EXM.2.AHL.TZ0.1

pestleMathematicsAIHLPaper 2EXM· ahl-5-17-phase-portraitsource ↗

Consider the system of paired differential equations

$\dot x = 3x + 2y$

$\dot y = 2x + 3y$ .

This represents the populations of two species of symbiotic toadstools in a large wood.

Time $t$ is measured in decades.

Use the eigenvalue method to find the general solution to this system of equations.

[10]

Given the initial conditions that when $t = 0$ , $x = 150$ , $y = 50$ , find the particular solution.

[3]

b.i.

Hence find the solution when $t = 1$ .

[1]

b.ii.

As $t \to \infty$ , find an asymptote to the trajectory of the particular solution found in (b)(i) and state if this trajectory will be moving towards or away from the origin.

[4]

Markscheme / solution

The characteristic equation is given by

$\left| {\begin{array}{*{20}{c}} {3 - \lambda }&2 \\ 2&{3 - \lambda } \end{array}} \right| = 0 \Rightarrow {\lambda ^2} - 6\lambda + 5 = 0 \Rightarrow \lambda = 1{\text{ or 5}}$ M1A1A1A1

$\lambda = 1{\text{ }}\left( {\begin{array}{*{20}{c}} 2&2 \\ 2&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right){\text{ gives an eigenvector of form }}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ M1A1

$\lambda = 5{\text{ }}\left( {\begin{array}{*{20}{c}} { - 2}&2 \\ 2&{ - 2} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right){\text{ gives an eigenvector of form }}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ M1A1

General solution is $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = A{e^t}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) + B{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ A1A1

[10 marks]

Require $A + B = 150,{\text{ }} - A + B = 50 \Rightarrow A = 50,{\text{ B = 100}}$ M1A1

Particular solution is $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = 50{e^t}\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) + 100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ A1

[3 marks]

b.i.

$t = 1 \Rightarrow \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {15000} \\ {14700} \end{array}} \right){\text{ }}\left( {3sf} \right)$ A1

[1 mark]

b.ii.

The dominant term is $100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ so as $t \to \infty$ , $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) \simeq 100{e^{5t}}\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)$ M1A1

Giving the asymptote as $y = x$ A1

The trajectory is moving away from the origin. A1

[4 marks]

Examiners’ report

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b.i.

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b.ii.

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