^{-7x}$ | A1 | |
| | **6** | |

---

Show LaTeX source

1 Find the solution of the differential equation

$$\frac { d y } { d x } + 5 y = e ^ { - 7 x }$$

for which $y = 0$ when $x = 0$. Give your answer in the form $y = f ( x )$.\\

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

This paper (8 questions)

View full paper

Q1 6 Q2 7 Q3 8 Q4 8 Q5 11 Q6 10 Q7 11 Q8 14