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Date November 2018 Marks available 4 Reference code 18N.3.AHL.TZ0.Hca_4
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number Hca_4 Adapted from N/A

Question

Consider the differential equation  d y d x = 1 + y x , where x 0 .

Consider the family of curves which satisfy the differential equation  d y d x = 1 + y x , where x 0 .

Given that  y ( 1 ) = 1 , use Euler’s method with step length  h = 0.25 to find an approximation for y ( 2 ) . Give your answer to two significant figures.

[4]
a.

Solve the equation  d y d x = 1 + y x for y ( 1 ) = 1 .

[6]
b.

Find the percentage error when  y ( 2 ) is approximated by the final rounded value found in part (a). Give your answer to two significant figures.

[3]
c.

Find the equation of the isocline corresponding to  d y d x = k , where  k 0 k R .

[1]
d.i.

Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

[4]
d.ii.

Markscheme

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

attempt to apply Euler’s method         (M1)

x n + 1 = x n + 0.25 ; y n + 1 = y n + 0.25 × ( 1 + y n x n )

     (A1)(A1)

Note: Award A1 for correct x values, A1 for first three correct y values.

 

y = 3.3      A1

 

[4 marks]

a.

METHOD 1

I ( x ) = e 1 x d x        (M1)

= e ln x

= 1 x        (A1)

1 x d y d x y x 2 = 1 x        (M1)

d d x ( y x ) = 1 x

y x = ln | x | + C        A1

y ( 1 ) = 1 C = 1        M1

y = x ln | x | + x        A1

 

METHOD 2

v = y x        M1

d v d x = 1 x d y d x 1 x 2 y        (A1)

v + x d v d x = 1 + v        M1

1 d v = 1 x d x

v = ln | x | + C

y x = ln | x | + C        A1

y ( 1 ) = 1 C = 1        M1

y = x ln | x | + x        A1

 

[6 marks]

b.

y ( 2 ) = 2 ln 2 + 2 = 3.38629

percentage error  = 3.38629 3.3 3.38629 × 100       (M1)(A1)

= 2.5%       A1

 

[3 marks]

c.

d y d x = k 1 + y x = k       A1

y = ( k 1 ) x

 

[1 mark]

d.i.

gradient of isocline equals gradient of normal        (M1)

k 1 = 1 k or  k ( k 1 ) = 1       A1

k 2 k + 1 = 0        A1

Δ = 1 4 < 0        R1

no solution       AG

Note: Accept alternative reasons for no solutions.

 

[4 marks]

d.ii.

Examiners report

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

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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