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+2y−50
˙y=2x+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.
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.
Extend this method to conjecture the limit of the ratio yx as t→∞.
Show how using the substitution X=x−10,Y=y−20 transforms the system of differential equations into ˙X=X+2Y˙Y=2X+Y.
Solve this system of equations by the eigenvalue method and hence find the general solution for (xy) of the original system.
Find the particular solution to the original system, given the initial conditions of part (b).
Hence find the exact values of x and y when t=1, giving the answers to 4 significant figures.
Use part (f) to find limit of the ratio yx as t→∞.
With the initial conditions as given in part (b) state if the equilibrium point is stable or unstable.
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.
With the initial conditions as given in part (j), determine if the equilibrium point is stable or unstable.
Markscheme
˙x=0⇒x+2y−50=0˙y=0⇒2x+y−40=0 ⇒x=10,y=20 M1A1
[2 marks]
Using xn+1=xn+0.1(xn+2yn−50)yn+1=yn+0.1(2xn+yn−40)tn+1=tn+0.1
Gives x(1)≃848,y(1)≃837(3sf) M1A1A1
[3 marks]
By extending the table, conjecture that limt→∞yx=1 M1A1
[2 marks]
X=x−10,Y=y−20⇒˙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]
|1−λ221−λ|=0⇒(1−λ)2−4=0⇒λ=−1or3 M1A1A1
λ=−1 (2222)(pq)=(00)⇒q=−p an eigenvector is (1−1)
λ=3 (−222−2)(pq)=(00)⇒q=p an eigenvector is (11) M1A1A1
(XY)=Ae−t(1−1)+Be3t(11)⇒(xy)=Ae−t(1−1)+Be3t(11)+(1020) A1A1
[8 marks]
100=A+B+1050=−A+B+20⇒A=30,B=60 M1
(xy)=30e−t(1−1)+60e3t(11)+(1020) A1
[2 marks]
x(1)=1226,y(1)=1214(4sf) A1A1
[2 marks]
Dominant term is 60e3t(11) so limt→∞yx=1 M1A1
[2 marks]
The equilibrium point is unstable. R1
[1 mark]
20=A+B+1010=−A+B+20⇒A=10,B=0 M1
(xy)=10e−t(1−1)+(1020) A1
[2 marks]
As e−t→0 as t→∞ the equilibrium point is stable. R1A1
[2 marks]