Euler's method; coupled systems practice questions.

Euler's method uses the recurrence yₙ₊₁ = yₙ + h·f(xₙ, yₙ) to approximate solutions of dy/dx = f(x, y). What does the term h·f(xₙ, yₙ) represent?

• The estimated change in y per step
• The exact change in y over one step
• The area under the curve on the step
• The change in x over one step

Answer: The estimated change in y per step

Euler's method follows the tangent: the estimated change in y over a step of width h is (gradient) × (step) = f(xₙ, yₙ)·h. It is only an approximation to the true change.

For a coupled system dx/dt = f(x, y), dy/dt = g(x, y), Euler's method updates both variables each step. Which pair of recurrences is correct?

• xₙ₊₁ = xₙ + h·f(xₙ, yₙ) and yₙ₊₁ = yₙ + h·g(xₙ, yₙ)
• xₙ₊₁ = xₙ + h·g(xₙ, yₙ) and yₙ₊₁ = yₙ + h·f(xₙ, yₙ)
• xₙ₊₁ = xₙ + f(xₙ, yₙ) and yₙ₊₁ = yₙ + g(xₙ, yₙ)
• xₙ₊₁ = xₙ·h·f(xₙ, yₙ) and yₙ₊₁ = yₙ·h·g(xₙ, yₙ)

Answer: xₙ₊₁ = xₙ + h·f(xₙ, yₙ) and yₙ₊₁ = yₙ + h·g(xₙ, yₙ)

Each variable advances by its own derivative times the step: x uses f, y uses g, both multiplied by h and added. Both are updated using the current (xₙ, yₙ).

A predator–prey model is dR/dt = 0.5R − 0.01RF (prey) and dF/dt = −0.4F + 0.005RF (predator). At R = 100, F = 20, what is dR/dt?

• 30
• 50
• 20
• 10

Answer: 30

dR/dt = 0.5(100) − 0.01(100)(20) = 50 − 20 = 30. Substitute both current populations into the prey equation.

Over a fixed interval, what is the approximate effect of halving the step size h in Euler's method?

• Halves the error but doubles the steps
• Doubles the error and halves the number of steps
• Has no effect on the error
• Removes all approximation error

Answer: Halves the error but doubles the steps

Euler's method has error roughly proportional to h, so halving h roughly halves the error, at the cost of twice as many computation steps.

A coupled system is dx/dt = x + y, dy/dt = 4x + y with x(0) = 1, y(0) = 0 and h = 0.1. Using one Euler step, what is the approximate value of y(0.1)?

• 0.4
• 0.1
• 1.1
• 0.5

Answer: 0.4

dy/dt at (1, 0) = 4·1 + 0 = 4, so y(0.1) = 0 + 0.1·4 = 0.4. (Meanwhile x(0.1) = 1 + 0.1·(1 + 0) = 1.1.)

Euler's method is applied to dy/dx = 2x + y with y(0) = 1 and step length h = 0.2. A student obtains y₁ = 1 + 0.2(0 + 1) = 1.2 and then y₂ = 1.2 + 0.2(0 + 1.2) = 1.44. Identify the error in the second step.

• The value of x was not advanced to 0.2 before the gradient was evaluated.
• The step length must be halved after the first step of the method.
• The value of y should have been kept at 1 for the second gradient.
• The gradient in the second step should be evaluated at x = 0.4 instead.

Answer: The value of x was not advanced to 0.2 before the gradient was evaluated.

Both variables advance together: the second gradient is f(0.2, 1.2) = 0.4 + 1.2 = 1.6, giving y₂ = 1.2 + 0.2(1.6) = 1.52. The gradient is taken at the start of the step, x = 0.2, not at its end, x = 0.4.