I am trying to show that, for the equation
$$y'+\alpha y=0$$
alternating between a forward Euler method step for $y_{2n}$ and a backward Euler step for $y_{2n+1}$ with time-step $h$ is equivalent to
$$y_{n+1}=\frac{1-\alpha h}{1+\alpha h}y_n$$
I have that
$$y_1 = y_0 + h(-\alpha y_0) = (1-\alpha h)y_0$$
How exactly $(1+\alpha h)$ works into this I don't see. If I just press forward anyway, using the trapezoid rule for the backward Euler step next, and call $\hat{y}_n$ the approximation of $y_n$ using the forward Euler method, I get
$$y_2 = y_1 + h\left(\frac{y_1 + \hat{y}_2}{2}\right)$$
I'm kind of doing this a priori but that seems right to me: the trapezoid rule is to multiply the change in the $x$-axis by the average of the bases which are $y_1$ and $y_2$. This becomes
$$y_2 = y_1 + h\left(\frac{y_1+(1-\alpha h)y_1}{2}\right) = \left( 1 + h\left(\frac{2-\alpha h}{2}\right)\right)y_1$$
This doesn't seem to be shaping up to the form I'm supposed to be getting. Am I doing something wrong?
Let $f(z) = -\alpha z$, so $y' = f(y)$.
The forward Euler method is
$y_{n+1} = y_n + hf(y_n) =y_n -h\alpha y_n$, and the the backwards Euler method is $y_{n+1} =y_n + hf(y_{n+1}) =y_n -h\alpha y_{n+1}$
So we have $y_2 = y_1 + hf(y_2) = y_1-h\alpha y_2$ and $y_1 =y_0 + hf(y_0) = y_0 -h\alpha y_0$
So putting this to gether we get $(1+h\alpha)y_2 = y_1 = (1-h\alpha )y_0$, or
$$y_2 = \frac{1-h\alpha}{1+h\alpha} y_0$$
Now you can generalize this and fin
$$y_{2n} = \frac{1-h\alpha}{1+h\alpha} y_{2(n-1)}$$
so in your solution $h$ probably only refers to half a time step.