## Calculus with Applications (10th Edition)

$$y = \frac{{{x^2} + 1000}}{{{e^{{x^3}}}}}$$
\eqalign{ & \frac{{dy}}{{dx}} + 3{x^2}y - 2x{e^{ - {x^3}}} = 0,\,\,\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 1000 \cr & {\text{add }}2x{e^{ - {x^3}}} \cr & \frac{{dy}}{{dx}} + 3{x^2}y = 2x{e^{ - {x^3}}} \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 = 2x{e^{ - {x^3}}}{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can see that }} \cr & P\left( x \right){\text{ is}}\,\,\,\,3{x^2}.\,\,\,{\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 = 2x{e^{ - {x^3}}}{\text{ equation by }}{e^{{x^3}}} \cr & {e^{{x^3}}}\frac{{dy}}{{dx}} + 3{x^2}y{e^{{x^3}}} = 2x \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] = 2x \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{{x^3}}}y = \int {2x} dx \cr & {e^{{x^3}}}y = {x^2} + C \cr & y = \frac{{{x^2}}}{{{e^{{x^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) = 1000 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & 1000 = \frac{0}{{{e^0}}} + \frac{C}{{{e^0}}} \cr & C = 1000 \cr & {\text{substitute }}C = 1000{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = \frac{{{x^2}}}{{{e^{{x^3}}}}} + \frac{{1000}}{{{e^{{x^3}}}}} \cr & y = \frac{{{x^2} + 1000}}{{{e^{{x^3}}}}} \cr}