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Euler's Methods for Approximate Solution of First Order Differential Equations

Example: Use Euler's Method twice to find the approximate solution to

dy/dx = y + 1 where y(0) = 1 in the interval [0, 0.5]

First use the step size, h, = 0.25 and then use 0.1. Compare your results with the

Exact solution y(x) = 2 e^x ˗ 1 for x = 0.5.

Strategy: Apply Euler's Algorithm to obtain successive approximations

y_n+1= y_n + h f(x_n, y_n)

Case 1: h = 0.25 f(x, y(x)) = y + 1, y₀ = y(0) = 1

y₁ = y₀ + h f(x₀, y₀) = 1 + 0.25 [ 1 + 1 ] = 1.5 (result when x = 0.25)

y₂ = y₁ + h f(x₁, y₁) = 1.5 + 0.25 [ 1.5 + 1 ] = 2.125 (result when x = 0.5)

Case 2: h = 0.1 f(x, y(x)) = y + 1, y₀ = y(0) = 1

y₁ = y₀ + h f(x₀, y₀) = 1 + 0.1 [ 1 + 1 ] = 1.2

y₂ = y₁ + h f(x₁, y₁) = 1.2 + 0.1 [ 1.2 + 1 ] = 1.42 (result when x = 0.2)

y₃ = y₂ + h f(x₂, y₂) = 1.42 + 0.1 [ 1.42 + 1 ] = 1.662 (result when x = 0.3)

y₄ = y₃ + h f(x₃, y₃) = 1.662 + 0.1 [ 1.662 + 1 ] = 1.9282 (result when x = 0.4)

y₅ = y₄ + h f(x₄, y₄) = 1.9282 + 0.1 [ 1.9282 + 1 ] = 2.221 (result when x = 0.5)

Case 3: Exact Solution

y(x) = 2 e^x ˗ 1 for x = 0.5 : y(0.5) = 2 e^0.5˗ 1 = 2.297 (result)

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