Answer
$V(t)=\frac{ kP_0e^{-mt}}{m}+V_0-\frac{ kP_0}{m} $
Work Step by Step
$V{'}(t)=-kP(t)$...................... eq (1)
$P(t)=P_0e^{-mt}$ ......................... eq (2)
From eq(1) and eq(2)
$V^{'}(t)=-kP_0e^{-mt}$
Taking antiderivative with respect to t
$V(t)=\frac{ -kP_0e^{-mt}}{-m}+C$
$V(t)=\frac{ kP_0e^{-mt}}{m}+C$ ....................... eq (3)
Putting t=0
$V(o)=\frac{ kP_0e^{-m(0)}}{m}+C$
$V_0=\frac{ kP_0}{m}+C$
$V_0-\frac{ kP_0}{m}=C$
Putting in eq (3)
$V(t)=\frac{ kP_0e^{-mt}}{m}+V_0-\frac{ kP_0}{m} $
