AQA Further Paper 1 2019 June — Question 11 7 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2019
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeStandard linear first order - variable coefficients
DifficultyChallenging +1.2 This is a first-order linear ODE requiring integrating factor method (standard Further Maths technique) followed by integration involving completing the square under a square root. While the integration is somewhat involved, the overall approach is methodical and the 7-mark allocation reflects routine application of known techniques rather than requiring novel insight or particularly complex manipulation.
Spec4.10c Integrating factor: first order equations
Find the general solution of the differential equation $$x \frac{dy}{dx} - 2y = \frac{x^3}{\sqrt{4 - 2x - x^2}}$$ where \(0 < x < \sqrt{5} - 1\) [7 marks]
Question 11:
AnswerMarks Guidance
11Divides through by x AO1.1a
− =
dx x 4−2x−x2
2
∫Pdx=−∫ dx=−2lnx
x
Integrating factor
∫Pdx
=e = e −2lnx = x −2
1 dy 2y 1
− =
x2 dx x3 4−2x− x2
d  y  1
=
 
dx x2  4−2x− x2
To find
1
∫ dx
4−2x−x2
4−2x−x2 =5−( x+1 )2
1 1
∫ dx = ∫ dx
4−2x−x2 5−( x+1 )2
  x+1 
∴y = x2 sin−1 +c
 
  5  
Recognises that the
Integrating Factor
Method can be applied
and finds correct
integrating factor, accept
AnswerMarks Guidance
e−2lnxAO3.1a M1
Multiplies equation by
AnswerMarks Guidance
their integrating factorAO1.1a M1
Integrates LHS to obtain
y
AnswerMarks Guidance
x2AO1.1b A1
Recognises the need to
complete the square
AnswerMarks Guidance
inside the square root.AO3.1a M1
Correctly uses the
appropriate inverse sine,
inverse cosh or inverse
sinh function to integrate
AnswerMarks Guidance
all (or part of) their RHS.AO3.1a M1
Finds correct solution
including constant of
integration. ACF
y
Accept =...
AnswerMarks Guidance
x2AO1.1b A1
Total7
QMarking Instructions AO 
Question 11:
11 | Divides through by x | AO1.1a | M1 | dy 2y x2
− =
dx x 4−2x−x2
2
∫Pdx=−∫ dx=−2lnx
x
Integrating factor
∫Pdx
=e = e −2lnx = x −2
1 dy 2y 1
− =
x2 dx x3 4−2x− x2
d  y  1
=
 
dx x2  4−2x− x2
To find
1
∫ dx
4−2x−x2
4−2x−x2 =5−( x+1 )2
1 1
∫ dx = ∫ dx
4−2x−x2 5−( x+1 )2
  x+1 
∴y = x2 sin−1 +c
 
  5  
Recognises that the
Integrating Factor
Method can be applied
and finds correct
integrating factor, accept
e−2lnx | AO3.1a | M1
Multiplies equation by
their integrating factor | AO1.1a | M1
Integrates LHS to obtain
y
x2 | AO1.1b | A1
Recognises the need to
complete the square
inside the square root. | AO3.1a | M1
Correctly uses the
appropriate inverse sine,
inverse cosh or inverse
sinh function to integrate
all (or part of) their RHS. | AO3.1a | M1
Finds correct solution
including constant of
integration. ACF
y
Accept =...
x2 | AO1.1b | A1
Total | 7
Q | Marking Instructions | AO | Marks | Typical solution
Find the general solution of the differential equation
$$x \frac{dy}{dx} - 2y = \frac{x^3}{\sqrt{4 - 2x - x^2}}$$
where $0 < x < \sqrt{5} - 1$
[7 marks]

\hfill \mbox{\textit{AQA Further Paper 1 2019 Q11 [7]}}
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