Answer
a) $u=u_0e^{-kt/m}$
b) $u=u_0e^{-kt/m}$
Work Step by Step
a) The standard form of the given equation is:
$u'+(k/m)u=0$ ....(1)
The integrating factor is: $e^{\int (k/m) dt}=e^{kt/m}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [e^{kt/m}u]'=\int 0 dt $
or, $u=ce^{-kt/m}$ ...(2)
Applying the initial conditions in the above equation, we get
$ c=u_0$
Thus, equation (2) becomes: $u=u_0e^{-kt/m}$
b)
In order to determine the general solution, separate the variables, factor and integrate both sides.
$\int \dfrac{1}{u} du=\int \dfrac{-k}{m} dt$
or, $\ln |u|=\dfrac{-kt}{m}+c'$
This implies that $u=e^ce^{-kt/m}$
Applying the initial conditions in the above equation, we get
$ c=u_0$
Thus, equation (2) becomes: $u=u_0e^{-kt/m}$
