Discussion
Let \(\dfrac{dy}{dx} = f(x, y)\) be an initial value problem with \(y(x_{0}) = y_{0}\). Then in Runge-Kutta method of order 4, the value of K3 to find first approximation is :
Answer: C
The third slope is K3 = h f(x0 + h/2, y0 + k2/2).
The classical fourth order Runge-Kutta method uses
\[ \begin{aligned} k_{1} &= hf(x_{0}, y_{0}) \\ k_{2} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{1}}{2}) \\ k_{3} &= hf(x_{0} + \tfrac{h}{2}, y_{0} + \tfrac{k_{2}}{2}) \\ k_{4} &= hf(x_{0} + h, y_{0} + k_{3}) \end{aligned} \]
and \(y_{1} = y_{0} + \frac{1}{6}(k_{1} + 2k_{2} + 2k_{3} + k_{4})\). The option \(hf(x_{0}, y_{0})\) is \(k_{1}\), and the step to \(x_{0} + h\) belongs to \(k_{4}\) (with \(k_{3}\), not h, added to y). The ratio \(k_{1}/k_{2}\) appears nowhere in the method.
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