Chapter 7 - Exponential Functions - 7.6 Models Involving y'=k(y-b) - Exercises - Page 360: 4

$$y= 2+8 e^{-6t+12}.$$

Since $y'+6y=12 \Longrightarrow y'=-6(y-2)$ then the general solution is $$y=2+c e^{-6t}.$$ When $y(2)=10$, then $10=2+ce^{-12}$, i.e. $c=8e^{12}$. In this case $$y=2+8e^{12} e^{-6t}=2+8 e^{-6t+12}.$$