Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 646: 17

$y = {x^2} + \frac{3}{x}$

$$\eqalign{ & xy' + y = 3{x^2},{\text{ }}y\left( 1 \right) = 4 \cr & {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr & x\frac{{dy}}{{dx}} + y = 3{x^2} \cr & \frac{{dy}}{{dx}} + \frac{1}{x}y = 3x \cr & {\text{The differential equation is in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = \frac{1}{x}{\text{ and }}Q\left( x \right) = 3x \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{\int {\frac{1}{x}dx} }} = {e^{\ln \left| x \right|}} = x \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & x\left( {\frac{{dy}}{{dx}} + \frac{1}{x}y} \right) = 3{x^2} \cr & \frac{{dy}}{{dx}} + y = 3{x^2} \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {xy} \right] = 3{x^2} \cr & d\left[ {xy} \right] = 3{x^2}dx \cr & {\text{Integrate both sides}} \cr & xy = \int {3{x^2}} dx{\text{ }} \cr & xy = {x^3} + C \cr & {\text{Solve for }}y \cr & y = {x^2} + \frac{C}{x}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}y\left( 1 \right) = 4 \cr & 4 = {\left( 1 \right)^2} + \frac{C}{{\left( 1 \right)}} \cr & C = 3 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & y = {x^2} + \frac{3}{x} \cr} $$