## Calculus with Applications (10th Edition)

$$y = - \frac{{{x^3}}}{2} + Cx$$
\eqalign{ & y - x\frac{{dy}}{{dx}} = {x^3},\,\,\,\,\,\,x > 0 \cr & {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{dividing both sides of the equation by }} - x \cr & - \frac{1}{x}y + \frac{{dy}}{{dx}} = - {x^2} \cr & or \cr & \frac{{dy}}{{dx}} - \frac{1}{x}y = - {x^2} \cr & {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{we can note that }}P\left( x \right){\text{ is }} - \frac{1}{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{1}{x}} dx}} = {e^{ - \ln x}} = {e^{\ln {x^{ - 1}}}} = {x^{ - 1}} = \frac{1}{x} \cr & {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} - \frac{1}{x}y = - {x^2}{\text{ by }}\frac{1}{x} \cr & {\text{ }}\frac{1}{x}\frac{{dy}}{{dx}} - \frac{1}{{{x^2}}}y = - x \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}} \right] = - x \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & \frac{y}{x} = \int {\left( { - x} \right)} dx \cr & \frac{y}{x} = - \frac{{{x^2}}}{2} + C \cr & y = - \frac{{{x^3}}}{2} + Cx \cr}