$\dfrac{dr}{dt} = -4rt$

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

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

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

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

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

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

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*

## Comments

## My Ph.D. instructor told me

My Ph.D. instructor told me "how did you get ln c?", how can I explain this?

## $r(t) = r_0$ is also a

$r(t) = r_0$ is also a solution

## Whre did u get lnc? Im

Whre did u get lnc? Im confused