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Date	May Example question	Marks available	1	Reference code	EXM.3.AHL.TZ0.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	State	Question number	2	Adapted from	N/A

Question

This question will investigate the solution to a coupled system of differential equations when there is only one eigenvalue.

It is desired to solve the coupled system of differential equations

$\dot x = 3x + y$

$\dot y = - x + y.$

The general solution to the coupled system of differential equations is hence given by

$\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = A\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}} + B\left( {\begin{array}{*{20}{c}} t \\ { - t + 1} \end{array}} \right){e^{2t}}$

As $t \to \infty$ the trajectory approaches an asymptote.

Show that the matrix $\left( {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&1 \end{array}} \right)$ has (sadly) only one eigenvalue. Find this eigenvalue and an associated eigenvector.

[7]

Hence, verify that $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}}$ is a solution to the above system.

[5]

Verify that $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} t \\ { - t + 1} \end{array}} \right){e^{2t}}$ is also a solution.

[5]

If initially at $t = 0{\text{,}}\,\,x = 20{\text{,}}\,\,y = 10$ find the particular solution.

[3]

Find the values of $x$ and $y$ when $t = 2$ .

[2]

Find the equation of this asymptote.

[3]

f.i.

State the direction of the trajectory, including the quadrant it is in as it approaches this asymptote.

[1]

f.ii.

Markscheme

$\left| {\begin{array}{*{20}{c}} {3 - \lambda }&1 \\ { - 1}&{1 - \lambda } \end{array}} \right| = 0 \Rightarrow \left( {3 - \lambda } \right)\left( {1 - \lambda } \right) + 1 = 0$ M1A1

${\lambda ^2} - 4\lambda + 4 = 0 \Rightarrow {\left( {\lambda - 2} \right)^2} = 0$ A1A1

So only one solution $\lambda = 2$ AGA1

$\left( {\begin{array}{*{20}{c}} 1&1 \\ { - 1}&{ - 1} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right) \Rightarrow p + q = 0$ M1

So an eigenvector is $\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$ A1

[7 marks]

$\left( {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) = 2\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)$

So $\left( {\begin{array}{*{20}{c}} {3x + y} \\ { - x + y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}} = 2\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}}$ M1A1A1

and $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}} \Rightarrow \left( {\begin{array}{*{20}{c}} {\dot x} \\ {\dot y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right)2{e^{2t}}$ M1A1

showing that $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}}$ is a solution AG

[5 marks]

$\left( {\begin{array}{*{20}{c}} {3x + y} \\ { - x + y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {3t - t + 1} \\ { - t - t + 1} \end{array}} \right){e^{2t}} = \left( {\begin{array}{*{20}{c}} {2t + 1} \\ { - 2t + 1} \end{array}} \right){e^{2t}}$ M1A1

$\left( {\begin{array}{*{20}{c}} {\dot x} \\ {\dot y} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{e^{2t}} + t2{e^{2t}}} \\ { - {e^{2t}} + ( - t + 1)2{e^{2t}}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {2t + 1} \\ { - 2t + 1} \end{array}} \right){e^{2t}}$ M1A1A1

Verifying that $\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} t \\ { - t + 1} \end{array}} \right){e^{2t}}$ is also a solution AG

[5 marks]

Require $\left( {\begin{array}{*{20}{c}} {20} \\ {10} \end{array}} \right) = A\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right) + B\left( {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right) \Rightarrow A = 20{\text{,}}\,\,B = 30$ M1A1

$\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = 20\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \end{array}} \right){e^{2t}} + 30\left( {\begin{array}{*{20}{c}} t \\ { - t + 1} \end{array}} \right){e^{2t}}$ A1