Chapter 10 - Differential Equations - 10.2 Linear First-Order Differential Equations - 10.2 Exercises - Page 544: 17

$$y = \frac{{{x^2}}}{7} + \frac{{2560}}{{7{x^5}}}$$

$$\eqalign{ & x\frac{{dy}}{{dx}} + 5y = {x^2},\,\,\,\,\,\,\,\,\,\,\,y\left( 2 \right) = 12 \cr & {\text{divide by x}} \cr & \frac{{dy}}{{dx}} + \frac{5}{x}y = x \cr & {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{comparing }}\frac{{dy}}{{dx}} + \frac{5}{x}y = x{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }}\frac{5}{x}{\text{ }} \cr & {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {P\left( x \right)} dx}} = {e^{\int {\frac{5}{x}dx} }} = {e^{5\operatorname{lnx} }} = {x^5} \cr & {\text{multiplying both sides of the differential equation by }}{e^x} \cr & {x^5}\frac{{dy}}{{dx}} + 5{x^4}y = {x^6} \cr & {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr & {D_x}\left[ {{x^5}y} \right] = {x^6} \cr & {\text{solve for }}y{\text{ integrating both sides}} \cr & {x^5}y = \int {{x^6}} dx \cr & {x^5}y = \frac{{{x^7}}}{7} + C \cr & y = \frac{{{x^2}}}{7} + \frac{C}{{{x^5}}}\,\,\,\,\,\left( {\bf{1}} \right) \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 2 \right) = 12 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & 12 = \frac{{{{\left( 2 \right)}^2}}}{7} + \frac{C}{{{{\left( 2 \right)}^5}}} \cr & \frac{{80}}{7} = \frac{C}{{32}} \cr & C = \frac{{2560}}{7} \cr & {\text{substitute }}C = \frac{{2560}}{7}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = \frac{2}{3} + \frac{{22}}{3}{x^3} \cr & y = \frac{{{x^2}}}{7} + \frac{{2560}}{{7{x^5}}} \cr} $$