## Thomas' Calculus 13th Edition

$$y = - \cos \left( {{e^t} - 2} \right) + 1$$
\eqalign{ & \frac{{dy}}{{dt}} = {e^t}\sin \left( {{e^t} - 2} \right),\,\,\,\,\,\,\,y\left( {\ln 2} \right) = 0 \cr & {\text{Separate the variables}} \cr & dy = {e^t}\sin \left( {{e^t} - 2} \right)dt \cr & {\text{Integrate both sides}} \cr & y = \int {{e^t}\sin \left( {{e^t} - 2} \right)dt} \cr & {\text{Use substitution}}\cr &{\text{Let }}u = {e^t} - 2;{\text{ then }}du = {e^t}dt \cr & y = \int {{e^t}\sin u\left( {\frac{{du}}{{{e^t}}}} \right)} \cr & y = \int {\sin u} du \cr & y = - \cos u + C\,\,\,\left( {\bf{1}} \right) \cr & {\text{Write the integral in terms of }}t \cr & y = - \cos \left( {{e^t} - 2} \right) + C \cr & {\text{Use the initial condition }}y\left( {\ln 2} \right) = 0 \cr & 0 = - \cos \left( {{e^{\ln 2}} - 2} \right) + C \cr & 0 = - \cos \left( 0 \right) + C \cr & C = 1 \cr & \cr & {\text{Then}}{\text{, substituting }}C = 1{\text{ in }}\left( {\bf{1}} \right) \cr & y = - \cos \left( {{e^t} - 2} \right) + 1 \cr}