Chapter 7 - Exponential Functions - 7.6 Models Involving y'=k(y-b) - Exercises - Page 359: 2

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Since $y=12+Ce^{-3t}$, then by differentiation, we have $$y'=-3Ce^{-3t}=-3(y-12).$$ Now, when $y(0)=20$, then $20=12+C$, i.e. $C=20-12=8$. In this case $$y=12+8 e^{-3t}.$$ When $y(0)=0$, then $0=12+C$, i.e. $C=0-12=-12$. In this case $$y=12-12e^{-3t}.$$ See the graphs below.