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Date November Example questions Marks available 9 Reference code EXN.2.AHL.TZ0.12
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Solve and Show that Question number 12 Adapted from N/A

Question

Consider the differential equation

dydx=fyx, x>0

The curve y=fx for x>0 has a gradient function given by

dydx=y2+3xy+2x2x2.

The curve passes through the point 1,-1.

Use the substitution y=vx to show that dvfv-v=lnx+C where C is an arbitrary constant.

[3]
a.

By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation y=xtanlnx-1.

[9]
b.

The curve has a point of inflexion at x1,y1 where e-π2<x1<eπ2. Determine the coordinates of this point of inflexion.

[6]
c.

Use the differential equation dydx=y2+3xy+2x2x2 to show that the points of zero gradient on the curve lie on two straight lines of the form y=mx where the values of m are to be determined.

[4]
d.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

y=vxdydx=v+xdvdx       M1

v+xdvdx=fv       A1

dvfv-v=dxx       A1

integrating the RHS, dvfv-v=lnx+C       AG

 

[3 marks]

a.

EITHER

attempts to find fv       M1

fv=v2+3v+2       (A1)

substitutes their fv into dvfv-v       M1

dvfv-v=dvv2+2v+2

attempts to complete the square       (M1)

dvv+12+1       A1

arctanv+1 =lnx+C       A1

 

OR

attempts to find fv       M1

v+xdvdx=v2+3v+2       A1

dvv2+2v+2=dxx       M1

attempts to complete the square       (M1)

dvv+12+1=dxx       A1

arctanv+1 =lnx+C       A1

 

THEN

when x=1v=-1 (or y=-1) and so C=0       M1

substitutes for v into their expression       M1

arctanyx+1=lnx

yx+1=tanlnx       A1

so y=xtanlnx-1       AG

 

[9 marks]

b.

METHOD 1

EITHER

a correct graph of y=f'x (for approximately e-π2<x<eπ2) with a local minimum point below the x-axis        A2

 

Note: Award M1A1 for dydx=tanlnx+sec2lnx-1.

 

attempts to find the x-coordinate of the local minimum point on the graph of y=f'x        (M1)

OR

a correct graph of y=f''x (for approximately e-π2<x<eπ2) showing the location of the x-intercept        A2

 

Note: Award M1A1 for d2ydx2=sec2lnxx+2sec2lnxtanlnxx.

 

attempts to find the x-intercept        (M1)

THEN

x=0.629  =e-arctan12       A1

attempts to find f0.629 fe-arctan12        (M1)

the coordinates are 0.629,-0.943 e-arctan12,-32e-arctan12       A1

 

METHOD 2

attempts implicit differentiation on dydx to find d2ydx2        M1

d2ydx2=2y+3xxdydx-yx3 (or equivalent)

d2ydx2=0y=-3x2 (dydxyx)       A1

attempts to solve -3x2=xtanlnx-1 for x where e-π2<x<eπ2        M1

x=0.629  =e-arctan12       A1

attempts to find f0.629 f=e-arctan12        (M1)

the coordinates are 0.629,-0.943 e-arctan12,-32e-arctan12       A1

 

[6 marks]

c.

dydx=0y2+3xy+2x2=0        M1 

attempts to solve y2+3xy+2x2=0 for y        M1 

y+2xy+x=0 or y=-3x±3x2-42x22 =-3x±x2, x>0        A1

y=-2x and y=-x m=-2,-1        A1

 

Note: Award M1 for stating dydx=0M1 for substituting y=mx into dydx=0, A1 for m+2m+1=0 and A1 for m=-2,-1y=-2x and y=-x.

 

[4 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
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c.
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d.

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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