Chapter 10 - Differential Equations - 10.2 Linear First-Order Differential Equations - 10.2 Exercises - Page 544: 18

$$y = - \frac{5}{4} + \frac{{45}}{4}{e^{{x^2} - 1}}$$

$$\eqalign{ & 2\frac{{dy}}{{dx}} - 4xy = 5x,\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = 10 \cr & {\text{divide by 2}} \cr & \frac{{dy}}{{dx}} - 2xy = \frac{5}{2}x \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}} - 2xy = \frac{5}{2}x{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is}}\,\, - 2x{\text{ }} \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int { - 2xdx} }} = {e^{ - {x^2}}} \cr & {\text{multiplying both sides of the differential equation by }}{e^x} \cr & {e^{ - {x^2}}}\frac{{dy}}{{dx}} - 2xy{e^{ - {x^2}}} = \frac{5}{2}x{e^{ - {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^{ - {x^2}}}y} \right] = \frac{5}{2}x{e^{ - {x^2}}} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {e^{ - {x^2}}}y = \int {\frac{5}{2}x{e^{ - {x^2}}}} dx \cr & or \cr & {e^{ - {x^2}}}y = - \frac{5}{4}\int {\left( { - 2x} \right){e^{ - {x^2}}}} dx \cr & {e^{ - {x^2}}}y = - \frac{5}{4}{e^{ - {x^2}}} + C \cr & y = - \frac{5}{4} + \frac{C}{{{e^{ - {x^2}}}}} \cr & y = - \frac{5}{4} + C{e^{{x^2}}}\,\,\,\,\,\left( {\bf{1}} \right) \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = 10 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & 10 = - \frac{5}{4} + C{e^{{{\left( 1 \right)}^2}}} \cr & 10 = - \frac{5}{4} + C{e^{{{\left( 1 \right)}^2}}} \cr & \frac{{45}}{4} = Ce \cr & C = \frac{{45}}{{4e}} \cr & {\text{substitute }}C = \frac{{45}}{{4e}}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = - \frac{5}{4} + \left( {\frac{{45}}{{4e}}} \right){e^{{x^2}}} \cr & y = - \frac{5}{4} + \frac{{45}}{4}{e^{{x^2} - 1}} \cr} $$