## Calculus with Applications (10th Edition)

$$y = \frac{{{e^{x - 1}} - 3}}{{{e^{x - 1}} - 2}}$$
\eqalign{ & \frac{{dy}}{{dx}} = {\left( {y - 1} \right)^2}{e^{x - 1}};\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = 2 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{{{{\left( {y - 1} \right)}^2}}} = {e^{x - 1}}dx \cr & {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr & {\left( {y - 1} \right)^{ - 2}}dy = {e^{x - 1}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{{\left( {y - 1} \right)}^{ - 2}}dy} = \int {{e^{x - 1}}dx} \cr & {\text{integrate by using the power rule for integration}} \cr & \frac{{{{\left( {y - 1} \right)}^{ - 1}}}}{{ - 1}} = {e^{x - 1}} + C \cr & - \frac{1}{{y - 1}} = {e^{x - 1}} + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = 2 \cr & \,\,y\left( 1 \right) = 2{\text{ implies that }}y = 2{\text{ when }}x = 1 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & - \frac{1}{{2 - 1}} = {e^{1 - 1}} + C \cr & - 1 = 1 + C \cr & C = - 2 \cr & {\text{substitute }}C = - 2{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & - \frac{1}{{y - 1}} = {e^{x - 1}} - 2 \cr & {\text{Solve the equation for }}y \cr & y - 1 = - \frac{1}{{{e^{x - 1}} - 2}} \cr & y = 1 - \frac{1}{{{e^{x - 1}} - 2}} \cr & y = \frac{{{e^{x - 1}} - 2 - 1}}{{{e^{x - 1}} - 2}} \cr & y = \frac{{{e^{x - 1}} - 3}}{{{e^{x - 1}} - 2}} \cr}