Question

When the IVP \( \frac{dy}{dx} = y + x,\; y(0) = 1 \) is solved by Rung–Kutta fourth order method to find \( y(0.2) \), the approximate value of \( y(0.2) \) by taking step size \( h = 0.2 \) is:

• A. 1.3502
• B. 1.2428
• C. 1.3141
• D. 0.9998

Hint:

In RK4, carefully compute \( k_1, k_2, k_3, k_4 \) using the step size, and use the weighted average formula. It gives very accurate results for small \( h \).

Answer 1

The Correct Option is B

Given: \( \frac{dy}{dx} = y + x \), initial value \( y(0) = 1 \), and \( h = 0.2 \).
We apply the RK4 method: \[ k_1 = h \cdot f(x_0, y_0) = 0.2 \cdot (1 + 0) = 0.2 \] \[ k_2 = h \cdot f(x_0 + \frac{h}{2}, y_0 + \frac{k_1}{2}) = 0.2 \cdot \left(1 + \frac{0.2}{2} + \frac{0}{2} \right) = 0.2 \cdot (1.1) = 0.22 \] \[ k_3 = h \cdot f(x_0 + \frac{h}{2}, y_0 + \frac{k_2}{2}) = 0.2 \cdot (1 + 0.1 + \frac{0.22}{2}) = 0.2 \cdot (1.21) = 0.242 \] \[ k_4 = h \cdot f(x_0 + h, y_0 + k_3) = 0.2 \cdot (0.2 + 1 + 0.242) = 0.2 \cdot (1.442) = 0.2884 \] Now calculate: \[ y(0.2) \approx y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \] \[ = 1 + \frac{1}{6}(0.2 + 2 \cdot 0.22 + 2 \cdot 0.242 + 0.2884) = 1 + \frac{1}{6}(1.4124) = 1 + 0.2354 \approx 1.2428 \]