Say we have a function $y(t)$, that satisfies the ordinary differential equation $\frac{\mathrm{d}y}{\mathrm{d}t}= f(t,y) \quad\text{for } t\in (t_0, t_{\max}],$ where $t$ takes discrete values, $t_n$, with a constant step size $h = t_{n+1} - t_n$. We also are given the initial condition, for example, $y(0) = 1$.
If we were to use the Backward Euler Method: $$y_{n+1} \approx y_n + hf(t_{n+1}, y_{n+1}),$$ to approximate a solution for a function, say, $f(y) = -y$, I am told that it is possible to use this information to show that the Backward Euler Scheme is first order, by expanding the global error as a Taylor series in powers of $h$. (The global error at fixed $t$ is the difference between the approximate solution and the exact solution, $y(t)$.)
Can anyone figure out how this might be possible? I cannot find any resources online that dmeonstrate this idea.
You get $$ y_n=(1+h)^{-n}y_0=\exp(-n\ln(1+h))=\exp(-nh+\tfrac12nh^2+O(nh^3))=\exp(-t+\tfrac12th+t·O(h^2)) $$ so that the first error term is of order 1.