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Date May Example question Marks available 1 Reference code EXM.3.AHL.TZ0.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term State Question number 2 Adapted from N/A

Question

This question will investigate the solution to a coupled system of differential equations when there is only one eigenvalue.

It is desired to solve the coupled system of differential equations

x ˙ = 3 x + y

y ˙ = x + y .

The general solution to the coupled system of differential equations is hence given by

( x y ) = A ( 1 1 ) e 2 t + B ( t t + 1 ) e 2 t

As  t the trajectory approaches an asymptote.

Show that the matrix  ( 3 1 1 1 )  has (sadly) only one eigenvalue.  Find this eigenvalue and an associated eigenvector.

[7]
a.

Hence, verify that  ( x y ) = ( 1 1 ) e 2 t  is a solution to the above system.

[5]
b.

Verify that  ( x y ) = ( t t + 1 ) e 2 t is also a solution.

[5]
c.

If initially at  t = 0 , x = 20 , y = 10  find the particular solution.

[3]
d.

Find the values of x and y  when  t = 2 .

[2]
e.

 Find the equation of this asymptote.

[3]
f.i.

State the direction of the trajectory, including the quadrant it is in as it approaches this asymptote.

[1]
f.ii.

Markscheme

| 3 λ 1 1 1 λ | = 0 ( 3 λ ) ( 1 λ ) + 1 = 0       M1A1

λ 2 4 λ + 4 = 0 ( λ 2 ) 2 = 0       A1A1

So only one solution    λ = 2       AGA1

( 1 1 1 1 ) ( p q ) = ( 0 0 ) p + q = 0       M1

So an eigenvector is  ( 1 1 )       A1

[7 marks]

a.

( 3 1 1 1 ) ( 1 1 ) = 2 ( 1 1 )

So  ( 3 x + y x + y ) = ( 3 1 1 1 ) ( x y ) = ( 3 1 1 1 ) ( 1 1 ) e 2 t = 2 ( 1 1 ) e 2 t             M1A1A1

and  ( x y ) = ( 1 1 ) e 2 t ( x ˙ y ˙ ) = ( 1 1 ) 2 e 2 t       M1A1

showing that ( x y ) = ( 1 1 ) e 2 t  is a solution      AG

[5 marks]

b.

( 3 x + y x + y ) = ( 3 t t + 1 t t + 1 ) e 2 t = ( 2 t + 1 2 t + 1 ) e 2 t           M1A1

( x ˙ y ˙ ) = ( e 2 t + t 2 e 2 t e 2 t + ( t + 1 ) 2 e 2 t ) = ( 2 t + 1 2 t + 1 ) e 2 t       M1A1A1

Verifying that ( x y ) = ( t t + 1 ) e 2 t  is also a solution      AG

[5 marks]

c.

Require  ( 20 10 ) = A ( 1 1 ) + B ( 0 1 ) A = 20 , B = 30            M1A1

( x y ) = 20 ( 1 1 ) e 2 t + 30 ( t t + 1 ) e 2 t     A1

[3 marks]

d.

t = 2 x = 4370 , y = 2730 ( 3 s f )       A1A1

[2 marks]

e.

As  t , x 30 t e 2 t , y 30 t e 2 t       M1A1

so asymptote is y = x       A1

[3 marks]

f.i.

Will approach the asymptote in the 4th quadrant, moving away from the origin.      R1

[1 mark]

f.ii.

Examiners report

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Syllabus sections

Topic 5—Calculus » AHL 5.17—Phase portrait
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