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 $\frac{y}{x}$ as $t\to \mathrm{\infty }$.

Show how using the substitution $X=x-10\text{,}Y=y-20$ transforms the system of differential equations into $\begin{array}{c}\dot{X}=X+2Y\\ \dot{Y}=2X+Y\end{array}$.

Solve this system of equations by the eigenvalue method and hence find the general solution for $\left(\begin{array}{c}x\\ y\end{array}\right)$ 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 $\frac{y}{x}$ as $t\to \mathrm{\infty }$.

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 $\left(\begin{array}{c}x\\ y\end{array}\right)$ 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.

$\begin{array}{c}\dot{x}=0\Rightarrow x+2y-50=0\\ \dot{y}=0\Rightarrow 2x+y-40=0\end{array}$     $\Rightarrow x=10\text{,}y=20$      M1A1

[2 marks]

Using $\begin{array}{c}{x}_{n+1}=x_n+0.1\left(x_n+2y_n-50\right)\\ y_{n+1}=y_n+0.1\left(2x_n+y_n-40\right)\\ t_{n+1}=t_n+0.1\end{array}$

Gives $x\left(1\right)\simeq 848\text{,}y\left(1\right)\simeq 837\left(3sf\right)$      M1A1A1

[3 marks]

By extending the table, conjecture that $\underset{t\to \mathrm{\infty }}{\text{lim}}\frac{y}{x}=1$      M1A1

[2 marks]

$X=x-10\text{,}Y=y-20\Rightarrow \dot{X}=\dot{x}\text{,}\dot{Y}=\dot{y}$      R1

$\begin{array}{c}\dot{X}=\left(X+10\right)+2\left(Y+20\right)-50=X+2Y\\ \dot{Y}=2\left(X+10\right)+\left(Y+20\right)-40=2X+Y\end{array}$     M1A1AG

[3 marks]

     M1A1A1

$\lambda =-1$         an eigenvector is $\left(\begin{array}{c}1\\ -1\end{array}\right)$

$\lambda =3$           an eigenvector is $\left(\begin{array}{c}1\\ 1\end{array}\right)$     M1A1A1

$\left(\begin{array}{c}X\\ Y\end{array}\right)=A{e}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{e}^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)\Rightarrow \left(\begin{array}{c}x\\ y\end{array}\right)=A{e}^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+B{e}^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$      A1A1

[8 marks]

$\begin{array}{c}100=A+B+10\\ 50=-A+B+20\end{array}\Rightarrow A=30\text{,}B=60$     M1

$\left(\begin{array}{c}x\\ y\end{array}\right)=30e^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+60e^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$      A1

[2 marks]

$x\left(1\right)=1226\text{,}y\left(1\right)=1214\left(4sf\right)$     A1A1

[2 marks]

Dominant term is $60e^{3t}\left(\begin{array}{c}1\\ 1\end{array}\right)$ so $\underset{t\to \mathrm{\infty }}{\text{lim}}\frac{y}{x}=1$    M1A1

[2 marks]

The equilibrium point is unstable.               R1

[1 mark]

$\begin{array}{c}20=A+B+10\\ 10=-A+B+20\end{array}\Rightarrow A=10\text{,}B=0$            M1

$\left(\begin{array}{c}x\\ y\end{array}\right)=10e^{-t}\left(\begin{array}{c}1\\ -1\end{array}\right)+\left(\begin{array}{c}10\\ 20\end{array}\right)$             A1

[2 marks]

As $e^{-t}\to 0$ as $t\to \mathrm{\infty }$ the equilibrium point is stable.           R1A1

[2 marks]