Use Euler's method with step size 0.25 to compute the approximate y-values y1, y2, y3, y4 and 
 of the solution of the initial-value problem y'=2-2x-2y, y(1)=1


We are asked to use **Euler's method** to approximate the solution to the initial-value problem:

\[
y' = 2 - 2x - 2y, \quad y(1) = 1
\]

with a **step size** of \( h = 0.25 \). Euler's method is an iterative procedure given by the formula:

\[
y_{n+1} = y_n + h \cdot f(x_n, y_n)
\]

where \( f(x, y) = 2 - 2x - 2y \) is the differential equation \( y' \), \( x_n \) and \( y_n \) are the current values of \( x \) and \( y \), and \( h \) is the step size.

### Step-by-step solution:

#### Initial Values:
We start at \( x_0 = 1 \) and \( y_0 = 1 \).

#### Calculate \( y_1 \):
- \( x_0 = 1 \), \( y_0 = 1 \)
- \( f(x_0, y_0) = 2 - 2(1) - 2(1) = 2 - 2 - 2 = -2 \)
- Using Euler's method:
  \[
  y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.25 \cdot (-2) = 1 - 0.5 = 0.5
  \]

#### Calculate \( y_2 \):
- \( x_1 = 1 + 0.25 = 1.25 \), \( y_1 = 0.5 \)
- \( f(x_1, y_1) = 2 - 2(1.25) - 2(0.5) = 2 - 2.5 - 1 = -1.5 \)
- Using Euler's method:
  \[
  y_2 = y_1 + h \cdot f(x_1, y_1) = 0.5 + 0.25 \cdot (-1.5) = 0.5 - 0.375 = 0.125
  \]

#### Calculate \( y_3 \):
- \( x_2 = 1.25 + 0.25 = 1.5 \), \( y_2 = 0.125 \)
- \( f(x_2, y_2) = 2 - 2(1.5) - 2(0.125) = 2 - 3 - 0.25 = -1.25 \)
- Using Euler's method:
  \[
  y_3 = y_2 + h \cdot f(x_2, y_2) = 0.125 + 0.25 \cdot (-1.25) = 0.125 - 0.3125 = -0.1875
  \]

#### Calculate \( y_4 \):
- \( x_3 = 1.5 + 0.25 = 1.75 \), \( y_3 = -0.1875 \)
- \( f(x_3, y_3) = 2 - 2(1.75) - 2(-0.1875) = 2 - 3.5 + 0.375 = -1.125 \)
- Using Euler's method:
  \[
  y_4 = y_3 + h \cdot f(x_3, y_3) = -0.1875 + 0.25 \cdot (-1.125) = -0.1875 - 0.28125 = -0.46875
  \]

#### Summary of the results:
- \( y_0 = 1 \)
- \( y_1 = 0.5 \)
- \( y_2 = 0.125 \)
- \( y_3 = -0.1875 \)
- \( y_4 = -0.46875 \)

Let me know if you'd like more details or have any questions!

### Related Questions:
1. How does the choice of step size affect the accuracy of Euler's method?
2. What is the exact solution to this differential equation, and how does it compare to the Euler approximation?
3. How does the error in Euler's method change with each step?
4. Can we use a smaller step size to improve the approximation? What would the new results be for \( h = 0.1 \)?
5. What are some alternative numerical methods that can give more accurate results than Euler's method?

### Tip:
For greater accuracy, consider using **Runge-Kutta methods** (like the 4th-order method), which provide better approximations with fewer steps than Euler’s method.

Use Euler's method with step size 0.25 to compute the approximate y-values y1, y2, y3, y4 and of the solution of the initial-value problem y'=2-2x-2y, y(1)=1

Learn how to use Euler's method with a step size of 0.25 to approximate the solution to the initial-value problem y' = 2 - 2x - 2y, y(1) = 1. This problem covers key concepts in differential equations and numerical methods, suitable for students in Grades 11-12.

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