<s>[INST] Stability analysis of Ralston's method Ralston's method is given by:
$$y_{n+1} = y_n + \frac{h}3(f(t_n,y_n)+2f(t_n+\frac34h, y_n + \frac34h f(t_n,y_n)))$$
carry out a stability analysis of this method to determine the condition for stability based on the test problem:
$$\frac{dy}{dt} -\lambda y$$
From what I understand I need to use the test problem to determine rewrite the method without the $f$ function. I have that
$f(t_n,y_n) = -\lambda y_n$
$f(t_n+\frac34h, y_n + \frac34h f(t_n,y_n)) = f(t_n+\frac34h, y_n + \frac34h (-\lambda y_n))$
I'm not sure how to simplify this second $f$ equation though.
Any help would really be appreciated! 
 [/INST] Usually, stability of a method is studied on a test ODE
$$
y' = f(t,y) \quad\Leftrightarrow\quad y' - f(t,y) = 0
$$
with $f(t,y) = \lambda y,\; \lambda \in \mathbb{C}^{-}$.
So
$$
f(t_n, y_n) = \lambda y_n\\
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h f(t_n, y_n)) =
f(t_n + \tfrac{3}{4}h, y_n + \tfrac{3}{4}h \lambda y_n) =
\lambda( y_n + \tfrac{3}{4}h \lambda y_n) =
\lambda y_n(1 + \tfrac{3}{4}h \lambda)
$$
Now
$$
y_{n+1} = y_n + \tfrac{h}{3}\left(\lambda y_n + 2 \lambda y_n(1 + \tfrac{3}{4}h\lambda)\right) = y_n\left(1 + \lambda h + \frac{(\lambda h)^2}{2}\right) = y_n r(\lambda h)\\
r(z) = 1 + z + \frac{z^2}{2}.
$$
The function $r(z)$ is the stability function of the method.
Also, one can note that the given method falls in Runge-Kutta family with following Butcher's tableau
$$
\begin{array}{c|c}
\mathbf{c} & \mathbf{A}\\
\hline
& \mathbf{b}^\top
\end{array}
=
\begin{array}{c|cc}
0 & 0 & 0 \\
\tfrac{3}{4} & \tfrac{3}{4} & 0\\
\hline
& \tfrac{1}{3} & \tfrac{2}{3}
\end{array}
$$
and for Runge-Kutta methods the stability function can be found directly from their Butcher's tableau by
$$
r(z) = \frac{\operatorname{det}(\mathbf{E} - z\mathbf{A} + z\mathbf{e}\mathbf{b}^\top)}{\operatorname{det}(\mathbf{E} - z\mathbf{A})} = 
\frac{\begin{vmatrix}1 +\tfrac{z}{3} & \tfrac{2z}{3}\\-\tfrac{3z}{4}+\tfrac{z}{3}& 1+\tfrac{2z}{3}\end{vmatrix}}{\begin{vmatrix}1 & 0\\-\tfrac{3z}{4}& 1\end{vmatrix}} = \left(1 + \frac{z}{3}\right)\left(1 + \frac{2z}{3}\right) + \frac{5z}{12} \frac{2z}{3} = \\ = 1 + z + \frac{2z^2}{9} + \frac{5z^2}{18} = 1 + z + \frac{z^2}{2}
$$
It is easy to check that this RK method is an explicit two-stage second order method. From the former one can deduce that $r(z)$ is some quadratic polynomial in $z$, and from the latter 
$$
r(z) = e^z + o(z^2).
$$
That gives the only possibility for $r(z) = 1 + z + \tfrac{z^2}{2}$ which is the stability function of every explicit two-stage second order Runge-Kutta method.
</s>