Answer
See details below.
Work Step by Step
The general solution of $y'=-2 y+8 =-2(y-4)$ is
$$y=4+c e^{-2t}.$$
When $y(0)=3$, then $3=4+c$, i.e. $c=3-4=-1$. In this case
$$y=4- e^{-2t}.$$
When $y(0)=4$, then $4=4+c$, i.e. $c=0$. In this case
$$y=4.$$
See the graphs below.
