AQA FP3 2008 January — Question 5 9 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2008
SessionJanuary
Marks9
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TopicFirst order differential equations (integrating factor)
TypeStandard linear first order - variable coefficients
DifficultyStandard +0.3 This is a standard integrating factor question from Further Maths with a straightforward structure: identify P(x) = 4x/(x²+1), compute the integrating factor e^(2ln(x²+1)) = (x²+1)², multiply through, integrate both sides, and apply the initial condition. While it's a Further Maths topic, the execution is mechanical with no conceptual surprises, making it slightly easier than average overall but routine for FP3 students.
Spec4.10c Integrating factor: first order equations
5 By using an integrating factor, find the solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 x } { x ^ { 2 } + 1 } y = x$$ given that \(y = 1\) when \(x = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).
AnswerMarks Guidance
IF is \(e^{\int \frac{4x}{x^2+1}dx}\)M1
\(= e^{2\ln(x^2+1)}\)A1
\(= e^{\ln(x^2+1)^2} = (x^2+1)^2\)A1/
\(\frac{d}{dx}[y(x^2+1)^2] = x(x^2+1)^2\)M1, A1/
\(y(x^2+1)^2 = \int x(x^2+1)^2\,dx\)
\(y(x^2+1)^2 = \frac{1}{6}(x^2+1)^3 + c\)M1, A1
\(y(0) = 1 \Rightarrow c = \frac{5}{6}\)m1
\(y = \frac{1}{6}(x^2+1) + \frac{5}{6(x^2+1)^2}\)A1 Total: 9 
IF is $e^{\int \frac{4x}{x^2+1}dx}$ | M1 | |
$= e^{2\ln(x^2+1)}$ | A1 | |
$= e^{\ln(x^2+1)^2} = (x^2+1)^2$ | A1/ | | Ft on $e^{p\ln(x^2+1)}$
$\frac{d}{dx}[y(x^2+1)^2] = x(x^2+1)^2$ | M1, A1/ | | LHS as d/dx(xcand's IF) PI and also RHS of form $kx(x^2+1)^p$
$y(x^2+1)^2 = \int x(x^2+1)^2\,dx$ | | |
$y(x^2+1)^2 = \frac{1}{6}(x^2+1)^3 + c$ | M1, A1 | | Use of suitable substitution to find RHS or reaching $k(x^2+1)^3$ OE. Condone missing c
$y(0) = 1 \Rightarrow c = \frac{5}{6}$ | m1 | |
$y = \frac{1}{6}(x^2+1) + \frac{5}{6(x^2+1)^2}$ | A1 | Total: 9 | Accept other forms of f(x), e.g. $y = \frac{x^6/6 + 2x^4/4 + x^2/2 + 1}{(x^2+1)^2}$
5 By using an integrating factor, find the solution of the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + \frac { 4 x } { x ^ { 2 } + 1 } y = x$$

given that $y = 1$ when $x = 0$. Give your answer in the form $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{AQA FP3 2008 Q5 [9]}}
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