Answer
$$P = - \frac{N}{k} + \left( {{P_0} + \frac{N}{k}} \right){e^{kt}}$$
Work Step by Step
$$\eqalign{
& {\text{We have the differential equation }} \cr
& \frac{{dP}}{{dt}} = kP + N \cr
& \frac{{dP}}{{dt}} - kP = N,{\text{ }}N{\text{ is constant}} \cr
& {\text{The differential equation has the form }}\frac{{dP}}{{dt}} + S\left( t \right)y = Q\left( t \right) \cr
& {\text{With }}S\left( t \right) = - k{\text{ }}Q\left( t \right) = N \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {S\left( t \right)} dt}} \cr
& I\left( t \right) = {e^{ - \int {kdt} }} = {e^{ - kt}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{ - kt}}\left( {\frac{{dP}}{{dt}} - kP} \right) = N{e^{ - kt}} \cr
& {e^{ - kt}}\frac{{dP}}{{dt}} - {e^{ - kt}}kP = N{e^{ - kt}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dt}}\left[ {I\left( t \right)P} \right] \cr
& \frac{d}{{dt}}\left[ {{e^{ - kt}}P} \right] = N{e^{ - kt}} \cr
& d\left[ {{e^{ - kt}}P} \right] = N{e^{ - kt}}dt \cr
& {\text{Integrate both sides}} \cr
& {e^{ - kt}}y = \int {N{e^{ - kt}}} dt \cr
& {e^{ - kt}}y = \frac{N}{{ - k}}{e^{ - kt}} + C \cr
& {\text{Solve for }}y \cr
& P = - \frac{N}{k} + C{e^{kt}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}P\left( 0 \right) = {P_0} \cr
& {P_0} = - \frac{N}{k} + C{e^0} \cr
& C = {P_0} + \frac{N}{k} \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& P = - \frac{N}{k} + \left( {{P_0} + \frac{N}{k}} \right){e^{kt}} \cr} $$
