## Calculus with Applications (10th Edition)

$$y = - \frac{1}{2}{x^2} - \frac{1}{2}x - \frac{1}{4} + C{e^{2x}}$$
\eqalign{ & 2xy + {x^3} = x\frac{{dy}}{{dx}} \cr & {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{subtracting }}2xy{\text{ from each side}} \cr & {x^3} = x\frac{{dy}}{{dx}} - 2xy \cr & {\text{dividing both sides of the equation by }}x \cr & {x^2} = \frac{{dy}}{{dx}} - 2y \cr & or \cr & \frac{{dy}}{{dx}} - 2y = {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 }} - 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 2 dx}} = {e^{ - 2x}} \cr & {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} - 2y = {x^2}{\text{ by }}{e^{ - 2x}} \cr & {e^{ - 2x}}\frac{{dy}}{{dx}} - 2{e^{ - 2x}}y = {x^2}{e^{ - 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^{ - 2x}}y} \right] = {x^2}{e^{ - 2x}} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{ - 2x}}y = \int {{x^2}{e^{ - 2x}}} dx \cr & {\text{integrating the right side using tabular integration we obtain }} \cr & {e^{ - 2x}}y = - \frac{1}{2}{x^2}{e^{ - 2x}} - \frac{1}{2}x{e^{ - 2x}} - \frac{1}{4}{e^{ - 2x}} + C \cr & y = - \frac{1}{2}{x^2} - \frac{1}{2}x - \frac{1}{4} + \frac{C}{{{e^{ - 2x}}}} \cr & y = - \frac{1}{2}{x^2} - \frac{1}{2}x - \frac{1}{4} + C{e^{2x}} \cr}