Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 646: 18

$y = - 2x + {x^2}{\text{ }}$

$$\eqalign{ & xy' - 2y = 2x,{\text{ }}y\left( 2 \right) = 0 \cr & {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr & x\frac{{dy}}{{dx}} - 2y = 2x \cr & \frac{{dy}}{{dx}} - \frac{2}{x}y = 2 \cr & {\text{The differential equation is in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = - \frac{2}{x}{\text{ and }}Q\left( x \right) = 2 \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{ - \int {\frac{2}{x}dx} }} = {e^{ - 2\ln \left| x \right|}} = {x^{ - 2}} = \frac{1}{{{x^2}}} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & \left( {\frac{{dy}}{{dx}} - \frac{2}{x}y} \right)\frac{1}{{{x^2}}} = \frac{2}{{{x^2}}} \cr & \frac{1}{{{x^2}}}\frac{{dy}}{{dx}} - \frac{2}{{{x^3}}}y = \frac{2}{{{x^2}}} \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {\frac{y}{{{x^2}}}} \right] = \frac{2}{{{x^2}}} \cr & d\left[ {\frac{y}{{{x^2}}}} \right] = \frac{2}{{{x^2}}}dx \cr & {\text{Integrate both sides}} \cr & \frac{y}{{{x^2}}} = \int {\frac{2}{{{x^2}}}} dx{\text{ }} \cr & \frac{y}{{{x^2}}} = - \frac{2}{x} + C \cr & {\text{Solve for }}y \cr & \frac{y}{{{x^2}}} = - \frac{{2{x^2}}}{x} + C{x^2} \cr & y = - 2x + C{x^2}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}y\left( 2 \right) = 0 \cr & 0 = - 2\left( 2 \right) + C{\left( 2 \right)^2} \cr & 0 = - 4 + C\left( 4 \right) \cr & C = 1 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & y = - 2x + {x^2}{\text{ }} \cr} $$