Answer
$$y = 5 - 4{e^{ - x}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 5 - y,{\text{ }}y\left( 0 \right) = 1 \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{5 - y}} = dx \cr
& {\text{Integrate both sides}} \cr
& - \ln \left| {5 - y} \right| = x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 1 \cr
& - \ln \left| {5 - 1} \right| = 0 + C \cr
& C = - \ln 4 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& - \ln \left| {5 - y} \right| = x - \ln 4 \cr
& {\text{Solve for }}y \cr
& {e^{ - \ln \left| {5 - y} \right|}} = {e^{x - \ln 4}} \cr
& {e^{\ln {{\left| {5 - y} \right|}^{ - 1}}}} = {e^x}{e^{ - \ln 4}} \cr
& \frac{1}{{5 - y}} = \frac{1}{4}{e^x} \cr
& 5 - y = 4{e^{ - x}} \cr
& y = 5 - 4{e^{ - x}} \cr
& \cr
& {\text{Graph}} \cr} $$
