Consider the initial value problem $\frac{dy}{dx} = x + y$ with $y(0) = 1$. Using the classical fourth-order Runge-Kutta (RK4) method with a step size of $h = 0.2$, the intermediate slopes $k_1$ and $k_2$ are given by:

(A) $k_1 = 1.000$ and $k_2 = 1.200$

(B) $k_1 = 1.000$ and $k_2 = 1.100$

(C) $k_1 = 1.200$ and $k_2 = 1.440$

(D) $k_1 = 0.200$ and $k_2 = 0.240$

Answer Key:

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### 1. Strategy & Theoretical Principles In the classical RK4 algorithm for $y' = f(x, y)$: $$k_1 = f(x_0, y_0)$$ $$k_2 = f\left(x_0 + \frac{h}{2}, y_0 + \frac{h}{2} k_1\right)$$ $$k_3 = f\left(x_0 + \frac{h}{2}, y_0 + \frac{h}{2} k_2\right)$$ $$k_4 = f(x_0 + h, y_0 + h k_3)$$ ### 2. Analytical Formulation Given $f(x, y) = x + y$, $x_0 = 0$, $y_0 = 1$, and $h = 0.2$: 1. First slope $k_1$: $$k_1 = f(0, 1) = 0 + 1 = 1.000$$ 2. Second slope $k_2$: $$x_0 + \frac{h}{2} = 0 + \frac{0.2}{2} = 0.1$$ $$y_0 + \frac{h}{2} k_1 = 1 + (0.1)(1.0) = 1.1$$ $$k_2 = f(0.1, 1.1) = 0.1 + 1.1 = 1.200$$ ### 3. Distractor Forensics - **Option A ($k_1 = 1.000, k_2 = 1.200$):** Exactly matches the RK4 formulas. - **Option B ($k_1 = 1.000, k_2 = 1.100$):** Forgot to add $x = 0.1$ in $f(x, y) = x + y$. - **Option C:** Multiplied slopes by $h$ prematurely. ### 4. Exam Trap & Standard Reference - **Trap Alert:** In RK4, remember to evaluate $f$ at the midpoint $x_0 + h/2$, not just update $y$. - **Standard Reference:** Chapra, S.C. & Canale, R.P., *Numerical Methods for Engineers*, Chapter 25.

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