Chapter 6 - Differential Equations - 6.4 Exercises - Page 428: 26

$$A = \frac{P}{r}\left( {{e^{rt}} - 1} \right)$$

$$\eqalign{ & {\text{We have the differential equation }} \cr & \frac{{dA}}{{dt}} = rA + P \cr & \frac{{dA}}{{dt}} - rA = P,{\text{ }}P{\text{ is constant}} \cr & {\text{The differential equation has the form }}\frac{{dA}}{{dt}} + S\left( t \right)A = Q\left( t \right) \cr & {\text{With }}S\left( t \right) = - r{\text{ }}Q\left( t \right) = P \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {S\left( t \right)} dt}} \cr & I\left( t \right) = {e^{ - \int {rdt} }} = {e^{ - rt}} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & {e^{ - rt}}\left( {\frac{{dA}}{{dt}} - rA} \right) = P{e^{ - rt}} \cr & {e^{ - rt}}\frac{{dA}}{{dt}} - {e^{ - rt}}rA = P{e^{ - rt}} \cr & {\text{Write the left side in the form }}\frac{d}{{dt}}\left[ {I\left( t \right)P} \right] \cr & \frac{d}{{dt}}\left[ {A{e^{ - rt}}} \right] = P{e^{ - rt}} \cr & d\left[ {A{e^{ - rt}}} \right] = P{e^{ - rt}}dt \cr & {\text{Integrate both sides}} \cr & A{e^{ - rt}} = \int {P{e^{ - rt}}} dt \cr & A{e^{ - rt}} = \frac{P}{{ - r}}{e^{ - rt}} + C \cr & A{e^{ - rt}} = \frac{P}{{ - r}}{e^{ - rt}} + C \cr & {\text{Solve for }}A \cr & A = - \frac{P}{r} + C{e^{rt}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}A\left( 0 \right) = 0 \cr & 0 = - \frac{P}{r} + C{e^0}{\text{ }} \cr & C = \frac{P}{r} \cr & {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr & A = - \frac{P}{r} + \frac{P}{r}{e^{rt}} \cr & {\text{Factoring}} \cr & A = \frac{P}{r}\left( {{e^{rt}} - 1} \right) \cr} $$