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Date November 2019 Marks available 10 Reference code 19N.3.AHL.TZ0.Hca_4
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Solve Question number Hca_4 Adapted from N/A

Question

Consider the differential equation  d y d x = 4 x 2 + y 2 x y x 2 , with y = 2 when  x = 1 .

Use Euler’s method, with step length h = 0.1 , to find an approximate value of y when x = 1.4 .

[5]
a.

Sketch the isoclines for  d y d x = 4 .

[3]
b.

Express  m 2 2 m + 4  in the form  ( m a ) 2 + b , where  a b Z .

[1]
c.i.

Solve the differential equation, for x > 0 , giving your answer in the form y = f ( x ) .

[10]
c.ii.

Sketch the graph of  y = f ( x ) for  1 x 1.4  .

[1]
c.iii.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture f ( 1.4 ) will be less than, equal to, or greater than your answer in part (a).

[2]
c.iv.

Markscheme

       (M1)(A1)(A1)(A1)A1

y ( 1.4 ) 5.34

Note: Award A1 for each correct y value.
For the intermediate y values, accept answers that are accurate to 2 significant figures.
The final y value must be accurate to 3 significant figures or better.

[5 marks]

a.

attempt to solve  4 x 2 + y 2 x y x 2 = 4         (M1)

y 2 x y = 0

y ( y x ) = 0

y = 0   or  y = x

        A1A1

[3 marks]

b.

m 2 2 m + 4 = ( m 1 ) 2 + 3 ( a = 1 , b = 3 )         A1

[1 mark]

c.i.

recognition of homogeneous equation,
let  y = v x              M1

the equation can be written as

v + x d v d x = 4 + v 2 v          (A1)

x d v d x = v 2 2 v + 4

1 v 2 2 v + 4 d v = 1 x d x              M1

Note: Award M1 for attempt to separate the variables.

1(v1)2+3dv=1xdx from part (c)(i)             M1

1 3 arctan ( v 1 3 ) = ln x ( + c )           A1A1

x = 1 , y = 2 v = 2

1 3 arctan ( 1 3 ) = ln 1 + c              M1

Note: Award M1 for using initial conditions to find c .

c = π 6 3 ( = 0.302 )         A1

arctan ( v 1 3 ) = 3 ln x + π 6

substituting  v = y x              M1

Note: This M1 may be awarded earlier.

y = x ( 3 tan ( 3 ln x + π 6 ) + 1 )         A1

[10 marks]

c.ii.

curve drawn over correct domain       A1

 

[1 mark]

c.iii.

the sketch shows that f is concave up       A1

Note: Accept f is increasing.

this means the tangent drawn using Euler’s method will give an underestimate of the real value, so f ( 1.4 ) > estimate in part (a)       R1

Note: The R1 is dependent on the A1.

[2 marks]

c.iv.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
c.iii.
[N/A]
c.iv.

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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