Consider IVP $y'(x) = - \lambda y(x)$ and $y(0) = 1 $ , $\lambda > 0$ and the numerical method $$ y_n = y_{n-2} + \frac{1}{3} h [ f(x_n,y_n) + 4 f(x_{n-1},y_{n-1}) + f(x_{n-2},y_{n-2}) ]$$
$$y_0 = 1$$
Find the accuracy of this method. For small $h$, discuss stability of the method. As $h \to 0$, to what solution does the numerical mthod converge?
Attempt
Let $y(t)$ be the exact solution ${\bf restricted} $ to the grid $(y(t_n) = Y_n$) and upon subtituting into the method, we obtain
$$ T_k = Y_n - Y_{n-2} - \frac{1}{3} h [ f(x_n,Y_n) + 4 f(x_{n-1},Y_{n-1}) + f(x_{n-2},Y_{n-2}) ]$$
Now, we perform taylor expansion as follows
$$ Y_{n-2} = Y_n -2h Y'_n + \frac{ (2h)^2 }{2} Y''_n +...$$ $$ f(x_{n-1},Y_{n-1} ) = Y'_{n-1} = Y'_n - h Y''_n + \frac{h^2}{2} Y'''_n + ... $$
$$ Y_{n-2}' = Y'_n - 2h Y''_n + 2 h^2 Y'''_n + ... $$
So, upon substitution into equation above we obtain
$$ T_k = \frac{h}{3} Y'_n - \frac{4 h^3 }{3} Y_n'' + O(h^4) $$
method is consisten as $T_k \to 0$, but I only get first-order accuracy. Is this correct? Im suspicious as implicit methods are usually higher order schemes. For stability, we have
$$y_n = y_{n-2} - \frac{h}{3} \lambda [ y_n + 4 y_{n-1} + y_{n-2} ] $$
or
$$ \left( 1 + \frac{h \lambda }{3} \right) y_n = - \frac{4 h \lambda }{3} y_{n-1} - \left( \frac{\lambda h }{3} -1\right) y_{n-2} $$
Am I on the right track?