Chapter 10 - Differential Equations - 10.2 Linear First-Order Differential Equations - 10.2 Exercises - Page 543: 7

$$y = - \frac{1}{2} + C{e^{\frac{1}{2}{x^2}}}$$

$$\eqalign{ & 2\frac{{dy}}{{dx}} - 2xy - x = 0 \cr & {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{adding }}x{\text{ to both sides we obtain}} \cr & \frac{{dy}}{{dx}} - 2xy = x \cr & {\text{now}}{\text{, divide both sides of the equation by }}2 \cr & \frac{{dy}}{{dx}} - xy = \frac{x}{2} \cr & {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{ we can note that }}P\left( x \right){\text{ is }} - x \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int { - x} dx}} = {e^{ - \frac{1}{2}{x^2}}} \cr & {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} - xy = \frac{x}{2}{\text{ by }}{e^{ - \frac{1}{2}{x^2}}} \cr & {e^{ - \frac{1}{2}{x^2}}}\frac{{dy}}{{dx}} - xy{e^{ - \frac{1}{2}{x^2}}} = \frac{x}{2}{e^{ - \frac{1}{2}{x^2}}} \cr & {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr & {D_x}\left[ {{e^{ - \frac{1}{2}{x^2}}}y} \right] = \frac{x}{2}{e^{ - \frac{1}{2}{x^2}}} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{ - \frac{1}{2}{x^2}}}y = \int {\frac{x}{2}{e^{ - \frac{1}{2}{x^2}}}} dx \cr & {e^{ - \frac{1}{2}{x^2}}}y = - \frac{1}{2}\int {{e^{ - \frac{1}{2}{x^2}}}\left( { - x} \right)} dx \cr & {e^{ - \frac{1}{2}{x^2}}}y = - \frac{1}{2}{e^{ - \frac{1}{2}{x^2}}} + C \cr & y = - \frac{1}{2} + \frac{C}{{{e^{ - \frac{1}{2}{x^2}}}}} \cr & {\text{or}} \cr & y = - \frac{1}{2} + C{e^{\frac{1}{2}{x^2}}} \cr} $$