Use Taylor's series method to solve the initial value problem: dy/dx =x^2+y^2 for x=0.25 and 0.5 given that y(0) =1.

To solve the initial value problem \( \frac{dy}{dx} = x^2 + y^2 \) using the **Taylor series method**, we will proceed step-by-step.

### Step 1: Write the general Taylor series for \( y(x) \)
The Taylor series expansion of \( y(x) \) around \( x = 0 \) is given by:

\[
y(x) = y(0) + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y^{(3)}(0)}{3!} x^3 + \cdots
\]

Where \( y(0) \) and the derivatives \( y'(0), y''(0), y^{(3)}(0), \dots \) are evaluated at \( x = 0 \).

### Step 2: Use the initial conditions
Given that \( y(0) = 1 \), we have:

\[
y(x) = 1 + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y^{(3)}(0)}{3!} x^3 + \cdots
\]

### Step 3: Compute the derivatives of \( y(x) \)
The differential equation is:

\[
\frac{dy}{dx} = x^2 + y^2
\]

So, at \( x = 0 \):

- \( y'(x) = x^2 + y^2 \). At \( x = 0 \), \( y(0) = 1 \), hence:
  \[
  y'(0) = 0^2 + 1^2 = 1
  \]

- Next, differentiate \( y'(x) = x^2 + y^2 \) to find \( y''(x) \):
  \[
  y''(x) = 2x + 2y \cdot y'
  \]
  At \( x = 0 \), we know \( y(0) = 1 \) and \( y'(0) = 1 \), so:
  \[
  y''(0) = 2(0) + 2(1)(1) = 2
  \]

- Differentiate again to find \( y^{(3)}(x) \):
  \[
  y^{(3)}(x) = 2 + 2 \left( y' \right)^2 + 2 y \cdot y''
  \]
  At \( x = 0 \), \( y(0) = 1 \), \( y'(0) = 1 \), and \( y''(0) = 2 \), so:
  \[
  y^{(3)}(0) = 2 + 2(1)^2 + 2(1)(2) = 2 + 2 + 4 = 8
  \]

- Compute \( y^{(4)}(x) \):
  \[
  y^{(4)}(x) = 2 y' y^{(3)} + 2 y''^2 + 2 y \cdot y^{(3)}
  \]
  At \( x = 0 \), using known values:
  \[
  y^{(4)}(0) = 2(1)(8) + 2(2)^2 + 2(1)(8) = 16 + 8 + 16 = 40
  \]

### Step 4: Write the Taylor series for \( y(x) \)
Now, using the values for the derivatives, we can write the Taylor series expansion up to the fourth derivative:

\[
y(x) = 1 + 1 \cdot x + \frac{2}{2!} x^2 + \frac{8}{3!} x^3 + \frac{40}{4!} x^4 + \cdots
\]

Simplifying the coefficients:

\[
y(x) = 1 + x + x^2 + \frac{4}{3} x^3 + \frac{5}{3} x^4 + \cdots
\]

### Step 5: Approximate \( y(x) \) for \( x = 0.25 \) and \( x = 0.5 \)
Now, substitute \( x = 0.25 \) and \( x = 0.5 \) into the series expansion to approximate the value of \( y(x) \).

- For \( x = 0.25 \):
  \[
  y(0.25) \approx 1 + 0.25 + (0.25)^2 + \frac{4}{3} (0.25)^3 + \frac{5}{3} (0.25)^4
  \]
  \[
  y(0.25) \approx 1 + 0.25 + 0.0625 + \frac{4}{3} (0.015625) + \frac{5}{3} (0.00390625)
  \]
  \[
  y(0.25) \approx 1 + 0.25 + 0.0625 + 0.02083 + 0.00651 \approx 1.33984
  \]

- For \( x = 0.5 \):
  \[
  y(0.5) \approx 1 + 0.5 + (0.5)^2 + \frac{4}{3} (0.5)^3 + \frac{5}{3} (0.5)^4
  \]
  \[
  y(0.5) \approx 1 + 0.5 + 0.25 + \frac{4}{3} (0.125) + \frac{5}{3} (0.0625)
  \]
  \[
  y(0.5) \approx 1 + 0.5 + 0.25 + 0.16667 + 0.10417 \approx 2.02084
  \]

### Final Results:
- \( y(0.25) \approx 1.33984 \)
- \( y(0.5) \approx 2.02084 \)

Would you like more details on any step?

### Related Questions:
1. How does the Taylor series compare to other methods for solving differential equations?
2. What is the error bound in the Taylor series expansion for this problem?
3. How does increasing the number of terms in the Taylor series affect the accuracy?
4. Can you apply the Taylor series method to non-polynomial differential equations?
5. How does the Taylor series method relate to Euler's method for solving ODEs?

**Tip:** Using more terms in the Taylor series improves the accuracy but can make the computation more complex.

Use Taylor's series method to solve the initial value problem: dy/dx = x^2 + y^2 for x=0.25 and 0.5 given that y(0) = 1.

Learn how to solve the initial value problem dy/dx = x^2 + y^2 using the Taylor series method, with step-by-step calculations for x = 0.25 and 0.5. This solution involves calculating successive derivatives and using the Taylor expansion to approximate the values of y(x). Suitable for university-level students studying calculus and differential equations.

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