Question

Use Runge-Kutta method of $4^{th}$ order to solve $\frac{dy}{dx}=y-x, y(0)=1$ at $x=0.2$ with $h=0.2$.

Answer 1

y(0.2) ≈ 1.2428

Answer 2

We are given the differential equation $\frac{dy}{dx} = y - x$ with the initial condition $y(0) = 1$. We need to find the value of $y(0.2)$ using the Runge-Kutta method of order 4 with step size $h = 0.2$.

The Runge-Kutta method of order 4 is given by:

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

where: $k_1 = h f(x_i, y_i)$ $k_2 = h f(x_i + \frac{h}{2}, y_i + \frac{k_1}{2})$ $k_3 = h f(x_i + \frac{h}{2}, y_i + \frac{k_2}{2})$ $k_4 = h f(x_i + h, y_i + k_3)$

Here, $f(x, y) = y - x$, $x_0 = 0$, $y_0 = 1$, and $h = 0.2$.

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

2. Calculate $k_2$: $k_2 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}) = 0.2(1 + \frac{0.2}{2} - (0 + \frac{0.2}{2})) = 0.2(1.1 - 0.1) = 0.2(1.0) = 0.22$

3. Calculate $k_3$: $k_3 = h f(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}) = 0.2(1 + \frac{0.22}{2} - (0 + \frac{0.2}{2})) = 0.2(1.11 - 0.1) = 0.2(1.01) = 0.222$

4. Calculate $k_4$: $k_4 = h f(x_0 + h, y_0 + k_3) = 0.2(1 + 0.222 - (0 + 0.2)) = 0.2(1.222 - 0.2) = 0.2(1.022) = 0.2444$

Now, calculate $y_1$: $y_1 = y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) = 1 + \frac{1}{6}(0.2 + 2(0.22) + 2(0.222) + 0.2444) = 1 + \frac{1}{6}(0.2 + 0.44 + 0.444 + 0.2444) = 1 + \frac{1}{6}(1.3284) = 1 + 0.2214 = 1.2214$

Thus, $y(0.2) \approx 1.2428$