Date | May 2021 | Marks available | 6 | Reference code | 21M.2.AHL.TZ2.7 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
Consider the following system of coupled differential equations.
dxdt=-4x
dydt=3x-2y
Find the value of dydx
Find the eigenvalues and corresponding eigenvectors of the matrix (-403-2).
Hence, write down the general solution of the system.
Determine, with justification, whether the equilibrium point (0, 0) is stable or unstable.
(i) at (4, 0).
(ii) at (-4, 0).
Sketch a phase portrait for the general solution to the system of coupled differential equations for −6≤x≤6, −6≤y≤6.
Markscheme
|-4-λ03-2-λ|=0 (M1)
(-4-λ)(-2-λ)=0 (A1)
λ=-4 OR λ=-2 A1
λ=-4
(-403-2)(xy)=(-4x-4y) (M1)
Note: This M1 can be awarded for attempting to find either eigenvector.
3x-2y=-4y
3x=-2y
possible eigenvector is (-23) (or any real multiple) A1
λ=-2
(-403-2)(xy)=(-2x-2y)
x=0, y=1
possible eigenvector is (01) (or any real multiple) A1
[6 marks]
(xy)=Ae-4t(-23)+Be-2t(01) (M1)A1
Note: Award M1A1 for x=-2Ae-4t, y=3Ae-4t+Be-2t, M1A0 if LHS is missing or incorrect.
[2 marks]
two (distinct) real negative eigenvalues R1
(or equivalent (eg both e-4t→0, e-2t→0 as t→∞))
⇒ stable equilibrium point A1
Note: Do not award R0A1.
[2 marks]
dydx=3x-2y-4x (M1)
(i) (4, 0)⇒dydx=-34 A1
(ii) (-4, 0)⇒dydx=-34 A1
[3 marks]
A1A1A1A1
Note: Award A1 for a phase plane, with correct axes (condone omission of labels) and at least three non-overlapping trajectories. Award A1 for all trajectories leading to a stable node at (0, 0). Award A1 for showing gradient is negative at x=4 and -4. Award A1 for both eigenvectors on diagram.
[4 marks]