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Given the scalar IVP $\dot{y}(t) = f(t, y(t))$ , $y(0)=y_0$, and the butcher Tableau $$ \begin{array} {c|cccc} 0 & 0 & 0 & 0\\ \frac{1}{2} & \frac{1}{2} & 0 & 0\\ 1& -1& 2& 0\\ \hline & \frac{1}{6} &\frac{2}{3} &\frac{1}{6} \end{array} $$ We're asked to find the stability function of that Runge Kutta Method.

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$$k_1=f(t_j, y_j) = \lambda y_j$$ $$k_2= f(t_j + \frac{1}{2}, y_j + \frac{1}{2}hk_1) = \lambda(y_j + \frac{1}{2}hk_1)$$ $$k_3= f(t_j + 1, y_j - hk_1 + 2hk_2) = \lambda (y_j - hk_1 + 2hk_2)$$ $$y_{j+1}=y_j +\frac{h}{6}(k_1 + 4k_2 + k_3)$$

Thus $$y_{j+1}=y_j +\frac{h}{6}(\lambda y_j + 4\lambda(y_j + \frac{1}{2}h(\lambda y_j)) + \lambda (y_j - h(\lambda y_j) + 2h(\lambda(y_j + \frac{1}{2}h(\lambda y_j))))) $$

Which simplifies to $$ y_{j+1}= y_j(1 + h\lambda + \frac{(h \lambda)^2}{2} + \frac{(h \lambda)^3}{6}) $$ If we let $z := h\lambda$, then our stability function is $$g(z) = 1 + z + \frac{z^2}{2} + \frac{z^3}{6}$$

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