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Date November 2021 Marks available 7 Reference code 21N.1.AHL.TZ0.8
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Solve Question number 8 Adapted from N/A

Question

Solve the differential equation dydx=ln2xx2-2yx, x>0, given that y=4 at x=12.

Give your answer in the form y=fx.

Markscheme

dydx+2yx=ln2xx2                 (M1)

attempt to find integrating factor                 (M1)

e2xdx=e2lnx=x2                 (A1)

x2dydx+2xy=ln2x

ddxx2y=ln2x

x2y=ln2xdx

attempt to use integration by parts                 (M1)

x2y=xln2x-x+c                 A1

y=ln2xx-1x+cx2

substituting x=12, y=4 into an integrated equation involving c                 M1

4=0-2+4c

c=32

y=ln2xx-1x+32x2                 A1

 

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.16—Integration by substitution, parts and repeated parts
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Topic 5 —Calculus

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