CAIE Further Paper 2 2021 November — Question 2 7 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeStandard linear first order - variable coefficients
DifficultyStandard +0.3 This is a standard integrating factor question with a straightforward structure: the integrating factor is immediately recognizable as e^(ln(x^4+5)) = x^4+5, and the right-hand side integrates cleanly. While it requires multiple steps (identify method, find integrating factor, integrate both sides, apply initial condition), each step follows a well-practiced algorithm with no conceptual surprises. Slightly easier than average due to the clean algebraic form throughout.
Spec4.10c Integrating factor: first order equations

Find the solution of the differential equation $$\frac{dy}{dx} + \frac{4x^3y}{x^4 + 5} = 6x$$ for which \(y = 1\) when \(x = 1\). Give your answer in the form \(y = f(x)\). [7]

Question 2:
AnswerMarks Guidance
2( )
e ln x4+5 =x4 +5M1 A1 Finds integrating factor.
d ( ( ))
y x4 +5 =6x5 +30x
AnswerMarks Guidance
dxM1 Writes in correct form.
y(x4 +5)=x6 +15x2 +CA1
6=16+CM1 Substitutes initial conditions.
x6 +15x2 −10
y=
AnswerMarks Guidance
x4 +5M1 A1 Division through by coefficient of y.
7
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | ( )
e ln x4+5 =x4 +5 | M1 A1 | Finds integrating factor.
d ( ( ))
y x4 +5 =6x5 +30x
dx | M1 | Writes in correct form.
y(x4 +5)=x6 +15x2 +C | A1
6=16+C | M1 | Substitutes initial conditions.
x6 +15x2 −10
y=
x4 +5 | M1 A1 | Division through by coefficient of y.
7
Question | Answer | Marks | Guidance
Find the solution of the differential equation

$$\frac{dy}{dx} + \frac{4x^3y}{x^4 + 5} = 6x$$

for which $y = 1$ when $x = 1$. Give your answer in the form $y = f(x)$. [7]

\hfill \mbox{\textit{CAIE Further Paper 2 2021 Q2 [7]}}