Chapter 6 - Differential Equations - 6.2 Exercises - Page 412: 23

$$\eqalign{ & y = 5{e^{ - \ln \root 4 \of {\frac{2}{5}} }}{e^{\ln \root 4 \of {\frac{2}{5}} t}} \cr & or \cr & y \approx 6.2872{e^{ - 0.2291t}} \cr} $$

$$\eqalign{ & {\text{The exponential function is in the form }}y = C{e^{kt}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Where }}C{\text{ and }}k{\text{ are constants}} \cr & \cr & {\text{From the graph we know the points }}\left( {1,5} \right){\text{ and }}\left( {5,2} \right) \cr & {\text{*Substituing the point }}\underbrace {\left( {1,5} \right)}_{\left( {t,y} \right)}{\text{ in the equation }}\left( {\bf{1}} \right){\text{we obtain:}} \cr & 5 = C{e^{k\left( 1 \right)}} \to 5 = C{e^k}{\text{, }}\left( {\bf{2}} \right) \cr & {\text{*Substituing the point }}\underbrace {\left( {5,2} \right)}_{\left( {t,y} \right)}{\text{ in the equation }}\left( {\bf{1}} \right){\text{we obtain:}} \cr & 2 = C{e^{k\left( 5 \right)}} \to 2 = C{e^{5k}}{\text{, }}\left( {\bf{3}} \right) \cr & \cr & {\text{*Solve the equation }}\left( {\bf{2}} \right){\text{ for }}C \cr & C = \frac{5}{{{e^k}}} \cr & {\text{Substitute the previous result in the equation }}\left( {\bf{3}} \right) \cr & 2 = \left( {\frac{5}{{{e^k}}}} \right){e^{5k}} \cr & \frac{2}{5} = {e^{4k}} \cr & {\text{Solve for }}k \cr & \ln \left( {\frac{2}{5}} \right) = 4k \cr & k = \frac{1}{4}\ln \left( {\frac{2}{5}} \right) \cr & k = \ln \root 4 \of {\frac{2}{5}} \cr & {\text{Substitute }}k\,{\text{into the equation }}\left( {\bf{2}} \right){\text{ or }}\left( {\bf{3}} \right) \cr & 5 = C{e^{\ln \root 4 \of {\frac{2}{5}} }} \cr & {\text{Solve for }}C \cr & 5 = C{e^{\ln \root 4 \of {\frac{2}{5}} }} \cr & C = 5{e^{ - \ln \root 4 \of {\frac{2}{5}} }} \cr & \cr & {\text{Substitute }}k = \ln \root 4 \of {\frac{2}{5}} {\text{ and }}C = 5{e^{ - \ln \root 4 \of {\frac{2}{5}} }}{\text{ into }}\left( {\bf{1}} \right) \cr & y = 5{e^{ - \ln \root 4 \of {\frac{2}{5}} }}{e^{\ln \root 4 \of {\frac{2}{5}} t}} \cr & or \cr & y \approx 6.2872{e^{ - 0.2291t}} \cr} $$