I have $$y(n)-3y(n-1)+2y(n-2)=4x(n)-2x(n-1)$$ that is the equation for a causal, discrete time LTI system. Using the Laplace Transform I rewrote it as: $$Y(s)-3e^{-s}Y(s)+2e^{-2s}Y(s)=4X(s)-2e^{-s}X(s)$$ And then I tried calculating $$H(s)=\frac{Y(s)}{X(s)}=\frac{4-2e^{-s}}{1-3e^{-s}+2e^{-2s}}$$ But the correct response for the transfer function is $$H(z)=\frac{4z(z-1/2)}{z^2-3z+2}$$ so I think I'm missing something here.
Then I have another problem, given what was calculated before, what should be the initial conditions so that the system's response to the unit step function are $y(0)=1$ and $y(1)=3$. The answer is $$y(-1)=1$$ and $$y(-2)=3$$ and I have no idea how I'm even supposed to calculate that, I'm assuming it has something to do with the Unilateral LaPlace Transform, but I'm very new to this and I don't even know how to start.