I am trying to understand why a system with a single pole inside the unit circle is stable. For example, take a system with one pole at $z=\frac{1}{2}$. The literature says the system is stable. As a Physics major with not that much intuition for pole-zero plots I first tried to solve the differential equation that this transfer function represents:
$\frac{Y(z)}{X(z)}=H(z)=\frac{z}{z-\frac{1}{2}}$
Which, I re-write to find
$Y(z)\cdot\big(z-\frac{1}{2}\big)=X(z)\cdot z$
Which I transformed to the time domain giving
$\dot{y}-\frac{1}{2}y=\dot{x}$
For the impulse response $\dot{x}=\dot{\delta}(t)=0$, so the solution to my equation looks like
$y_{\delta}(t)=c_1e^{\frac{1}{2}t}$
And this looks quite unstable to me!
Of course, I also tried doing the same by doing a long-tail division of the transfer function, so I got
$H(z)=\sum\limits_{n=0}^{\infty}2^{-n}z^{-n}$
The inverse transform of this gives
$y_{\delta}(t)=2^{-t} = \big(\frac{1}{2}\big)^t$
This, I agree, is a stable system.
I've tried looking all over but I haven't found an example of how to go from the differential equation to a stable system. I have seen that some other questions had answers that included something like "look at the characteristic equation, when $\lambda < 1$ then the system is stable. I suppose I would disagree there - since $e^{\lambda t}$ would only be stable for negative $\lambda$.
What am I missing?