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I'm given the 1-step implicit scheme $$y_{n+1} = y_n + \frac{h}{6}[4f(t_n, y_n) + 2f(t_{n+1}, y_{n+1}) + hf'(t_n, y_n)],$$ where $y'(t) = f(t, y)$, and I'm seeking the scheme's local truncation error. I've been able to use a Taylor series to get that $$e_n(h) = hy'(t_{n+1}) - \frac{h^2}{2}y''(t_{n+1}) + O(h^3) - \frac{2h}{3}f(t_n, y(t_n)) + \frac{h}{3}f(t_{n+1}, y(t_{n+1})) + \frac{h^2}{6}f'(t_n, y(t_n))$$, but I'm having trouble understanding what the $f(t_n, y(t_n))$ terms even represent. If $f(t, y) = y'(t),$ how can the expression $f(t_n, y(t_n))$ make much sense?

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You insert the exact solution on both sides so that $y'(t_{n+1})=f(t_{n+1},y(t_{n+1}))$ and $y''(t_{n})=f'(t_{n},y(t_{n}))$. Thus \begin{align} O(h^{p+1})=g(h)&=-y(t+h)+y(t)+\frac{h}{6}[4y'(t)+2y'(t+h)+hy''(t)]\\ &=\left[y(t)+\frac{2h}3y'(t)\right]-\left[y(t+h)-\frac{h}3y'(t+h)\right]+\frac{h^2}6y''(t) \end{align} Then compute derivatives until the last is generally not zero for $h=0$. \begin{align} g'(h)&=\left[\frac{2}3y'(t)\right]-\left[\frac23y'(t+h)-\frac{h}3y''(t+h)\right]+\frac{h}3y''(t)\\ g''(h)&=-\left[\frac13y''(t+h)-\frac{h}3y'''(t+h)\right]+\frac{1}3y''(t)\\ g'''(h)&=-\left[-\frac{h}3y^{(4)}(t+h)\right]\\ \end{align} Thus $g(h)=O(h^4)$ and $p=3$ as order of the method.

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