Q:

Finite Differences Method

I am confused when following the instructions where i have gone wrong and why its not coming out right when considering a first order ODE
$ y′(x) = y(x), \quad \forall x ∈ [0, 1] $
$ y(0) = 1.$
I am trying to prove $ y_i = (1+h)^i $ for $ i=0, ... , n $ where $h=1/n$.So far i have done the following working.

A:

You're almost there.  $y_{i+1}=y_i(1+h)$ as you have shown.  So $y_{1}=y_0(1+h)$, $y_{2}=y_1(1+h)=y_0(1+h)^2$ etc.  Applying your relation again and again in this way, you get $y_{i}=y_0(1+h)^i$.

