Discussion
For a given initial value problem
\(y' = y - x\), \(y(0) = 2\)
the value of \(y(0.1)\) by Runge-Kutta second order method is :
Answer: C
\(y(0.1) \approx\) 2.2050.
With \(f(x, y) = y - x\), \(h = 0.1\) and the second-order Runge-Kutta (Heun) method:
\[ \begin{aligned} k_{1} &= hf(0, 2) = 0.1 \times 2 = 0.2 \\ k_{2} &= hf(0.1, 2.2) = 0.1 \times 2.1 = 0.21 \\ y_{1} &= 2 + \tfrac{1}{2}(0.2 + 0.21) = 2.205 \end{aligned} \]
The midpoint form of RK2 gives the same value here. The exact solution \(y = x + 1 + e^{x}\) gives 2.2052, so the answer is accurate. 2.2100 adds \(k_{2}\) alone (an Euler step with the end slope), and 2.0050 and 2.1900 do not follow from the method.
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