Chapter 7 - Integration Techniques - 7.9 Introduction to Differential Equations - 7.9 Exercises - Page 589: 4

$$y = -5{e^{ - 3t}} + 10$$

$$\eqalign{ & y = C{e^{ - 3t}} + 10 \cr & {\text{The initial condition is }}y\left( 0 \right) = 5,{\text{ then}} \cr & y\left( 0 \right) = C{e^{ - 3\left( 0 \right)}} + 10 \cr & 5 = C\left( {{e^0}} \right) + 10 \cr & 5 = C + 10 \cr & C = -5 \cr & {\text{The solution is }} \cr & \underbrace {y = C{e^{ - 3t}} + 10}_{C = -5} \cr & y = -5{e^{ - 3t}} + 10 \cr} $$