IBDP MAI : Topic 5 Calculus - AHL 5.17 coupled differential equations AI HL Paper 3
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]