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=ax+by
˙y=cx+dy.
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+4bc>0.
Find two conditions that must be satisfied by a, b, c, d.
Explain why b and c must have opposite signs.
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.
Markscheme
The characteristic equation is given by
|a−λbcd−λ|=0⇒(a−λ)(d−λ)−bc=0⇒λ2−(a+d)λ+(ad−bc)=0 M1A1A1
λ=a+d±√(a+d)2−4(ad−bc)2
For two distinct real roots require (a+d)2−4(ad−bc)>0 R1
⇒a2+2ad+d2−4ad+4bc>0⇒a2−2ad+d2+4bc>0 A1A1
⇒(a−d)2+4bc>0 AG
[6 marks]
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+4bc<0 R1A1A1
⇒d=−a and ⇒a2+bc<0 A1A1
[5 marks]
a2+bc<0⇒bc<0 so b and c must have opposite signs M1AG
[1 mark]
When crossing the x axes, y=0 so ˙y=cx 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=by 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]