14.4. Example Problem

Use Milne’s method to compute 𝑦(0.4) given the differential equation:

$\frac{dy}{dx} = x + y, \quad y(0) = 1$

Use step size ℎ = 0.1, and the values:

• 𝑦₀ = 1.0000 (𝑥 = 0.0)
• 𝑦₁ = 1.1103 (𝑥 = 0.1)
• 𝑦₂ = 1.2428 (𝑥 = 0.2)
• 𝑦₃ = 1.3997 (𝑥 = 0.3)

In this problem, we are asked to compute the approximate value of 𝑦 when 𝑥 = 0.4 using Milne’s Predictor-Corrector Method. We are provided with a differential equation $dy/dx = x + y$ which describes the relationship between 𝑦 and its derivative. The initial condition given is $y(0) = 1$, meaning that at 𝑥 = 0, the value of 𝑦 is 1. Additionally, we're using a step size of ℎ = 0.1, and we already have known values of 𝑦 from 𝑥 = 0 to 0.3.

Think of this problem like tracking the position of a car moving in a straight line. At different points in time (𝑥 values), you know the car's location (𝑦 values). Now, you want to predict where the car will be after a small interval (ℎ) using the current and past positions, just like predicting 𝑦(0.4) using the previous 𝑦 values.

Step 1: Compute $f = x + y$ $f_0 = 0.0 + 1.0000 = 1.0000$ $f_1 = 0.1 + 1.1103 = 1.2103$ $f_2 = 0.2 + 1.2428 = 1.4428$ $f_3 = 0.3 + 1.3997 = 1.6997$

In this step, we compute the function values $f_i$ for each of the known points. The function is defined as $f = x + y$, so we substitute the known values of 𝑥 and 𝑦 into this equation. This helps us understand how the value of y changes in relation to its derivative at each step.

Think of this as checking the speed of a car at specific intervals (0 to 0.3). The speed (function value) at 0.0 is its current position's speed; you calculate it at each step to see how the speed varies as you move forward.

Step 2: Predict $y$ using: $y_4^{(p)} = y_0 + \frac{4\times h}{3} (2f_3 - f_2 + 2f_1)$ $= 1.0000 + \frac{4 \times 0.1}{3} (2 imes 1.6997 - 1.4428 + 2 imes 1.2103)$ $= 1.0000 + \frac{0.4}{3} (3.3994 - 1.4428 + 2.4206)$ $= 1.0000 + 0.5836 = 1.5836$

Using the predictor formula, we estimate the value of $y$ at the next step (when 𝑥 = 0.4). The formula takes a weighted average of the function values from previous steps. We plug in the function values we calculated in the previous step to make this prediction.

Imagine you're trying to guess where a moving car will be in about 10 seconds based on its speed in the last few seconds. You check its speed at previous intervals (like the function values) and predict its next location using that information.

Step 3: Compute $f_4^{(p)} = x_4 + y_4^{(p)}$ $f_4^{(p)} = 0.4 + 1.5836 = 1.9836$

Step 4: Correct using: $y_4^{(c)} = y_3 + h \frac{1}{3} (f_2 + 4f_3 + f_4^{(p)})$ $= 1.2428 + 0.1 \times \frac{1}{3} (1.4428 + 4 \times 1.6997 + 1.9836)$ $= 1.2428 + 0.1 \times \frac{10.2252}{3}$ $= 1.2428 + 0.3408 = 1.5836$

In step 3, we compute the function value at the predicted point, and in step 4, we refine our prediction using the corrector formula. This involves using the previously predicted value along with new function values to determine a more accurate estimate of $y$. The numerator in the corrector formula captures the contributions of the function behavior at several points.

Think of this like adjusting your guess about where the moving car will be after 10 seconds when you receive updated information about its speed right before it reaches that point. You refine your initial guess based on the latest speed readings to deliver a more accurate estimate.

Since predicted and corrected values match (or are very close), final answer is: $y(0.4) \approx 1.5836$

Finally, we check if the predicted and corrected values are close enough. Since they match, we conclude our calculation. This gives us an approximate value of $y(0.4) = 1.5836$, showcasing the effectiveness of the Milne's method in providing an accurate solution from numerical approximations.

Imagine you first guessed that the car would be at a certain point based on speed, then adjusted your guess as new speed data came in. When both guesses align closely, you can confidently say where the car will be after that time.

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