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Date November 2020 Marks available 9 Reference code 20N.2.AHL.TZ0.F_9
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Solve Question number F_9 Adapted from N/A

Question

Consider the differential equation dydx=y-xy+x, where x, y>0.

It is given that y=2 when x=1.

Solve the differential equation, giving your answer in the form fx,y=0.

[9]
a.

The graph of y against x has a local maximum between x=2 and x=3. Determine the coordinates of this local maximum.

[4]
b.

Show that there are no points of inflexion on the graph of y against x.

[4]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

puts y=vx so that dydx=v+xdvdx               M1

v+xdvdx=vx-xvx+x =v-1v+1            A1

attempts to express xdvdx as a single rational fraction in v

xdvdx=-v2+1v+1               M1

attempts to separate variables               M1

v+1v2+1dv=-1xdx

12lnv2+1+arctanv=-lnx+C            A1A1

substitutes y=2,x=1 and attempts to find the value of C               M1

C=12ln5+arctan2            A1

the solution is

12lny2x2+1+arctanyx+lnx-12ln5-arctan2=0            A1


[9 marks]

a.

at a maximum, dydx=0               M1

attempts to substitute y=x into their solution               M1

12ln2+arctan1+lnx=12ln5+arctan2

attempts to solve for x,y               (M1)

2.18,2.18  102earctan2-π4,102earctan2-π4              A1


Note:
Accept all answers that round to the correct 2sf answer.
Accept x=2.18,y=2.18.


[4 marks]

b.

METHOD 1

attempts (quotient rule) implicit differentiation               M1

d2ydx2=dydx-1y+x-y-xdydx+1y+x2

correctly substitutes dydx=y-xy+x into d2ydx2

=y-xy+x-1y+x-y-xy-xy+x+1y+x2              A1

=-2x2+y2x+y3              A1

this expression can never be zero therefore no points of inflexion              R1

 

METHOD 2

attempts implicit differentiation on y+xdydx=y-x               M1

dydx+1dydx+y+xd2ydx2=dydx-1              A1

y+xd2ydx2=dydx-1-dydx2-dydx

=-1-dydx2              A1

-1-dydx2<0 and x+y>0, d2ydx20 therefore no points of inflexion              R1


Note:
Accept putting d2ydx2=0 and obtaining contradiction.


[4 marks]

c.

Examiners report

[N/A]
a.
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b.
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c.

Syllabus sections

Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
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Topic 5 —Calculus

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