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The problem asks to use **Milne's Predictor-Corrector formula** to solve the initial value problem:

\[
\frac{dy}{dx} = x^3 + y^2, \quad y(0) = 1
\]

We are to find the solution at \( x = 0.4, 0.5, 0.6 \).

### Steps to solve this problem using **Milne's Predictor-Corrector Method**:

1. **Initial Setup**: The Milne method is a multi-step method that requires some starting values, typically obtained using a simpler method like Euler's or Runge-Kutta method.

2. **Predictor Formula** (Milne's Predictor):
   \[
   y_{n+1}^{(p)} = y_{n-3} + \frac{4h}{3} \left( 2f(x_{n}, y_n) - f(x_{n-1}, y_{n-1}) + 2f(x_{n-2}, y_{n-2}) \right)
   \]
   where \( f(x, y) = \frac{dy}{dx} = x^3 + y^2 \).

3. **Corrector Formula** (Milne's Corrector):
   \[
   y_{n+1}^{(c)} = y_{n-1} + \frac{h}{3} \left( f(x_{n+1}, y_{n+1}^{(p)}) + 4f(x_n, y_n) + f(x_{n-1}, y_{n-1}) \right)
   \]

4. **Steps**:
   - Use a method like Euler or Runge-Kutta to calculate the initial values \( y_1, y_2, y_3 \) at \( x = 0.1, 0.2, 0.3 \).
   - Then use the predictor and corrector formulas to find \( y_4, y_5, y_6 \) at \( x = 0.4, 0.5, 0.6 \).

I'll calculate this step-by-step using Milne's method. Let me first compute the initial values \( y_1, y_2, y_3 \) using a method like Runge-Kutta to get a good start for the predictor-corrector method.Using the Runge-Kutta method, we have the following initial values for \( y \) at the corresponding \( x \)-values:

\[
\begin{aligned}
y(0) &= 1, \\
y(0.1) &= 1.1111, \\
y(0.2) &= 1.2504, \\
y(0.3) &= 1.4310.
\end{aligned}
\]

With these initial values, we can now use **Milne's Predictor-Corrector method** to compute the values of \( y \) at \( x = 0.4, 0.5, 0.6 \).

I'll proceed with Milne's method to predict and correct the solutions at \( x = 0.4, 0.5, 0.6 \).Using Milne's Predictor-Corrector method, we computed the following solutions for the differential equation \( \frac{dy}{dx} = x^3 + y^2 \) with \( y(0) = 1 \) at the specified \( x \)-values:

\[
\begin{aligned}
y(0.4) &= 1.6749, \\
y(0.5) &= 2.0227, \\
y(0.6) &= 2.5559.
\end{aligned}
\]

Would you like further details or explanations on any of the steps?

Here are some related questions to consider:
1. How does Milne's method compare to other numerical methods like Euler's or Runge-Kutta?
2. What are the potential sources of error in predictor-corrector methods?
3. Can Milne’s method be generalized for systems of differential equations?
4. What is the importance of the step size \( h \) in numerical methods?
5. How can we check the accuracy of Milne's method?

**Tip**: In numerical methods, using smaller step sizes generally increases the accuracy but also increases computation time. Always balance precision with practicality depending on your problem.

Use Milne's predictor-corrector formula to find the solutions at x = 0.4, 0.5, 0.6 of the differential equation dy/dx = x^3 + y^2, y(0) = 1.

Learn how to solve the differential equation dy/dx = x^3 + y^2 using Milne's predictor-corrector method. This step-by-step guide includes the predictor and corrector formulas and walks you through solving the equation at x = 0.4, 0.5, and 0.6. Suitable for undergraduate students studying numerical methods and differential equations.

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