OCR H240/01 2020 November — Question 12 9 marks

Exam BoardOCR
ModuleH240/01 (Pure Mathematics)
Year2020
SessionNovember
Marks9
PaperDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - partial fractions
DifficultyStandard +0.3 This is a separable variables question requiring standard technique (separate, integrate both sides, solve for y). The main challenge is factorising the cubic and partial fractions for integration, but these are routine A-level skills. Slightly above average due to the algebraic manipulation required, but still a standard textbook-style question.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

12 Find the general solution of the differential equation \(\left( 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 20 x - 35 )\).
Give your answer in the form \(y = \mathrm { f } ( x )\). \section*{END OF QUESTION PAPER} \section*{OCR
Oxford Cambridge and RSA}

Question 12
AnswerMarks Guidance
AnswerMarks Guidance
\(\int \frac{1}{y}\,dy = \int \frac{20x-35}{2x^3 - 3x^2 - 11x + 6}\,dx\)M1 Separate variables; correct process to deal with algebraic fractions with BOD on integral notation
\(f(x) = 2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2+3x-2)\)M1 Attempt to factorise cubic; possibly BC, so correct factorised cubic implies M1A1
\(= (x-3)(x+2)(2x-1)\)A1 Correct factorised cubic
\(\frac{20x-35}{2x^3-3x^2-11x+6} = \frac{A}{x+2} + \frac{B}{x-3} + \frac{C}{2x-1}\)M1 Attempt partial fractions using their 3 linear factors; must be correct structure, attempting at least one numerator
\(= \frac{-3}{x+2} + \frac{1}{x-3} + \frac{4}{2x-1}\)A1 Obtain any one correct fraction www
A1Obtain fully correct partial fractions
\(\int \frac{1}{y}\,dy = \lny \)
\(-3\lnx+2 + \ln
\(y = \frac{A(x-3)(2x-1)^2}{(x+2)^3}\)A1 Obtain correct equation; any correct form not involving ln; may be \(e^c\) not \(A\), but A0 if fraction \(+c\); could have \((x+2)^{-3}\) in a product
[9]
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## Question 12

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{1}{y}\,dy = \int \frac{20x-35}{2x^3 - 3x^2 - 11x + 6}\,dx$ | M1 | Separate variables; correct process to deal with algebraic fractions with BOD on integral notation |
| $f(x) = 2x^3 - 3x^2 - 11x + 6 = (x-3)(2x^2+3x-2)$ | M1 | Attempt to factorise cubic; possibly BC, so correct factorised cubic implies M1A1 |
| $= (x-3)(x+2)(2x-1)$ | A1 | Correct factorised cubic |
| $\frac{20x-35}{2x^3-3x^2-11x+6} = \frac{A}{x+2} + \frac{B}{x-3} + \frac{C}{2x-1}$ | M1 | Attempt partial fractions using their 3 linear factors; must be correct structure, attempting at least one numerator |
| $= \frac{-3}{x+2} + \frac{1}{x-3} + \frac{4}{2x-1}$ | A1 | Obtain any one correct fraction www |
| | A1 | Obtain fully correct partial fractions |
| $\int \frac{1}{y}\,dy = \ln|y|$ | B1 | Correct integration of $\frac{1}{y}$; condone no modulus sign |
| $-3\ln|x+2| + \ln|x-3| + 2\ln|2x-1| + \ln A$ | A1FT | Obtain correct integral following their 3 linear partial fractions; condone no constant of integration; condone brackets and not modulus; FT from point that partial fractions were credited |
| $y = \frac{A(x-3)(2x-1)^2}{(x+2)^3}$ | A1 | Obtain correct equation; any correct form not involving ln; may be $e^c$ not $A$, but A0 if fraction $+c$; could have $(x+2)^{-3}$ in a product |
| | **[9]** | |

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12 Find the general solution of the differential equation\\
$\left( 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 20 x - 35 )$.\\
Give your answer in the form $y = \mathrm { f } ( x )$.

\section*{END OF QUESTION PAPER}
\section*{OCR \\
 Oxford Cambridge and RSA}

\hfill \mbox{\textit{OCR H240/01 2020 Q12 [9]}}