## Calculus (3rd Edition)

The general solution of $y'=4(y-12)$ is $$y=12+c e^{2t}.$$ When $y(0)=20$, then $20=12+c$, i.e. $c=20-12=8$. In this case $$y=12+8 e^{4t}.$$ When $y(0)=0$, then $0=12+c$, i.e. $c=-12$. In this case $$y=12-12 e^{4t}.$$ See the graphs below.