Answer
$$y = \frac{x}{{2\cosh x}} + \frac{1}{2}\sinh x + 2$$
Work Step by Step
$$\eqalign{
& y'\cosh x + y\sinh x = {\cosh ^2}x,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{initial condition }}y\left( 0 \right) = 2 \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \cosh x\frac{{dy}}{{dx}} + y\sinh x = {\cosh ^2}x \cr
& \frac{{dy}}{{dx}} + \frac{{\sinh x}}{{\cosh x}}y = \frac{{{{\cosh }^2}x}}{{\cosh x}} \cr
& \frac{{dy}}{{dx}} + \tanh xy = \cosh x \cr
& \cr
& {\text{The equation is in the form }}\frac{{dy}}{{dp}} + p\left( x \right)y = q\left( x \right) \cr
& {\text{with }}p\left( x \right) = \tanh x{\text{ and }}q\left( x \right) = \cosh x \cr
& \cr
& {\text{Use the method of the integrating factors}}{\text{, calculate }}\mu = {e^{\int {p\left( x \right)dx} }} \cr
& \mu = {e^{\int {p\left( x \right)dx} }} = {e^{\int {\tanh x} dx}} \cr
& \mu = {e^{\ln \left| {\cosh x} \right|}} = \cosh x \cr
& \cr
& {\text{Multiply both sides of the equation by }}\mu \cr
& \cosh x\frac{{dy}}{{dx}} + \cosh x\tanh xy = \cosh x\cosh x \cr
& \cosh x\frac{{dy}}{{dx}} + \sinh xy = {\cosh ^2}x \cr
& {\text{Express the result as }}\frac{d}{{dx}}\left[ {\mu y} \right] = {e^x}{\text{. Then}}{\text{,}} \cr
& \frac{d}{{dx}}\left[ {y\cosh x} \right] = {\cosh ^2}x \cr
& \cr
& {\text{Integrate both sides}} \cr
& y\cosh x = \int {{{\cosh }^2}x} dx \cr
& {\text{use the identity cos}}{{\text{h}}^2}x = \frac{{1 + \cosh 2x}}{2} \cr
& y\cosh x = \int {\left( {\frac{{1 + \cosh 2x}}{2}} \right)} dx \cr
& y\cosh x = \frac{1}{2}x + \frac{1}{4}\sinh 2x + C \cr
& y\cosh x = \frac{1}{2}x + \frac{1}{4}\left( {2\sinh x\cosh x} \right) + C \cr
& \cr
& {\text{Solve for }}y \cr
& y = \frac{x}{{2\cosh x}} + \frac{1}{2}\sinh x + C \cr
& \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 2 \cr
& 2 = \frac{0}{{2\cosh \left( 0 \right)}} + \frac{1}{2}\sinh \left( 0 \right) + C \cr
& C = 2 \cr
& \cr
& {\text{Then}}{\text{, the particular solution of the differential equation is}} \cr
& y = \frac{x}{{2\cosh x}} + \frac{1}{2}\sinh x + 2 \cr} $$
