Answer
$r(t) = 5(e^{t-t^2})$
Work Step by Step
$\frac{dr}{dt} + 2tr = r$
$\frac{dr}{dt} = r(1-2t)$
$\int \frac{dr}{r} = \int (1-2t)dt$
$ln(r) = t-t^2+c$
$r(t) = e^{t-t^2+c}$
At t=0, r=5:
$5 = e^{c}$
So, $c = ln5$
Thus, $r(t) = e^{t-t^2+ln5}$
(OR) $r(t) = 5(e^{t-t^2})$
