OCR H240/03 2018 June — Question 7 9 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (polynomial/exponential x-side)
DifficultyStandard +0.3 This is a straightforward separable variables question requiring standard technique: separate variables, integrate both sides (including a substitution for the left side), apply initial condition, and rearrange to match the given form. While it involves multiple steps and algebraic manipulation, it follows a completely standard pattern with no conceptual challenges or novel insights required, making it slightly easier than average.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)
7 The gradient of the curve \(y = \mathrm { f } ( x )\) is given by the differential equation $$( 2 x - 1 ) ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y ^ { 2 } = 0$$ and the curve passes through the point \(( 1,1 )\). By solving this differential equation show that $$f ( x ) = \frac { a x ^ { 2 } - a x + 1 } { b x ^ { 2 } - b x + 1 }$$ where \(a\) and \(b\) are integers to be determined.
Question 7:
\((2x-1)^3\dfrac{dy}{dx} + 4y^2 = 0\)
AnswerMarks Guidance
\(-\dfrac{1}{4}\displaystyle\int \frac{dy}{y^2} = \int \frac{dx}{(2x-1)^3}\)M1 2.5 Attempt to separate variables
\(\displaystyle\int \frac{dy}{y^2} = -\frac{1}{y}\)A1 1.1 —
\(\displaystyle\int \frac{dx}{(2x-1)^3} = \frac{(2x-1)^{-2}}{(2)(-2)}\)M1 A1 1.1 M1 for \(k(2x-1)^{-2}\)
\(\dfrac{1}{4y} = -\dfrac{1}{4(2x-1)^2} + c\), \((1,1) \Rightarrow c = \ldots\)M1 2.1 Use of \((1,1)\) to find \(c\) — dependent on previous two M marks, substituted into correct form
\(\dfrac{1}{y} = -\dfrac{1}{(2x-1)^2} + 2\)A1 2.2a Oe
\(\dfrac{1}{y} = \dfrac{2(2x-1)^2 - 1}{(2x-1)^2}\)M1 3.1a Correct method for combining both terms on rhs (dependent on previous M mark) before taking reciprocal; or re-write in terms of \(y\)
\(y = \dfrac{(2x-1)^2}{2(2x-1)^2 - 1}\)M1 1.1 Taking the reciprocal (dependent on previous M marks) and making \(y\) the subject; remove triple-decker fractions
\(y = \dfrac{4x^2 - 4x + 1}{8x^2 - 8x + 1}\)A1 2.2a [9] \(a = 4,\, b = 8\) 
## Question 7:

$(2x-1)^3\dfrac{dy}{dx} + 4y^2 = 0$

$-\dfrac{1}{4}\displaystyle\int \frac{dy}{y^2} = \int \frac{dx}{(2x-1)^3}$ | M1 2.5 | Attempt to separate variables

$\displaystyle\int \frac{dy}{y^2} = -\frac{1}{y}$ | A1 1.1 | —

$\displaystyle\int \frac{dx}{(2x-1)^3} = \frac{(2x-1)^{-2}}{(2)(-2)}$ | M1 A1 1.1 | M1 for $k(2x-1)^{-2}$

$\dfrac{1}{4y} = -\dfrac{1}{4(2x-1)^2} + c$, $(1,1) \Rightarrow c = \ldots$ | M1 2.1 | Use of $(1,1)$ to find $c$ — dependent on previous two M marks, substituted into correct form

$\dfrac{1}{y} = -\dfrac{1}{(2x-1)^2} + 2$ | A1 2.2a | Oe

$\dfrac{1}{y} = \dfrac{2(2x-1)^2 - 1}{(2x-1)^2}$ | M1 3.1a | Correct method for combining both terms on rhs (dependent on previous M mark) before taking reciprocal; or re-write in terms of $y$

$y = \dfrac{(2x-1)^2}{2(2x-1)^2 - 1}$ | M1 1.1 | Taking the reciprocal (dependent on previous M marks) and making $y$ the subject; remove triple-decker fractions

$y = \dfrac{4x^2 - 4x + 1}{8x^2 - 8x + 1}$ | A1 2.2a [9] | $a = 4,\, b = 8$
7 The gradient of the curve $y = \mathrm { f } ( x )$ is given by the differential equation

$$( 2 x - 1 ) ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y ^ { 2 } = 0$$

and the curve passes through the point $( 1,1 )$. By solving this differential equation show that

$$f ( x ) = \frac { a x ^ { 2 } - a x + 1 } { b x ^ { 2 } - b x + 1 }$$

where $a$ and $b$ are integers to be determined.

\hfill \mbox{\textit{OCR H240/03 2018 Q7 [9]}}
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