Question
This question will investigate the solution to a coupled system of differential equations and how to transform it to a system that can be solved by the eigenvector method.
It is desired to solve the coupled system of differential equations
$$
\begin{aligned}
& \dot{x}=x+2 y-50 \\
& \dot{y}=2 x+y-40
\end{aligned}
$$
where $x$ and $y$ represent the population of two types of symbiotic coral and $t$ is time measured in decades.
a. Find the equilibrium point for this system.
b. 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 $t=1$.
c. Extend this method to conjecture the limit of the ratio $\frac{y}{x}$ as $t \rightarrow \infty$.
d. Show how using the substitution $X=x-10, Y=y-20$ transforms the system of differential equations into $\begin{gathered}\dot{X}=X+2 Y \\ \dot{Y}=2 X+Y\end{gathered}$.
e. Solve this system of equations by the eigenvalue method and hence find the general solution for $\left(\begin{array}{l}x \\ y\end{array}\right)$ of the original system.
f. Find the particular solution to the original system, given the initial conditions of part (b).
g. Hence find the exact values of $x$ and $y$ when $t=1$, giving the answers to 4 significant figures.
h. Use part (f) to find limit of the ratio $\frac{y}{x}$ as $t \rightarrow \infty$.
i. With the initial conditions as given in part (b) state if the equilibrium point is stable or unstable.
j. If instead the initial conditions were given as $x=20$ and $y=10$, find the particular solution for $\left(\begin{array}{l}x \\ y\end{array}\right)$ of the original system, in this case.
k. With the initial conditions as given in part (j), determine if the equilibrium point is stable or unstable.
▶️Answer/Explanation
a.
$$
\begin{aligned}
& \dot{x}=0 \Rightarrow x+2 y-50=0 \\
& \dot{y}=0 \Rightarrow 2 x+y-40=0 \\
&
\end{aligned}
$$
$\Rightarrow x=10, y=20$
M1A1
[2 marks]
b. $x_{n+1}=x_n+0.1\left(x_n+2 y_n-50\right)$
Using $_{y_{n+1}}=y_n+0.1\left(2 x_n+y_n-40\right)$
$$
t_{n+1}=t_n+0.1
$$
Gives $x(1) \simeq 848, y(1) \simeq 837(3 s f) \quad$ M1A1A1
[3 marks]
c. By extending the table, conjecture that $\lim _{t \rightarrow \infty} \frac{y}{x}=1 \quad$ M1A1
[2 marks]
d. $X=x-10, Y=y-20 \Rightarrow \dot{X}=\dot{x}, \dot{Y}=\dot{y}$
R1
$$
\begin{aligned}
& \dot{X}=(X+10)+2(Y+20)-50=X+2 Y \\
& \dot{Y}=2(X+10)+(Y+20)-40=2 X+Y
\end{aligned}
$$
M1A1AG
[3 marks]
e. $\left|\begin{array}{cc}1-\lambda & 2 \\ 2 & 1-\lambda\end{array}\right|=0 \Rightarrow(1-\lambda)^2-4=0 \Rightarrow \lambda=-1$ or $3 \quad$ M1A1A1
$\lambda=-1 \quad\left(\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right)\left(\begin{array}{l}p \\ q\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right) \Rightarrow q=-p$ an eigenvector is $\left(\begin{array}{c}1 \\ -1\end{array}\right)$
$\lambda=3 \quad\left(\begin{array}{cc}-2 & 2 \\ 2 & -2\end{array}\right)\left(\begin{array}{l}p \\ q\end{array}\right)=\left(\begin{array}{l}0 \\ 0\end{array}\right) \Rightarrow q=p$ an eigenvector is $\left(\begin{array}{l}1 \\ 1\end{array}\right) \quad$ M1A1A1
$\left(\begin{array}{l}X \\ Y\end{array}\right)=A e^{-t}\left(\begin{array}{c}1 \\ -1\end{array}\right)+B e^{3 t}\left(\begin{array}{l}1 \\ 1\end{array}\right) \Rightarrow\left(\begin{array}{l}x \\ y\end{array}\right)=A e^{-t}\left(\begin{array}{c}1 \\ -1\end{array}\right)+B e^{3 t}\left(\begin{array}{l}1 \\ 1\end{array}\right)+\left(\begin{array}{l}10 \\ 20\end{array}\right) \quad$ A1A1
[8 marks]
f.
$$
100=A+B+10
$$
$$
\begin{aligned}
& 50=-A+B+20 \\
& \left(\begin{array}{l}
x \\
y
\end{array}\right)=30 e^{-t}\left(\begin{array}{c}
1 \\
-1
\end{array}\right)+60 e^{3 t}\left(\begin{array}{l}
1 \\
1
\end{array}\right)+\left(\begin{array}{l}
10 \\
20
\end{array}\right) \quad \text { A1 }
\end{aligned}
$$
[2 marks]
g. $x(1)=1226, y(1)=1214(4 s f) \quad$ A1A1
[2 marks]
h. Dominant term is $60 e^{3 t}\left(\begin{array}{l}1 \\ 1\end{array}\right)$ so $\lim _{t \rightarrow \infty} \frac{y}{x}=1 \quad$ M1A1
[2 marks]
i. The equilibrium point is unstable.
R1
[1 mark]
j.
$$
\begin{gathered}
20=A+B+10 \\
10=-A+B+20
\end{gathered} \Rightarrow A=10, B=0 \quad \text { M1 }
$$
A1
[2 marks]
k. As $e^{-t} \rightarrow 0$ as $t \rightarrow \infty$ the equilibrium point is stable.
R1A1
[2 marks]
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
$$
\begin{aligned}
& \dot{x}=3 x+y \\
& \dot{y}=-x+y .
\end{aligned}
$$
The general solution to the coupled system of differential equations is hence given by
$$
\left(\begin{array}{l}
x \\
y
\end{array}\right)=A\left(\begin{array}{c}
1 \\
-1
\end{array}\right) e^{2 t}+B\left(\begin{array}{c}
t \\
-t+1
\end{array}\right) e^{2 t}
$$
As $t \rightarrow \infty$ the trajectory approaches an asymptote.
a. Show that the matrix $\left(\begin{array}{cc}3 & 1 \\ -1 & 1\end{array}\right)$ has (sadly) only one eigenvalue. Find this eigenvalue and an associated eigenvector.
b. Hence, verify that $\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t}$ is a solution to the above system.
C. Verify that $\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}t \\ -t+1\end{array}\right) e^{2 t}$ is also a solution.
d. If initially at $t=0, x=20, y=10$ find the particular solution.
e. Find the values of $x$ and $y$ when $t=2$.
f.i. Find the equation of this asymptote.
f.ii. State the direction of the trajectory, including the quadrant it is in as it approaches this asymptote.
▶️Answer/Explanation
a. $\left|\begin{array}{cc}3-\lambda & 1 \\ -1 & 1-\lambda\end{array}\right|=0 \Rightarrow(3-\lambda)(1-\lambda)+1=0 \quad$ M1A1
$\lambda^2-4 \lambda+4=0 \Rightarrow(\lambda-2)^2=0$
A1A1
So only one solution $\quad \lambda=2 \quad$ AGA1
$$
\left(\begin{array}{cc}
1 & 1 \\
-1 & -1
\end{array}\right)\left(\begin{array}{l}
p \\
q
\end{array}\right)=\left(\begin{array}{l}
0 \\
0
\end{array}\right) \Rightarrow p+q=0 \quad \text { M1 }
$$
So an eigenvector is $\left(\begin{array}{c}1 \\ -1\end{array}\right) \quad \boldsymbol{A 1}$
[7 marks]
b. $\left(\begin{array}{cc}3 & 1 \\ -1 & 1\end{array}\right)\left(\begin{array}{c}1 \\ -1\end{array}\right)=2\left(\begin{array}{c}1 \\ -1\end{array}\right)$
So $\left(\begin{array}{c}3 x+y \\ -x+y\end{array}\right)=\left(\begin{array}{cc}3 & 1 \\ -1 & 1\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{cc}3 & 1 \\ -1 & 1\end{array}\right)\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t}=2\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t} \quad$ M1A1A1
and $\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t} \Rightarrow\left(\begin{array}{c}\dot{x} \\ \dot{y}\end{array}\right)=\left(\begin{array}{c}1 \\ -1\end{array}\right) 2 e^{2 t} \quad$ M1A1
showing that $\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t}$ is a solution $\quad \boldsymbol{A G}$
[5 marks]
c.
$$
\begin{aligned}
& \left(\begin{array}{l}
3 x+y \\
-x+y
\end{array}\right)=\left(\begin{array}{c}
3 t-t+1 \\
-t-t+1
\end{array}\right) e^{2 t}=\left(\begin{array}{c}
2 t+1 \\
-2 t+1
\end{array}\right) e^{2 t} \quad \text { M1A1 } \\
& \left(\begin{array}{l}
\dot{x} \\
\dot{y}
\end{array}\right)=\left(\begin{array}{c}
e^{2 t}+t 2 e^{2 t} \\
-e^{2 t}+(-t+1) 2 e^{2 t}
\end{array}\right)=\left(\begin{array}{c}
2 t+1 \\
-2 t+1
\end{array}\right) e^{2 t} \quad \text { M1A1A1 }
\end{aligned}
$$
Verifying that $\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}t \\ -t+1\end{array}\right) e^{2 t}$ is also a solution
AG
[5 marks]
d.
Require $\left(\begin{array}{l}20 \\ 10\end{array}\right)=A\left(\begin{array}{c}1 \\ -1\end{array}\right)+B\left(\begin{array}{l}0 \\ 1\end{array}\right) \Rightarrow A=20, B=30 \quad$ M1A1
$\left(\begin{array}{l}x \\ y\end{array}\right)=20\left(\begin{array}{c}1 \\ -1\end{array}\right) e^{2 t}+30\left(\begin{array}{c}t \\ -t+1\end{array}\right) e^{2 t} \quad$ A1
[3 marks]
e. $t=2 \Rightarrow x=4370, y=-2730(3 s f) \quad$ A1A1
[2 marks]
f.i. As $t \rightarrow \infty, x \simeq 30 t e^{2 t}, y \simeq-30 t e^{2 t} \quad$ M1A1
so asymptote is $y=-x \quad \boldsymbol{A 1}$
[3 marks]
f.ii. Will approach the asymptote in the 4th quadrant, moving away from the origin.
R1
[1 mark]
Question
Consider the system of paired differential equations
$$
\begin{aligned}
& \dot{x}=a x+b y \\
& \dot{y}=c x+d y .
\end{aligned}
$$
This system is going to be solved by using the eigenvalue method.
If the system has a pair of purely imaginary eigenvalues
a. Show that if the system has two distinct real eigenvalues then $(a-d)^2+4 b c>0$.
b.i. Find two conditions that must be satisfied by $a, b, c, d$.
b.iiExplain why $b$ and $c$ must have opposite signs.
c. 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.
▶️Answer/Explanation
a. The characteristic equation is given by
$$
\begin{aligned}
& \left|\begin{array}{cc}
a-\lambda & b \\
c & d-\lambda
\end{array}\right|=0 \Rightarrow(a-\lambda)(d-\lambda)-b c=0 \Rightarrow \lambda^2-(a+d) \lambda+(a d-b c)=0 \quad \text { M1A1A1 } \\
& \lambda=\frac{a+d \pm \sqrt{(a+d)^2-4(a d-b c)}}{2}
\end{aligned}
$$
For two distinct real roots require $(a+d)^2-4(a d-b c)>0 \quad \boldsymbol{R 1}$
$$
\begin{aligned}
& \Rightarrow a^2+2 a d+d^2-4 a d+4 b c>0 \Rightarrow a^2-2 a d+d^2+4 b c>0 \\
& \Rightarrow(a-d)^2+4 b c>0 \quad \text { AG }
\end{aligned}
$$
A1A1
[6 marks]
b.i.Using the working from part (a) (or using the characteristic equation) for a pair of purely imaginary eigenvalues require $a+d=0$ and $(a-d)^2+4 b c<0 \quad$ R1A1A1
$\Rightarrow d=-a$ and $\Rightarrow a^2+b c<0$
A1A1
[5 marks]
b.ii $a^2+b c<0 \Rightarrow b c<0$ so $b$ and $c$ must have opposite signs M1AG
[1 mark]
C. When crossing the $x$ axes, $y=0$ so $\dot{y}=c x \quad$ M1A1
When crossing the positive $x$ axes, $\dot{y}$ has the sign of $c$.
A1
When crossing the negative $x$ axes, $\dot{y}$ has the sign of $-c$.
A1
Hence if $c$ is positive the trajectory is anticlockwise and if $c$ is negative the trajectory is clockwise.
$R 1 R 1$
When crossing the $y$ axes, $x=0$ so $\dot{x}=b y$
M1A1
When crossing the positive $y$ axis, $\dot{x}$ has the sign of $b$.
A1
When crossing the negative $y$ axes, $\dot{x}$ 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]
Question
Alessia is an ecologist working for Mediterranean fishing authorities. She is interested in whether the mackerel population density is likely to fall below 5000 mackerel per $\mathrm{km}^3$, as this is the minimum value required for sustainable fishing. She believes that the primary factor affecting the mackerel population is the interaction of mackerel with sharks, their main predator.
The population densities of mackerel $\left(M\right.$ thousands per $\left.\mathrm{km}^3\right)$ and sharks $\left(S\right.$ per $\left.\mathrm{km}^3\right)$ in the Mediterranean Sea are modelled by the coupled differential equations:
$$
\begin{aligned}
& \frac{\mathrm{d} M}{\mathrm{~d} t}=\alpha M-\beta M S \\
& \frac{\mathrm{d} S}{\mathrm{~d} t}=\gamma M S-\delta S
\end{aligned}
$$
where $t$ is measured in years, and $\alpha, \beta, \gamma$ and $\delta$ are parameters.
This model assumes that no other factors affect the mackerel or shark population densities.
The term $\alpha M$ models the population growth rate of the mackerel in the absence of sharks.
The term $\beta M S$ models the death rate of the mackerel due to being eaten by sharks.
Suggest similar interpretations for the following terms.
An equilibrium point is a set of values of $M$ and $S$, such that $\frac{\mathrm{d} M}{\mathrm{~d} t}=0$ and $\frac{\mathrm{d} S}{\mathrm{~d} t}=0$.
Given that both species are present at the equilibrium point,
The equilibrium point found in part (b) gives the average values of $M$ and $S$ over time.
Use the model to predict how the following events would affect the average value of $M$. Justify your answers.
To estimate the value of $\alpha$, Alessia considers a situation where there are no sharks and the initial mackerel population density is $M_0$.
Based on additional observations, it is believed that
$$
\begin{aligned}
& \alpha=0.549, \\
& \beta=0.236, \\
& \gamma=0.244, \\
& \delta=1.39 .
\end{aligned}
$$
Alessia decides to use Euler’s method to estimate future mackerel and shark population densities. The initial population densities are estimated to be $M_0=5.7$ and $S_0=2$. She uses a step length of 0.1 years.
Alessia will use her model to estimate whether the mackerel population density is likely to fall below the minimum value required for sustainable fishing, 5000 per $\mathrm{km}^3$, during the first nine years.
a.i. $\gamma M S$
a.ii. $\delta S$
b.i. show that, at the equilibrium point, the value of the mackerel population density is $\frac{\delta}{\gamma}$;
b.iifind the value of the shark population density at the equilibrium point.
c.i. Toxic sewage is added to the Mediterranean Sea. Alessia claims this reduces the shark population growth rate and hence the value of $\gamma$ is halved. No other parameter changes.
c.ii.Global warming increases the temperature of the Mediterranean Sea. Alessia claims that this promotes the mackerel population growth rate and hence the value of $\alpha$ is doubled. No other parameter changes.
d.i. Write down the differential equation for $M$ that models this situation.
d.ii.Show that the expression for the mackerel population density after $t$ years is $M=M_0 \mathrm{e}^{\alpha t}$
d.iiiAlessia estimates that the mackerel population density increases by a factor of three every two years. Show that $\alpha=0.549$ to three significant figures.
e.i. Write down expressions for $M_{n+1}$ and $S_{n+1}$ in terms of $M_n$ and $S_n$.
e.ii.Use Euler’s method to find an estimate for the mackerel population density after one year.
f.i. Use Euler’s method to sketch the trajectory of the phase portrait, for $4 \leq M \leq 7$ and $1.5 \leq S \leq 3$, over the first nine years.
f.ii. Using your phase portrait, or otherwise, determine whether the mackerel population density would be sufficient to support sustainable fishing during the first nine years.
f.iii.State two reasons why Alessia’s conclusion, found in part (f)(ii), might not be valid.
▶️Answer/Explanation
a.i. population growth rate / birth rate of sharks (due to eating mackerel)
A1
[1 mark]
a.ii.(net) death rate of sharks
A1
[1 mark]
b.i. $\gamma M S-\delta S=0$
A1
since $S \neq 0$
R1
Note: Accept $S>0$.
getting to given answer without further error by either cancelling or factorizing
A1
$M=\frac{\delta}{\gamma} \quad \boldsymbol{A G}$
[3 marks]
b.ii. $\frac{\mathrm{d} M}{\mathrm{~d} t}=0$
$$
\alpha M-\beta M S=0
$$
(M1)
(since $M \neq 0$ ) $S=\frac{\alpha}{\beta}$
A1
[2 marks]
c.i. $M_{e q}=\frac{\delta}{\gamma} \Rightarrow \frac{\delta}{\frac{1}{2} \gamma}=2 M_{e q}$
M1
Note: Accept equivalent in words.
Doubles
A1
Note: Do not accept “increases”.
[2 marks]
c.ii. $M_{e q}=\frac{\delta}{\gamma}$ is not dependent on $\alpha$
R1
Note: Award RO for any contextual argument.
no change
A1
Note: Do not award ROA1.
[2 marks]
d.i. $\frac{\mathrm{d} M}{\mathrm{~d} t}=\alpha M$
A1
[1 mark]
d.ii. $\int \frac{1}{M} \mathrm{~d} M=\int \alpha \mathrm{d} t$
M1
Note: Award $\boldsymbol{M 1}$ is for an attempt to separate variables. This means getting to the point $\int f(M) \mathrm{d} M=\int g(t) \mathrm{d} t$ where the integral can be seen or implied by further work.
$$
\ln |M|=\alpha t+c
$$
A1
Note: Accept $\ln M$. Condone missing constant of integration for this mark.
$$
M=k \mathrm{e}^{\alpha t}
$$
when $t=0, M_0=k$
M1
Note: Award $\boldsymbol{M 1}$ for a clear attempt at using initial conditions to find a constant of integration. Only possible if the constant of integration exists. $t=0$ or “initially” or similar must be seen. Substitution may appear earlier, following the integration.
initial conditions and all other manipulations correct and clearly communicated to get to the final answer
A1
$$
M=M_0 \mathrm{e}^{\alpha t}
$$
AG
[4 marks]
d. iii $M=3 M_0$ seen anywhere
(A1)
substituting $t=2, M=3 M_0$ into equation $M=M_0 \mathrm{e}^{\alpha t}$
(M1)
$$
\begin{aligned}
& 3 M_0=M_0 \mathrm{e}^{2 \alpha} \\
& \alpha=\frac{1}{2} \ln 3 \text { OR } 0.549306 \ldots
\end{aligned}
$$
A1
Note: The $\boldsymbol{A 1}$ requires either the exact answer or an answer to at least 4 sf.
$$
\approx 0.549
$$
$A G$
[3 marks]
e.i. an attempt to set up one recursive equation
(M1)
Note: Must include two given parameters and $M_n$ and $S_n$ and $M_{n+1}$ or $S_{n+1}$ for the (M1) to be awarded.
$$
\begin{aligned}
& M_{n+1}=M_n+0.1\left(0.549 M_n-0.236 M_n S_n\right) \\
& S_{n+1}=S_n+0.1\left(0.244 M_n S_n-1.39 S_n\right)
\end{aligned}
$$
A1
[3 marks]
e.ii.EITHER
6. 12 (6.11609…)
A2
OR
6120 (6116.09…) (mackerel per $\mathrm{km}^3$ )
A2
[2 marks]
f.i.
spiral or closed loop shape
A1
approximately 1.25 rotations (can only be awarded if a spiral)
A1
correct shape, in approximately correct position (centred at approx. (5. 5, 2. 5))
A1
Note: Award AOAOAO for any plot of $S$ or $M$ against $t$.
[3 marks]
f.ii. EITHER
approximate minimum is (5.07223…) 5.07 (which is greater than 5)
A1
OR
the line $M=5$ clearly labelled on their phase portrait
A1
THEN
(the density will not fall below 5000) hence sufficient for sustainable fishing
A1
Note: Do not award AOA1. Only if the minimum point is labelled on the sketch then a statement here that “the mackerel population is always above $5000^{\prime \prime}$ would be sufficient. Accept the value 5.07 seen within a table of values.
[2 marks]
f.iii.Any two from:
A1A1
– Current values / parameters are only an estimate,
– The Euler method is only an approximate method / choosing $h=0.1$ might be too large.
– There might be random variation / the model has no stochastic component
– Conditions / parameters might change over the nine years,
– A discrete system is being approximated by a continuous system,
Allow any other sensible critique.
If a candidate identifies factors which the model ignores, award $\mathbf{A 1}$ per factor identified. These factors could include:
– Other predators
– Seasonality
– Temperature
– The effect of fishing
– Environmental catastrophe
– Migration
Note: Do not allow:
“You cannot have 5.07 mackerel”.
It is only a model (as this is too vague).
Some factors have been ignored (without specifically identifying the factors).
Values do not always follow the equation / model. (as this is too vague).
[2 marks]