Answer
$$y = - 4{e^{ - x/2}} + 5$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 2{e^{ - x/2}},{\text{ }}\left( {0,1} \right) \cr
& {\text{Separate the variables}} \cr
& dy = 2{e^{ - x/2}}dx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = 2\int {{e^{ - x/2}}} dx \cr
& y = 2\left( {\frac{1}{{ - 1/2}}} \right){e^{ - x/2}} + C \cr
& y = - 4{e^{ - x/2}} + C \cr
& {\text{Use the initial condition }}\left( {0,1} \right) \cr
& 1 = - 4{e^{ - \left( 0 \right)/2}} + C \cr
& C = 5 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - 4{e^{ - x/2}} + 5 \cr
& \cr
& {\text{Graph}} \cr} $$
