Answer
$$y = \sqrt {{e^t} - 1} $$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = \frac{{{e^t}}}{{2y}} \cr
& \frac{{dy}}{{dt}} = \frac{{{e^t}}}{{2y}} \cr
& {\text{separate}} \cr
& 2ydy = {e^t}dt \cr
& {\text{integrate both sides}} \cr
& \int {2ydy} = \int {{e^t}dt} \cr
& {y^2} = {e^t} + C \cr
& {\text{the initial condition }}y\left( {\ln 2} \right) = 1{\text{ implies that}} \cr
& {\left( 1 \right)^2} = {e^{\ln 2}} + C \cr
& 1 = 2 + C \cr
& C = - 1 \cr
& {y^2} = {e^t} - 1 \cr
& y = \sqrt {{e^t} - 1} \cr} $$
