To solve the given first-order differential equation using the Runge-Kutta method of fourth order (RK4), we'll follow these steps:

Problem Setup

The differential equation is:

$\frac{dy}{dx} = x - y$

with the initial condition $y(0) = 1$. We are tasked to find $y(0.1)$ and $y(0.2)$ using the RK4 method. Let the step size $h = 0.1$.

Runge-Kutta 4th Order Formula

Given $\frac{dy}{dx} = f(x, y)$, the RK4 method updates $y_n$ to $y_{n+1}$ using:

$y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$

where:

$k_1 = h \cdot f(x_n, y_n)$ $k_2 = h \cdot f\left(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}\right)$ $k_3 = h \cdot f\left(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}\right)$ $k_4 = h \cdot f(x_n + h, y_n + k_3)$

Step-by-Step Calculation

Step 1: From $x_0 = 0$, $y_0 = 1$, $h = 0.1$

1. Compute $k_1, k_2, k_3, k_4$:

   • $k_1 = h \cdot f(x_0, y_0) = 0.1 \cdot (0 - 1) = -0.1$
   • $k_2 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}\right) = 0.1 \cdot \left(0.05 - (1 - 0.05)\right) = 0.1 \cdot (0.05 - 0.95) = -0.09$
   • $k_3 = h \cdot f\left(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}\right) = 0.1 \cdot \left(0.05 - (1 - 0.045)\right) = -0.0895$
   • $k_4 = h \cdot f(x_0 + h, y_0 + k_3) = 0.1 \cdot \left(0.1 - (1 - 0.0895)\right) = 0.1 \cdot (0.1 - 0.9105) = -0.08105$

2. Update $y_1 = y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$: $y_1 = 1 + \frac{1}{6}(-0.1 + 2(-0.09) + 2(-0.0895) + (-0.08105))$ $y_1 = 1 - 0.08985 = 0.91015$

Thus, $y(0.1) \approx 0.91015$.

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Step 2: From $x_1 = 0.1$, $y_1 \approx 0.91015$

1. Compute $k_1, k_2, k_3, k_4$:

   • $k_1 = h \cdot f(x_1, y_1) = 0.1 \cdot (0.1 - 0.91015) = -0.081015$
   • $k_2 = h \cdot f\left(x_1 + \frac{h}{2}, y_1 + \frac{k_1}{2}\right) = 0.1 \cdot (0.15 - (0.91015 - 0.0405)) = 0.1 \cdot (0.15 - 0.86965) = -0.071965$
   • $k_3 = h \cdot f\left(x_1 + \frac{h}{2}, y_1 + \frac{k_2}{2}\right) = 0.1 \cdot (0.15 - (0.91015 - 0.0359825)) = 0.1 \cdot (0.15 - 0.8741675) = -0.07241675$
   • $k_4 = h \cdot f(x_1 + h, y_1 + k_3) = 0.1 \cdot (0.2 - (0.91015 - 0.07241675)) = 0.1 \cdot (0.2 - 0.83773325) = -0.063773325$

2. Update $y_2 = y_1 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$: $y_2 = 0.91015 + \frac{1}{6}(-0.081015 + 2(-0.071965) + 2(-0.07241675) + (-0.063773325))$ $y_2 = 0.91015 - 0.07192 = 0.83823$

Thus, $y(0.2) \approx 0.83823$.

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Final Results

1. $y(0.1) \approx 0.91015$
2. $y(0.2) \approx 0.83823$

Would you like further details or clarifications?

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Expanding Questions

1. How does the RK4 method compare to other numerical methods like Euler or Heun in accuracy?
2. Could you explain how the step size $h$ impacts the results of RK4?
3. How would the results change if the step size were halved to $h = 0.05$?
4. Are there alternative methods to solve this ODE analytically for comparison?
5. How is the error in RK4 quantified, and why is it so accurate?

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Tip

Always double-check your intermediate computations in RK4 to ensure accuracy in each step, especially for $k_2$ and $k_3$, as errors there propagate significantly.