Problem 01 | Separation of Variables

$\dfrac{dr}{r} = -4t\,dt$

$\displaystyle \int \dfrac{dr}{r} = -4 \int t\,dt$

$\ln r = -2t^2 + \ln c$

$\ln r = \ln e^{-2t^2} + \ln c$

$\ln r = \ln ce^{-2t^2}$

$r = ce^{-2t^2}$
 

when   $t = 0$,   $r = r_o$
$r_o = ce^{-2(0^2)}$

$c = r_o$
 

then,
$r = r_o \, e^{-2t^2}$

$r = r_o \, \exp (-2t^2)$           answer