Chapter 10 - Differential Equations - 10.2 Linear First-Order Differential Equations - 10.2 Exercises - Page 543: 8

$$y = - \frac{1}{6} + C{e^{ - {x^2}}}$$

$$\eqalign{ & 3\frac{{dy}}{{dx}} + 6xy + x = 0 \cr & {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{subtracting }}x{\text{ from both sides we obtain}} \cr & 3\frac{{dy}}{{dx}} + 6xy = - x \cr & {\text{now}}{\text{, divide both sides of the equation by }}3 \cr & \frac{{dy}}{{dx}} + 2xy = - \frac{x}{3} \cr & {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{ we can note that }}P\left( x \right){\text{ is }}2x \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {2x} dx}} = {e^{{x^2}}} \cr & {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} + 2xy = - \frac{x}{3}{\text{ by }}{e^{{x^2}}} \cr & {e^{{x^2}}}\frac{{dy}}{{dx}} + 2xy{e^{{x^2}}} = - \frac{{x{e^{{x^2}}}}}{3} \cr & {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr & {D_x}\left[ {{e^{{x^2}}}y} \right] = - \frac{{x{e^{{x^2}}}}}{3} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{{x^2}}}y = \int {\left( { - \frac{{x{e^{{x^2}}}}}{3}} \right)} dx \cr & {e^{{x^2}}}y = - \frac{1}{3}\int {x{e^{{x^2}}}} dx \cr & {e^{{x^2}}}y = - \frac{1}{6}{e^{{x^2}}} + C \cr & y = - \frac{1}{6} + \frac{C}{{{e^{{x^2}}}}} \cr & {\text{or}} \cr & y = - \frac{1}{6} + C{e^{ - {x^2}}} \cr} $$