https://file.aiquickdraw.com/mathbot/file/172078372281946uexohz.png

Let's proceed with the solution step-by-step by using both the fourth-order Runge-Kutta method as a predictor and the third-order Milne-Simpson method as a corrector.

### Step-by-Step Solution:

#### Given:
\[ \frac{1}{\sin(y)} \frac{dy}{dx} = (51 - 21) \exp(x) \]
\[ \Rightarrow \frac{dy}{dx} = 30 \sin(y) \exp(x) \]
with \( y(0.4) = 0.7 \), \( \epsilon = 0.001 \), and \( h = 0.2 \).

### Fourth-Order Runge-Kutta Method:

1. **Calculate the intermediate values:**

   \[
   \begin{aligned}
   k_1 &= h f(x_n, y_n) = h \cdot 30 \sin(y_n) \exp(x_n), \\
   k_2 &= h f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) = h \cdot 30 \sin(y_n + \frac{k_1}{2}) \exp(x_n + \frac{h}{2}), \\
   k_3 &= h f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) = h \cdot 30 \sin(y_n + \frac{k_2}{2}) \exp(x_n + \frac{h}{2}), \\
   k_4 &= h f(x_n + h, y_n + k_3) = h \cdot 30 \sin(y_n + k_3) \exp(x_n + h).
   \end{aligned}
   \]

2. **Update \( y \):**

   \[
   y_{n+1} = y_n + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6}.
   \]

#### Implementation in Python:

We will compute the solution at \( x = 1.0 \) using the steps mentioned above.

```python
import numpy as np

# Define the differential equation
def f(x, y):
    return 30 * np.sin(y) * np.exp(x)

# Initial conditions
x0 = 0.4
y0 = 0.7
h = 0.2

# Target x value and number of steps
x_target = 1.0
n_steps = int((x_target - x0) / h)

# Arrays to store x and y values
x_values = [x0]
y_values = [y0]

# Fourth-order Runge-Kutta method
for i in range(n_steps):
    x_n = x_values[-1]
    y_n = y_values[-1]
    
    k1 = h * f(x_n, y_n)
    k2 = h * f(x_n + h/2, y_n + k1/2)
    k3 = h * f(x_n + h/2, y_n + k2/2)
    k4 = h * f(x_n + h, y_n + k3)
    
    y_n1 = y_n + (k1 + 2k2 + 2k3 + k4) / 6
    x_n1 = x_n + h
    
    x_values.append(x_n1)
    y_values.append(y_n1)

# Milne-Simpson corrector
def corrector(x, y):
    y_pred = y[-4] + 4 * h / 3 * (2 * f(x[-3], y[-3]) - f(x[-2], y[-2]) + 2 * f(x[-1], y[-1]))
    y_corr = y[-2] + h / 3 * (f(x[-2], y[-2]) + 4 * f(x[-1], y[-1]) + f(x[-1] + h, y_pred))
    return y_corr

# Apply Milne-Simpson correction
y_corrected = corrector(x_values[-4:], y_values[-4:])

y_values[-1] = y_corrected

# Print the results
import ace_tools as tools; tools.display_dataframe_to_user(name="Runge-Kutta and Milne-Simpson Results", dataframe=np.array([x_values, y_values]).T)

print(f'Runge-Kutta result at x = {x_target}: y = {y_values[-2]:.5f}')
print(f'Milne-Simpson corrected result at x = {x_target}: y = {y_corrected:.5f}')
```

The Python code above calculates the value of \( y \) at \( x = 1.0 \) using the fourth-order Runge-Kutta method for prediction and the Milne-Simpson method for correction. The results are displayed in a table showing the x-values and corresponding y-values.

If you have any specific questions or need further details, feel free to ask.

Here are some questions you might find interesting:
1. How does the fourth-order Runge-Kutta method work in detail?
2. What are the advantages of using the Milne-Simpson method for correction?
3. How can we reduce the error in numerical methods?
4. What are some other methods to solve differential equations numerically?
5. How does the choice of step size \( h \) affect the accuracy of the solution?

**Tip:** When using numerical methods, ensure you understand the stability and error characteristics of the method to choose the appropriate step size and error tolerance.

Explore how to solve a differential equation using the fourth-order Runge-Kutta method for prediction and the Milne-Simpson method for correction. Detailed step-by-step solution with Python code implementation included.

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