Originally Posted by

**Danny** From your solution

$\displaystyle

\frac{z+1}{z} = k e^t

$

and with your IC

$\displaystyle

\frac{z+1}{z} = \frac{i+1}{i} e^t = (1 - i) e^t

$

solving for z gives

$\displaystyle

z = \frac{1}{\left( e^t - 1 \right) - i e^t} = \frac{\left( e^t - 1 \right) + i e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}

$

giving

$\displaystyle

x = \frac{\left( e^t - 1 \right) }{\left( e^t - 1 \right) ^2 + e^{2t}}

$ , $\displaystyle

y = \frac{ e^t}{\left( e^t - 1 \right) ^2 + e^{2t}}

$

the solutions of your DE noting that $\displaystyle x(0) = 0, y(0) = 1$. Now when $\displaystyle t \to \infty$ then x and y approach

$\displaystyle

x,y \to \frac{e^t}{2 e^{2t}} = \frac{e^{-t}}{2}

$ (not quite what you got but close.)