Answer
(a) $u=u_0e^{-kt/m}$
(b) $u=u_0e^{-kt/m}$$
Work Step by Step
a) Here, we have $u'+(\dfrac{k}{m})u=0$
The integrating factor is: $I=e^{\int (k/m) dt}=e^{kt/m}$
Now, $e^{kt/m}[u'+(\dfrac{k}{m})u]=0$
This gives $\int [e^{kt/m}u]'=\int 0 dt $
$\implies u=ce^{-kt/m}$
After applying the initial conditions, we have $ c=u_0$
So, $u=ce^{-kt/m} \implies u=u_0e^{-kt/m}$
(b) The general solution can be determined when we separate the variables factor and integrate them on the both sides.
Thus, we have $\int \dfrac{1}{u} du=\int \dfrac{-k}{m} dt$
$\implies \ln |u|=\dfrac{-kt}{m}+c'$
or, $u=e^ce^{-(kt/m)}$
After applying the initial conditions, we have: $ c=u_0$
So, $u=e^ce^{-(kt/m)} \implies u=u_0e^{-kt/m}$
