Chapter 6 - Differential Equations - 6.4 Exercises - Page 430: 47

$$\eqalign{ & y = \frac{1}{2}{x^3} - 4x{\text{ }}\left( {{\text{Particular solution 1}}} \right) \cr & y = \frac{1}{2}{x^3} + 2x{\text{ }}\left( {{\text{Particular solution 2}}} \right) \cr} $$

$$\eqalign{ & \frac{{dy}}{{dx}} - \frac{1}{x}y = {x^2} \cr & {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr & {\text{With }}P\left( x \right) = - \frac{1}{x}{\text{ and }}Q\left( x \right) = {x^2} \cr & {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr & I\left( x \right) = {e^{ - \int {\frac{1}{x}dx} }} = {e^{ - \ln x}} = \frac{1}{x} \cr & {\text{Multiply the differential equation by the integrating factor}} \cr & \left( {\frac{{dy}}{{dx}} - \frac{1}{x}y} \right) = \frac{1}{x}\left( {{x^2}} \right) \cr & \frac{1}{x}\frac{{dy}}{{dx}} - \frac{1}{{{x^2}}}y = x \cr & {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr & \frac{d}{{dx}}\left[ {\frac{y}{x}} \right] = x \cr & d\left[ {\frac{y}{x}} \right] = xdx \cr & {\text{Integrate both sides}} \cr & \frac{y}{x} = \int x dx \cr & \frac{y}{x} = \frac{1}{2}{x^2} + C \cr & {\text{Solve for }}y \cr & y = \frac{1}{2}{x^3} + Cx \cr & {\text{Use the initial condition }}\left( { - 2,4} \right) \cr & 4 = \frac{1}{2}{\left( { - 2} \right)^3} + C\left( { - 2} \right) \cr & C = - 4 \cr & y = \frac{1}{2}{x^3} - 4x{\text{ }}\left( {{\text{Particular solution 1}}} \right) \cr & {\text{Use the initial condition }}\left( {2,8} \right) \cr & 8 = \frac{1}{2}{\left( 2 \right)^3} + C\left( 2 \right) \cr & C = 2 \cr & y = \frac{1}{2}{x^3} + 2x{\text{ }}\left( {{\text{Particular solution 2}}} \right) \cr & \cr & {\text{Graph}} \cr} $$