Answer
$y=y_0e^{kt}$
Work Step by Step
The standard form of the given equation is:
$y'-ky=0$ ....(1)
The integrating factor is: $e^{\int -k dt}=e^{-kt}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [ye^{-kt}]'=\int e^{-kt} dy-\int e^{-kt} ky dt $
or, $y=ce^{kt}$ ...(2)
Applying the initial conditions in the above equation, we get
$ c=y_0$
Thus, equation (2) becomes: $y=y_0e^{kt}$
