Answer
$y=(\dfrac{-1}{\pi}) \tan(\pi e^{-\tau} ) +\dfrac{3}{\pi}$
Work Step by Step
Given: $\dfrac{dy}{dt}=e^{-\tau} \sec^2 (\pi e^{-\tau} )$
Here, $y=(-1/\pi)\int \sec^2 (\pi e^{-\tau} ) (-\pi e^{-\tau} ) d\tau$
Now, we have $y=(\dfrac{-1}{\pi}) \tan(\pi e^{-\tau} ) +c $
Apply the initial conditions $y[\ln (4)]=\dfrac{2}{\pi}$
$\dfrac{2}{\pi}=(\dfrac{-1}{\pi}) \tan(\pi e^{-\ln 4} ) +c \implies c=\dfrac{3}{\pi} $
Thus, we have $y=(\dfrac{-1}{\pi}) \tan(\pi e^{-\tau} ) +\dfrac{3}{\pi}$
