To solve the differential equation using Euler's method, we'll proceed step-by-step through the interval from $x = 0$ to $x = 2$ with a step size $h = 0.5$. The differential equation given is:

$\frac{dy}{dx} = 2 - 1.1yx$

with the initial condition:

$y(0) = 1$

Euler's Method

Euler's method approximates the solution to the differential equation by stepping forward from an initial point using the slope of the function. The formula for Euler's method is:

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

where $f(x, y) = \frac{dy}{dx} = 2 - 1.1yx$.

Step-by-Step Solution

1. Initial Condition:

   • $x_0 = 0$
   • $y_0 = 1$

2. Step Size:

   • $h = 0.5$

We will compute the values of $y$ at each step:

Step 1: From $x = 0$ to $x = 0.5$

$f(x_0, y_0) = 2 - 1.1 \cdot y_0 \cdot x_0 = 2 - 1.1 \cdot 1 \cdot 0 = 2$

$y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.5 \cdot 2 = 1 + 1 = 2$

Step 2: From $x = 0.5$ to $x = 1.0$

$x_1 = 0.5, \quad y_1 = 2$

$f(x_1, y_1) = 2 - 1.1 \cdot y_1 \cdot x_1 = 2 - 1.1 \cdot 2 \cdot 0.5 = 2 - 1.1 = 0.9$

$y_2 = y_1 + h \cdot f(x_1, y_1) = 2 + 0.5 \cdot 0.9 = 2 + 0.45 = 2.45$

Step 3: From $x = 1.0$ to $x = 1.5$

$x_2 = 1.0, \quad y_2 = 2.45$

$f(x_2, y_2) = 2 - 1.1 \cdot y_2 \cdot x_2 = 2 - 1.1 \cdot 2.45 \cdot 1 = 2 - 2.695 = -0.695$

$y_3 = y_2 + h \cdot f(x_2, y_2) = 2.45 + 0.5 \cdot (-0.695) = 2.45 - 0.3475 = 2.1025$

Step 4: From $x = 1.5$ to $x = 2.0$

$x_3 = 1.5, \quad y_3 = 2.1025$

$f(x_3, y_3) = 2 - 1.1 \cdot y_3 \cdot x_3 = 2 - 1.1 \cdot 2.1025 \cdot 1.5 = 2 - 3.473625 = -1.473625$

$y_4 = y_3 + h \cdot f(x_3, y_3) = 2.1025 + 0.5 \cdot (-1.473625) = 2.1025 - 0.7368125 = 1.3656875$

Summary of Results

Using Euler's method, the approximate values of $y(x)$ at each step are:

• $y(0.5) \approx 2$
• $y(1.0) \approx 2.45$
• $y(1.5) \approx 2.1025$
• $y(2.0) \approx 1.3656875$

Conclusion

The approximate solution of the differential equation using Euler's method over the interval $x = 0$ to $x = 2$ with a step size $h = 0.5$ has been calculated.

Would you like more details on the calculations, or do you have any questions?

Related Questions

1. How would the solution change if the step size $h$ were smaller, say $h = 0.1$?
2. What is the exact solution to the differential equation, and how does it compare to the Euler's method solution?
3. How does Euler's method compare to other numerical methods, such as the Runge-Kutta method?
4. What are the limitations of using Euler's method for solving differential equations?
5. How does the error in Euler's method change with different step sizes?

Tip

When using numerical methods like Euler's method, smaller step sizes generally lead to more accurate approximations, but they also require more computations.