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Date May Example question Marks available 2 Reference code EXM.3.AHL.TZ0.3
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 3 Adapted from N/A

Question

This question will investigate the solution to a coupled system of differential equations and how to transform it to a system that can be solved by the eigenvector method.

It is desired to solve the coupled system of differential equations

x ˙ = x + 2 y 50

y ˙ = 2 x + y 40

where x  and y  represent the population of two types of symbiotic coral and t  is time measured in decades.

Find the equilibrium point for this system.

[2]
a.

If initially x = 100 and y = 50  use Euler’s method with an time increment of 0.1 to find an approximation for the values of  x and  y when  t = 1 .

[3]
b.

Extend this method to conjecture the limit of the ratio  y x as  t .

 

[2]
c.

Show how using the substitution X = x 10 , Y = y 20 transforms the system of differential equations into X ˙ = X + 2 Y Y ˙ = 2 X + Y .

[3]
d.

Solve this system of equations by the eigenvalue method and hence find the general solution for  ( x y )  of the original system.

[8]
e.

Find the particular solution to the original system, given the initial conditions of part (b).

[2]
f.

Hence find the exact values of x and y  when t = 1 , giving the answers to 4 significant figures.

[2]
g.

Use part (f) to find limit of the ratio  y x as  t .

[2]
h.

With the initial conditions as given in part (b) state if the equilibrium point is stable or unstable.

[1]
i.

If instead the initial conditions were given as x = 20 and  y = 10 , find the particular solution for  ( x y )  of the original system, in this case.

[2]
j.

With the initial conditions as given in part (j), determine if the equilibrium point is stable or unstable.

[2]
k.

Markscheme

x ˙ = 0 x + 2 y 50 = 0 y ˙ = 0 2 x + y 40 = 0       x = 10 , y = 20       M1A1

[2 marks]

a.

Using x n + 1 = x n + 0.1 ( x n + 2 y n 50 ) y n + 1 = y n + 0.1 ( 2 x n + y n 40 ) t n + 1 = t n + 0.1

Gives  x ( 1 ) 848 , y ( 1 ) 837 ( 3 s f )       M1A1A1

[3 marks]

b.

By extending the table, conjecture that  lim t y x = 1       M1A1

[2 marks]

c.

X = x 10 , Y = y 20 X ˙ = x ˙ , Y ˙ = y ˙       R1

X ˙ = ( X + 10 ) + 2 ( Y + 20 ) 50 = X + 2 Y Y ˙ = 2 ( X + 10 ) + ( Y + 20 ) 40 = 2 X + Y      M1A1AG

[3 marks]

d.

| 1 λ 2 2 1 λ | = 0 ( 1 λ ) 2 4 = 0 λ = 1 or 3      M1A1A1

λ = 1        ( 2 2 2 2 ) ( p q ) = ( 0 0 ) q = p    an eigenvector is  ( 1 1 )

λ = 3         ( 2 2 2 2 ) ( p q ) = ( 0 0 ) q = p    an eigenvector is  ( 1 1 )      M1A1A1

( X Y ) = A e t ( 1 1 ) + B e 3 t ( 1 1 ) ( x y ) = A e t ( 1 1 ) + B e 3 t ( 1 1 ) + ( 10 20 )       A1A1

 

[8 marks]

e.

100 = A + B + 10 50 = A + B + 20 A = 30 , B = 60      M1

( x y ) = 30 e t ( 1 1 ) + 60 e 3 t ( 1 1 ) + ( 10 20 )       A1

[2 marks]

f.

x ( 1 ) = 1226 , y ( 1 ) = 1214 ( 4 s f )      A1A1

[2 marks]

g.

Dominant term is 60 e 3 t ( 1 1 ) so  lim t y x = 1     M1A1

[2 marks]

h.

The equilibrium point is unstable.               R1

[1 mark]

i.

20 = A + B + 10 10 = A + B + 20 A = 10 , B = 0             M1

( x y ) = 10 e t ( 1 1 ) + ( 10 20 )              A1

[2 marks]

j.

As  e t 0 as  t  the equilibrium point is stable.           R1A1

[2 marks]

k.

Examiners report

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Syllabus sections

Topic 5—Calculus » AHL 5.17—Phase portrait
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