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

Question

Consider the system of paired differential equations

x ˙ = a x + b y

y ˙ = c x + d y .

This system is going to be solved by using the eigenvalue method.

 

If the system has a pair of purely imaginary eigenvalues

Show that if the system has two distinct real eigenvalues then  ( a d ) 2 + 4 b c > 0 .

[6]
a.

Find two conditions that must be satisfied by a , b , c , d .

[5]
b.i.

Explain why b  and c  must have opposite signs.

[1]
b.ii.

In the case when there is a pair of purely imaginary eigenvalues you are told that the solution will form an ellipse.  You are also told that the initial conditions are such that the ellipse is large enough that it will cross both the positive and negative x  axes and the positive and negative y  axes.

By considering the differential equations at these four crossing point investigate if the trajectory is in a clockwise or anticlockwise direction round the ellipse. Give your decision in terms of b  and c . Using part (b) (ii) show that your conclusions are consistent.

[1]
c.

Markscheme

The characteristic equation is given by

| a λ b c d λ | = 0 ( a λ ) ( d λ ) b c = 0 λ 2 ( a + d ) λ + ( a d b c ) = 0       M1A1A1

λ = a + d ± ( a + d ) 2 4 ( a d b c ) 2

For two distinct real roots require  ( a + d ) 2 4 ( a d b c ) > 0        R1

a 2 + 2 a d + d 2 4 a d + 4 b c > 0 a 2 2 a d + d 2 + 4 b c > 0       A1A1

( a d ) 2 + 4 b c > 0         AG

[6 marks]

a.

Using the working from part (a) (or using the characteristic equation) for a pair of purely imaginary eigenvalues require

a + d = 0 and  ( a d ) 2 + 4 b c < 0     R1A1A1

d = a and  a 2 + b c < 0       A1A1

[5 marks]

b.i.

a 2 + b c < 0 b c < 0 so  b  and  c  must have opposite signs      M1AG

[1 mark]

b.ii.

When crossing the  x  axes,  y = 0 so  y ˙ = c x       M1A1

When crossing the positive  x  axes,  y ˙ has the sign of  c .       A1

When crossing the negative  x  axes,  y ˙ has the sign of  c .       A1

Hence if c  is positive the trajectory is anticlockwise and if c  is negative the trajectory is clockwise.       R1R1

When crossing the  y  axes,  x = 0 so  x ˙ = b y       M1A1

When crossing the positive y  axis, x ˙  has the sign of b .       A1

When crossing the negative  y  axes,  x ˙ has the sign of  b .       A1

Hence if b  is positive the trajectory is clockwise and if b  is negative the trajectory is anticlockwise.       R1R1

Since by (b)(ii), b  and c  have opposite signs the above conditions agree with each other.       R1

[13 marks]

c.

Examiners report

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

Syllabus sections

Topic 5—Calculus » AHL 5.17—Phase portrait
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