Question 5 - A-Level Maths

OCR FP3 2015 June — Question 5 8 marks

Exam Board	OCR
Module	FP3 (Further Pure Mathematics 3)
Year	2015
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Standard linear first order - variable coefficients
Difficulty	Standard +0.3 This is a standard linear first order ODE requiring the integrating factor method—a core Further Maths technique. The question is straightforward: divide by x to get standard form, find integrating factor x³, integrate, and apply initial conditions. While it's a Further Maths topic (inherently harder), it's a textbook application with no tricks or novel insights required, making it slightly easier than average overall.
Spec	4.10c Integrating factor: first order equations

5 Find the particular solution of the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } + x$$ for which $y = 1$ when $x = 1$, giving $y$ in terms of $x$.

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Answer	Marks	Guidance
$\frac{dy}{dx} + \frac{3}{x}y = x + 1$	B1	Divide both sides by $x$
$I = \exp\left(\int\frac{3}{x}dx\right) = e^{3\ln x} = x^3$	M1
$x^3\frac{dy}{dx} + 3x^2 y = x^4 + x^3$	M1	Multiply and recognise derivative
$\frac{d}{dx}(x^3 y) = \cdots$	M1
$\cdots = x^4 + x^3$	M1	Integrate both sides (their two term polynomial)
$x^3 y = \frac{1}{5}x^5 + \frac{1}{4}x^4 + A$	A1	condone absent $A$ at this stage
$x = 1, y = 1 \Rightarrow A = \frac{11}{20}$	M1	Use condition
$y = \frac{1}{5}x^2 + \frac{1}{4}x + \frac{11}{20}x^{-3}$	A1
[8]

$\frac{dy}{dx} + \frac{3}{x}y = x + 1$ | B1 | Divide both sides by $x$
$I = \exp\left(\int\frac{3}{x}dx\right) = e^{3\ln x} = x^3$ | M1 |
$x^3\frac{dy}{dx} + 3x^2 y = x^4 + x^3$ | M1 | Multiply and recognise derivative
$\frac{d}{dx}(x^3 y) = \cdots$ | M1 |
$\cdots = x^4 + x^3$ | M1 | Integrate both sides (their two term polynomial)
$x^3 y = \frac{1}{5}x^5 + \frac{1}{4}x^4 + A$ | A1 | condone absent $A$ at this stage
$x = 1, y = 1 \Rightarrow A = \frac{11}{20}$ | M1 | Use condition
$y = \frac{1}{5}x^2 + \frac{1}{4}x + \frac{11}{20}x^{-3}$ | A1 |
| [8] |

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5 Find the particular solution of the differential equation

$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 y = x ^ { 2 } + x$$

for which $y = 1$ when $x = 1$, giving $y$ in terms of $x$.

\hfill \mbox{\textit{OCR FP3 2015 Q5 [8]}}

This paper (8 questions)

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Q1 8 Q2 8 Q3 11 Q4 9 Q5 8 Q6 7 Q7 9 Q8 12

Answer	Marks	Guidance
\(\frac{dy}{dx} + \frac{3}{x}y = x + 1\)	B1	Divide both sides by \(x\)
\(I = \exp\left(\int\frac{3}{x}dx\right) = e^{3\ln x} = x^3\)	M1
\(x^3\frac{dy}{dx} + 3x^2 y = x^4 + x^3\)	M1	Multiply and recognise derivative
\(\frac{d}{dx}(x^3 y) = \cdots\)	M1
\(\cdots = x^4 + x^3\)	M1	Integrate both sides (their two term polynomial)
\(x^3 y = \frac{1}{5}x^5 + \frac{1}{4}x^4 + A\)	A1	condone absent \(A\) at this stage
\(x = 1, y = 1 \Rightarrow A = \frac{11}{20}\)	M1	Use condition
\(y = \frac{1}{5}x^2 + \frac{1}{4}x + \frac{11}{20}x^{-3}\)	A1
[8]