## Calculus with Applications (10th Edition)

$$y = k{e^{ - {e^{ - x}}}} + 3$$
\eqalign{ & \frac{{dy}}{{dx}} = \frac{{3 - y}}{{{e^x}}} \cr & {\text{Separating variables leads to}} \cr & \frac{1}{{3 - y}}dy = \frac{1}{{{e^x}}}dx \cr & \frac{1}{{3 - y}}dy = {e^{ - x}}dx \cr & {\text{multiply both sides by }} - 1 \cr & \frac{1}{{y - 3}}dy = - {e^{ - x}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{{y - 3}}} = \int {\left( { - {e^{ - x}}} \right)dx} \cr & {\text{integrating we obtain}} \cr & \ln \left| {y - 3} \right| = - {e^{ - x}} + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| {y - 3} \right|}} = {e^{ - {e^{ - x}} + C}} \cr & {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5 in the section 10}}{\text{.1}}} \right) \cr & {e^{\ln \left| {y - 3} \right|}} = {e^C}{e^{ - {e^{ - x}}}} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & {e^{\ln \left| {y - 3} \right|}} = k{e^{ - {e^{ - x}}}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y - 3 = k{e^{ - {e^{ - x}}}} \cr & y = k{e^{ - {e^{ - x}}}} + 3 \cr}