Answer
$$ y = - \frac{1}{\pi }\tan \left( {\pi {e^{ - t}}} \right) + \frac{3}{\pi }$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dt}} = {e^{ - t}}se{c^2}\left( {\pi {e^{ - t}}} \right),\,\,\,\,\,\,\,y\left( {\ln 4} \right) = \frac{2}{\pi } \cr
& {\text{Separate the variables}} \cr
& dy = {e^{ - t}}se{c^2}\left( {\pi {e^{ - t}}} \right)dt \cr
& {\text{Integrate both sides}} \cr
& y = \int {{e^{ - t}}se{c^2}\left( {\pi {e^{ - t}}} \right)dt} \cr
& {\text{Use substitution}}\cr
&{\text{Let }}u = \pi {e^{ - t}};{\text{ then }}du = - \pi {e^{ - t}}dt \cr
& y = \int {{e^{ - t}}se{c^2}u\left( {\frac{{du}}{{ - \pi {e^{ - t}}}}} \right)} \cr
& y = \int {se{c^2}u\left( {\frac{{du}}{{ - \pi }}} \right)} \cr
& y = - \frac{1}{\pi }\int {se{c^2}udu} \cr
& y = - \frac{1}{\pi }\tan u + C \cr
& {\text{Write the integral in terms of }}t \cr
& y = - \frac{1}{\pi }\tan \left( {\pi {e^{ - t}}} \right) + C\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{Use initial condition }}y\left( {\ln 4} \right) = \frac{2}{\pi } \cr
& \frac{2}{\pi } = - \frac{1}{\pi }\tan \left( {\pi {e^{ - \ln 4}}} \right) + C \cr
& \frac{2}{\pi } = - \frac{1}{\pi }\tan \left( {\frac{\pi }{4}} \right) + C \cr
& C = \frac{2}{\pi } + \frac{1}{\pi } \cr
& C = \frac{3}{\pi } \cr
& \cr
& {\text{Then}}{\text{, substituting }}C = \frac{3}{\pi }{\text{ in }}\left( {\bf{1}} \right) \cr
& y = - \frac{1}{\pi }\tan \left( {\pi {e^{ - t}}} \right) + \frac{3}{\pi } \cr} $$
