Solve the differential problem given the initial values 
Solve the initial value problem :
$$y' +ye^t = 0 ,\quad \quad y(0)=e$$
$(a.)  \quad  y =e^2 e^{-e^t} $
$(b.)  \quad  y =e^4 e^{e^t}$
$(c.)  \quad  y =e^{-4t}$
$(d.)  \quad  y =e^4$
$(e.)  \quad  y =e^2 e^{-e^{2t}} $

 A: Since $dy/y=-e^tdt$, $y\propto\exp(-e^t)$. Since $y(0)=\exp1$, $y=\exp(2-e^t)$. This agrees with $A$.
A: Set
$\dot y = \dfrac{dy}{dt} \tag 1$
for notational convenience (easier to type $\dot y$ than $dy/dt$!); then our equation is
$\dot y + e^t y = 0, \; y(0) = e; \tag 2$
note that with the given initial condition it is safe to assume that
$y(t) \ne 0 \tag 3$
everywhere; otherwise, by uniqueness of solutions, the only solution is $y(t) = 0$; thus it is safe to write
$\dfrac{\dot y}{y} = -e^t; \tag 4$
this leads us to
$\dfrac{d\ln y}{dt} = -e^t, \tag 5$
whence, upon integration 'twixt $0$ an $t$,
$\ln y(t) - \ln y(0) = \displaystyle \int_0^t \dfrac{d\ln y(s)}{ds} \; ds = -\int_0^t e^s \; ds = -(e^t - e^0) = 1 - e^t; \tag 6$
with
$y(0) = e \tag 7$
(6) becomes
$\ln y(t) - 1 = 1 - e^t, \tag 8$
or
$\ln y(t) = 2 - e^t, \tag 9$
from which
$y(t) = e^{2 - e^t} = e^2e^{-e^t}. \tag{10}$
The correct choice is thus (a).
Note Added in Edit, Wednesday 27 November 2019 9:26 AM PST: To keep every jot and tittle in place, it should be observed that the transition from (4) to (5) technically requires $y(t) > 0$; but this can be handled by if necesssry reversing the sign of $y(t)$, i.e. via the transformation $y(t) \longleftrightarrow -y(t)$, valid since the given equation (2) is linear. End of Note.
