Answer
$$y = \frac{{{e^{2{x^3}}} + 5{e^2}}}{{6{x^4}}}$$
Work Step by Step
$$\eqalign{
& {x^2}\frac{{dy}}{{dx}} + 4xy - {e^{2{x^3}}} = 0;\,\,\,\,\,\,\,\,\,y\left( 1 \right) = {e^2} \cr
& {\text{Adding }}{e^{2{x^3}}}{\text{ to each side}} \cr
& {x^2}\frac{{dy}}{{dx}} + 4xy = {e^{2{x^3}}} \cr
& {\text{divide by }}{x^2} \cr
& \frac{{dy}}{{dx}} + \frac{4}{x}y = \frac{{{e^{2{x^3}}}}}{{{x^2}}} \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{4}{x}y = \frac{{{e^{2{x^3}}}}}{{{x^2}}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can see that }}P\left( x \right){\text{ is }}\frac{4}{x} \cr
& \,{\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{4}{x}} dx}} = {e^{4\ln \left| x \right|}} = {e^{\ln {x^4}}} = {x^4} \cr
& {\text{multiplying both sides of the differential }}\frac{{dy}}{{dx}} + \frac{4}{x}y = \frac{{{e^{2{x^3}}}}}{{{x^2}}}{\text{ equation by }}{x^4} \cr
& {x^4}\frac{{dy}}{{dx}} + 4{x^3}y = {x^2}{e^{2{x^3}}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{x^4}y} \right] = {x^2}{e^{2{x^3}}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {x^4}y = \int {{x^2}{e^{2{x^3}}}} dx \cr
& {x^4}y = \frac{1}{6}\int {{e^{2{x^3}}}\left( {6{x^2}} \right)} dx \cr
& {x^4}y = \frac{1}{6}{e^{2{x^3}}} + C \cr
& y = \frac{1}{{6{x^4}}}{e^{2{x^3}}} + \frac{C}{{{x^4}}}\,\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = {e^2} \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& {e^2} = \frac{1}{{6{{\left( 1 \right)}^4}}}{e^{2{{\left( 1 \right)}^3}}} + \frac{C}{{{1^4}}} \cr
& {e^2} = \frac{1}{6}{e^2} + C \cr
& C = \frac{5}{6}{e^2} \cr
& {\text{substitute }}C = \frac{5}{6}{e^2}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = \frac{1}{{6{x^4}}}{e^{2{x^3}}} + \frac{{5{e^2}}}{{6{x^4}}} \cr
& y = \frac{{{e^{2{x^3}}} + 5{e^2}}}{{6{x^4}}} \cr} $$
