## Calculus with Applications (10th Edition)

$$y = \frac{1}{3} + \frac{5}{{3{e^{{x^3}}}}}$$
\eqalign{ & \frac{{dy}}{{dx}} + 3{x^2}y = {x^2};\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 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}} + 3{x^2}y = {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 }}3{x^2} \cr & \,{\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {3{x^2}} dx}} = {e^{{x^3}}} \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^3}}}\frac{{dy}}{{dx}} + 3{x^2}{e^{{x^3}}}y = {x^2}{e^{{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[ {{e^{{x^3}}}y} \right] = {x^2}{e^{{x^3}}} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{{x^3}}}y = \int {{x^2}{e^{{x^3}}}} dx \cr & {e^{{x^3}}}y = \frac{1}{3}\int {{e^{{x^3}}}\left( {3{x^2}} \right)} dx \cr & {e^{{x^3}}}y = \frac{1}{3}{e^{{x^3}}} + C \cr & y = \frac{1}{3} + \frac{C}{{{e^{{x^3}}}}}\,\,\,\,\,\left( {\bf{1}} \right) \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 0 \right) = 2 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & 2 = \frac{1}{3} + \frac{C}{{{e^0}}} \cr & C = 2 - \frac{1}{3} \cr & C = \frac{5}{3} \cr & {\text{substitute }}C = \frac{5}{3}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = \frac{1}{3} + \frac{5}{{3{e^{{x^3}}}}} \cr}