To solve the differential equation $\frac{dy}{dx} = y - x$ using the second-order Runge-Kutta method (also known as the midpoint method) to find $y(0.2)$, with the initial condition $y(0) = 2$ and step size $h = 0.1$, we'll follow these steps:

The Second-Order Runge-Kutta Method (Midpoint Method):

Given the differential equation $\frac{dy}{dx} = f(x, y)$ and initial condition $y(x_0) = y_0$, the update formula using the second-order Runge-Kutta method is:

$y_{n+1} = y_n + k_2$ where: $k_1 = h \cdot f(x_n, y_n)$ $k_2 = h \cdot f\left(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}\right)$

Step 1: Initialize the values

• $x_0 = 0$
• $y_0 = 2$
• $h = 0.1$
• $f(x, y) = y - x$

Step 2: Calculate $y(0.1)$

1. Calculate $k_1$: $k_1 = h \cdot f(x_0, y_0) = 0.1 \cdot (2 - 0) = 0.1 \cdot 2 = 0.2$

2. Calculate $k_2$: $k_2 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}\right) = 0.1 \cdot f\left(0 + 0.05, 2 + 0.1\right)$ $f\left(0.05, 2.1\right) = 2.1 - 0.05 = 2.05$ $k_2 = 0.1 \cdot 2.05 = 0.205$

3. Update $y_1$ to find $y(0.1)$: $y_1 = y_0 + k_2 = 2 + 0.205 = 2.205$

Step 3: Calculate $y(0.2)$

1. Update the values for the next step: $x_1 = 0.1, \quad y_1 = 2.205$

2. Calculate $k_1$ for the next step: $k_1 = h \cdot f(x_1, y_1) = 0.1 \cdot (2.205 - 0.1) = 0.1 \cdot 2.105 = 0.2105$

3. Calculate $k_2$ for the next step: $k_2 = h \cdot f\left(x_1 + \frac{h}{2}, y_1 + \frac{k_1}{2}\right) = 0.1 \cdot f\left(0.15, 2.205 + 0.10525\right)$ $f\left(0.15, 2.31025\right) = 2.31025 - 0.15 = 2.16025$ $k_2 = 0.1 \cdot 2.16025 = 0.216025$

4. Update $y_2$ to find $y(0.2)$: $y_2 = y_1 + k_2 = 2.205 + 0.216025 = 2.421025$

Final Answer:

Thus, the value of $y(0.2)$ is approximately $2.421$.

Would you like further clarification on any of these steps?

Related Questions

1. How does the second-order Runge-Kutta method compare to the fourth-order Runge-Kutta method?
2. Can you apply this method to a system of differential equations?
3. How would the solution change if the step size $h$ were smaller?
4. What is the global error in the second-order Runge-Kutta method?
5. How can we use the Runge-Kutta method to approximate higher-order derivatives?

Tip

When using numerical methods like Runge-Kutta, smaller step sizes $h$ generally lead to more accurate results but increase computational effort.