Answer
$$y = \frac{{2 + 22{x^3}}}{3}$$
Work Step by Step
$$\eqalign{
& x\frac{{dy}}{{dx}} - 3y + 2 = 0,\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = 8 \cr
& {\text{subtacting 2 from each side}} \cr
& x\frac{{dy}}{{dx}} - 3y = - 2 \cr
& {\text{divide by x}} \cr
& \frac{{dy}}{{dx}} - \frac{3}{x}y = - \frac{2}{x} \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} - \frac{3}{x}y = - \frac{2}{x}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }} - \frac{3}{x}{\text{ }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int { - \frac{3}{x}dx} }} = {e^{ - 3\operatorname{lnx} }} = {e^{\ln {x^{ - 3}}}} = \frac{1}{{{x^3}}} \cr
& {\text{multiplying both sides of the differential equation by }}{e^x} \cr
& \frac{1}{{{x^3}}}\frac{{dy}}{{dx}} - \frac{3}{{{x^4}}}y = - \frac{2}{{{x^4}}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {\frac{y}{{{x^3}}}} \right] = - \frac{2}{{{x^4}}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& \frac{y}{{{x^3}}} = \int { - \frac{2}{{{x^4}}}} dx \cr
& \frac{y}{{{x^3}}} = - 2\int {{x^{ - 4}}} dx \cr
& \frac{y}{{{x^3}}} = - 2\left( {\frac{{{x^{ - 3}}}}{{ - 3}}} \right) + C \cr
& \frac{y}{{{x^3}}} = \frac{2}{{3{x^3}}} + C \cr
& y = \frac{{2{x^3}}}{{3{x^3}}} + C{x^3} \cr
& y = \frac{2}{3} + C{x^3}\,\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = 8 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& 8 = \frac{2}{3} + C{\left( 1 \right)^3} \cr
& C = 8 - \frac{2}{3} \cr
& C = \frac{{22}}{3} \cr
& {\text{substitute }}C = \frac{{22}}{3}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = \frac{2}{3} + \frac{{22}}{3}{x^3} \cr
& y = \frac{{2 + 22{x^3}}}{3} \cr} $$
