AI HL · AHL 9
0 of 66 skills done
Euler's method for a coupled system: graphs and phase portraits
AHL 9.12
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
Choose a system and a step length h. Each Next takes one Euler step for x and y together. The left picture shows x and y against t, and the right picture shows the path of (x, y). Tick the box to compare with the true solution.
x (ink) and y (teal) against t
The path in the x-y plane
| t | x | y |
|---|
Press Next to take the first step.
The differential equations \(\dfrac{dx}{dt}=2x - 3y,\ \dfrac{dy}{dt}=3x - y\) have \(x=1\) and \(y=0\) when \(t=0\). Use Euler's method with a step length of 0.1 to sketch a graph of x against t for \(0\le t\le5\). Label the first t-intercept and the first minimum point.
Have a go first. Then open the working one line at a time.
\(x_{n+1}=x_n+0.1(2x_n - 3y_n)\), \(y_{n+1}=y_n+0.1(3x_n - y_n)\), \(t_n=0.1n\)
Always write these down. Each new x and y uses the old x and the old y.
n = 0: \(t=0,\ x=1,\ y=0\); n = 1: \(t=0.1,\ x=1.2,\ y=0.3\); n = 2: \(t=0.2,\ x=1.35,\ y=0.63\)
By hand: \(x_1=1+0.1(2(1)-3(0))=1.2\), \(y_1=0+0.1(3(1)-0)=0.3\). Then \(x_2=1.35\).
From the calculator: a spreadsheet with columns t, x and y, filled down to \(t=5\)
Type the starting values in row 1 and the recurrences in row 2, then fill down.
From the calculator: scatter plot of x against t. It crosses the t-axis at about \(t=0.86\) and has a minimum at about \((1.6,\ -3.96)\)
Read the intercept from the table, where x changes sign, and the minimum from the lowest value.
An oscillation with growing amplitude
x swings above and below zero and each swing is bigger.
For the same system \(\dfrac{dx}{dt}=2x - 3y,\ \dfrac{dy}{dt}=3x - y\) with \(x=1,\ y=0\) when \(t=0\), sketch the phase portrait of y against x for \(0\le t\le4\) using Euler's method with step length 0.1.
Have a go first. Then open the working one line at a time.
\(x_{n+1}=x_n+0.1(2x_n - 3y_n)\), \(y_{n+1}=y_n+0.1(3x_n - y_n)\)
The columns \(x_n\) and \(y_n\) are the same as before. Only the plot changes.
From the calculator: scatter plot with \(x_n\) on the horizontal axis and \(y_n\) on the vertical axis
Time does not appear on the axes. Each point is one row of the table.
From the calculator: a spiral out from \((1,0)\). It crosses the y-axis at about \(y=2\) and the negative x-axis at about \(x=-2.8\)
The curve goes round the origin and gets further from it each time.
y increases at the start, so the spiral is anticlockwise
Add arrowheads. Euler's method gives the direction as well.
For the system \(\dfrac{dx}{dt}=-x - 4y,\ \dfrac{dy}{dt}=2x\) with \(x=4,\ y=1\) when \(t=0\), use Euler's method with step length 0.1 to sketch x against t and y against x for \(0\le t\le10\). Describe what you see.
Have a go first. Then open the working one line at a time.
\(x_{n+1}=x_n+0.1(-x_n - 4y_n)\), \(y_{n+1}=y_n+0.1(2x_n)\)
Same method: spreadsheet, fill down to \(t=10\).
From the calculator: x oscillates about 0 and the swings get smaller. First t-intercept about \(t=0.38\), first minimum about \((0.9,\ -4.13)\)
An oscillation with decreasing amplitude.
From the calculator: a spiral going in towards \((0,0)\), anticlockwise
Each loop is smaller than the one before.
\(\lambda=-0.5\pm2.78i\)
The real part is negative, so the origin is a stable focus. The output agrees.