. Using Taylor’s series method, Compute 𝑦 0.12 for the initial value problem: , (0) 1 3  x  y y  dx dy taking h=0.02 .. Using Taylor’s series method, Compute 𝑦 0.12 for the initial value problem: , (0) 1 3  x  y y  dx dy taking h=0.02 .

To compute \( y(0.12) \) using Taylor’s series method, we are working with the initial value problem:

\[
\frac{dy}{dx} = x + y
\]
with \( y(0) = 1 \) and using a step size \( h = 0.02 \).

### Step-by-Step Breakdown

#### Step 1: Derivatives Calculation

We need to calculate higher-order derivatives of \( y \).

- **First derivative:** \( y' = x + y \)
- **Second derivative:** \( y'' = 1 + y' = 1 + (x + y) \)
- **Third derivative:** \( y^{(3)} = 1 + y' = 1 + (x + y) \)
- **Fourth derivative:** \( y^{(4)} = 1 + y' = 1 + (x + y) \)

#### Step 2: Taylor Series Expansion Formula

The Taylor series expansion for \( y(x) \) around \( x_0 = 0 \) is:

\[
y(x+h) = y(x) + h \cdot y'(x) + \frac{h^2}{2!} \cdot y''(x) + \frac{h^3}{3!} \cdot y^{(3)}(x) + \frac{h^4}{4!} \cdot y^{(4)}(x)
\]

#### Step 3: Iterative Calculation

We will calculate \( y(x) \) at successive points \( x = 0.02, 0.04, \dots, 0.12 \) using the Taylor expansion, including terms up to the 4th derivative.

Let's go step by step to compute \( y(0.12) \).# Let's go step by step again with detailed output to track each step

# Resetting initial condition
y = 1
x = 0
h = 0.02

# Function to calculate Taylor series up to the 4th derivative
def taylor_step_verbose(x, y, h):
    y1 = x + y  # First derivative y'
    y2 = 1 + x + y  # Second derivative y''
    y3 = 1 + x + y  # Third derivative y'''
    y4 = 1 + x + y  # Fourth derivative y''''
    
    # Taylor series expansion calculation
    y_next = y + h * y1 + (h**2 / math.factorial(2)) * y2 + (h**3 / math.factorial(3)) * y3 + (h**4 / math.factorial(4)) * y4
    
    # Output the intermediate values for each step
    return y_next, y1, y2, y3, y4

# Compute y(x) for x = 0.02, 0.04, ..., 0.12 with detailed steps
verbose_results = []

for step in steps:
    y, y1, y2, y3, y4 = taylor_step_verbose(x, y, h)
    x += h
    verbose_results.append((x, y, y1, y2, y3, y4))

verbose_resultsHere is the detailed calculation for each step of the Taylor series method, starting with \( y(0) = 1 \) and a step size of \( h = 0.02 \):

1. **At \( x = 0.02 \):**
   - \( y(0.02) = 1.0204 \)
   - \( y' = 1 \), \( y'' = 2 \), \( y^{(3)} = 2 \), \( y^{(4)} = 2 \)

2. **At \( x = 0.04 \):**
   - \( y(0.04) = 1.0416 \)
   - \( y' = 1.0404 \), \( y'' = 2.0404 \), \( y^{(3)} = 2.0404 \), \( y^{(4)} = 2.0404 \)

3. **At \( x = 0.06 \):**
   - \( y(0.06) = 1.0637 \)
   - \( y' = 1.0816 \), \( y'' = 2.0816 \), \( y^{(3)} = 2.0816 \), \( y^{(4)} = 2.0816 \)

4. **At \( x = 0.08 \):**
   - \( y(0.08) = 1.0866 \)
   - \( y' = 1.1237 \), \( y'' = 2.1237 \), \( y^{(3)} = 2.1237 \), \( y^{(4)} = 2.1237 \)

5. **At \( x = 0.10 \):**
   - \( y(0.10) = 1.1103 \)
   - \( y' = 1.1666 \), \( y'' = 2.1666 \), \( y^{(3)} = 2.1666 \), \( y^{(4)} = 2.1666 \)

6. **At \( x = 0.12 \):**
   - \( y(0.12) = 1.1350 \)
   - \( y' = 1.2103 \), \( y'' = 2.2103 \), \( y^{(3)} = 2.2103 \), \( y^{(4)} = 2.2103 \)

Thus, the approximate value of \( y(0.12) \) is **1.1350**.

If you have any questions or need further clarification, feel free to ask!

Here are some related questions to consider:
1. How does the accuracy of the Taylor series method depend on the number of derivatives used?
2. Why do we need to compute multiple derivatives in the Taylor series method?
3. How does the Taylor series method compare to other numerical methods like Runge-Kutta?
4. What happens if we increase the step size \( h \) in the Taylor series method?
5. Can the Taylor series method handle non-polynomial functions effectively?

**Tip:** Always balance the computational cost and accuracy when deciding how many derivatives to include in the Taylor series.

Learn how to compute y(0.12) using Taylor’s series method for an initial value problem involving ordinary differential equations (ODEs). Detailed step-by-step breakdown with calculations up to the 4th derivative. Suitable for advanced undergraduate level and beyond.

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