#### Answer

$$y = \frac{{3 + 27{e^{{x^2}}}}}{2}$$

#### Work Step by Step

$$\eqalign{
& x\frac{{dy}}{{dx}} - 2{x^2}y + 3{x^2} = 0;\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 15 \cr
& {\text{subtracting }}3{x^2}{\text{ from each side}} \cr
& x\frac{{dy}}{{dx}} - 2{x^2}y = - 3{x^2} \cr
& {\text{divide by }}x \cr
& \frac{{dy}}{{dx}} - 2xy = - 3x \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} - 2xy = - 3x{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can see that }}P\left( x \right){\text{ is }} - 2x \cr
& \,{\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int { - 2x} dx}} = {e^{ - {x^2}}} \cr
& {\text{multiplying both sides of the differential }}\frac{{dy}}{{dx}} + 3{x^2}y = {x^2}{\text{ equation by }}{e^{{x^3}}} \cr
& {e^{ - {x^2}}}\frac{{dy}}{{dx}} - 2x{e^{ - {x^2}}}y = - 3x{e^{ - {x^2}}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{e^{ - {x^2}}}y} \right] = - 3x{e^{ - {x^2}}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {e^{ - {x^2}}}y = \int { - 3x{e^{ - {x^2}}}} dx \cr
& {e^{{x^3}}}y = \frac{3}{2}{e^{ - {x^2}}} + C \cr
& y = \frac{3}{2} + C{e^{{x^2}}}\,\,\,\,\,\left( {\bf{1}} \right) \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 15 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& 15 = \frac{3}{2} + C{e^{{0^2}}} \cr
& C = 15 - \frac{3}{2} \cr
& C = \frac{{27}}{2} \cr
& {\text{substitute }}C = \frac{{27}}{2}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = \frac{3}{2} + \frac{{27}}{2}{e^{{x^2}}} \cr
& y = \frac{{3 + 27{e^{{x^2}}}}}{2} \cr} $$