Answer
$$y = x\ln \left| x \right| - 2 + 12x$$
Work Step by Step
$$\eqalign{
& xdy = \left( {x + y + 2} \right)dx,{\text{ }}y\left( 1 \right) = 10 \cr
& {\text{Divide by }}dx \cr
& x\frac{{dy}}{{dx}} = x + y + 2 \cr
& {\text{Divide by }}x \cr
& \frac{{dy}}{{dx}} = 1 + \frac{1}{x}y + \frac{2}{x} \cr
& \frac{{dy}}{{dx}} - \frac{1}{x}y = 1 + \frac{2}{x} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = - \frac{1}{x}{\text{ }}Q\left( x \right) = 1 + \frac{2}{x} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{ - \int {\frac{1}{x}dx} }} = {e^{ - \ln \left| x \right|}} = \frac{1}{x} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& \frac{1}{x}\left( {\frac{{dy}}{{dx}} - \frac{1}{x}y} \right) = \frac{1}{x}\left( {1 + \frac{2}{x}} \right) \cr
& \frac{1}{x}\frac{{dy}}{{dx}} - \frac{1}{{{x^2}}}y = \frac{1}{x} + \frac{2}{{{x^2}}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {\frac{1}{x}y} \right] = \frac{1}{x} + \frac{2}{{{x^2}}} \cr
& d\left[ {\frac{1}{x}y} \right] = \left( {\frac{1}{x} + \frac{2}{{{x^2}}}} \right)dx \cr
& {\text{Integrate both sides}} \cr
& \frac{1}{x}y = \int {\left( {\frac{1}{x} + \frac{2}{{{x^2}}}} \right)} dx \cr
& \frac{1}{x}y = \ln \left| x \right| - \frac{2}{x} + C \cr
& {\text{Solve for }}y \cr
& y = x\ln \left| x \right| - 2 + Cx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 1 \right) = 10 \cr
& 10 = \ln \left| 1 \right| - 2 + C \cr
& C = 12 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = x\ln \left| x \right| - 2 + 12x \cr} $$
