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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+2y50

˙y=2x+y40

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 yx as t.

 

[2]
c.

Show how using the substitution X=x10,Y=y20 transforms the system of differential equations into ˙X=X+2Y˙Y=2X+Y.

[3]
d.

Solve this system of equations by the eigenvalue method and hence find the general solution for (xy) 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 yx 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 (xy) 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=0x+2y50=0˙y=02x+y40=0     x=10,y=20      M1A1

[2 marks]

a.

Using xn+1=xn+0.1(xn+2yn50)yn+1=yn+0.1(2xn+yn40)tn+1=tn+0.1

Gives x(1)848,y(1)837(3sf)      M1A1A1

[3 marks]

b.

By extending the table, conjecture that limtyx=1      M1A1

[2 marks]

c.

X=x10,Y=y20˙X=˙x,˙Y=˙y      R1

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

[3 marks]

d.

|1λ221λ|=0(1λ)24=0λ=1or3     M1A1A1

λ=1      (2222)(pq)=(00)q=p   an eigenvector is (11)

λ=3        (2222)(pq)=(00)q=p   an eigenvector is (11)     M1A1A1

(XY)=Aet(11)+Be3t(11)(xy)=Aet(11)+Be3t(11)+(1020)      A1A1

 

[8 marks]

e.

100=A+B+1050=A+B+20A=30,B=60     M1

(xy)=30et(11)+60e3t(11)+(1020)      A1

[2 marks]

f.

x(1)=1226,y(1)=1214(4sf)     A1A1

[2 marks]

g.

Dominant term is 60e3t(11) so limtyx=1    M1A1

[2 marks]

h.

The equilibrium point is unstable.               R1

[1 mark]

i.

20=A+B+1010=A+B+20A=10,B=0            M1

(xy)=10et(11)+(1020)             A1

[2 marks]

j.

As et0 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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