Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 9

$$y = k{e^{{x^3} - {x^2}}}$$

$$\eqalign{ & \frac{{dy}}{{dx}} = 3{x^2}y - 2xy \cr & {\text{factor the right side of the equation}} \cr & \frac{{dy}}{{dx}} = \left( {3{x^2} - 2x} \right)y \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = \left( {3{x^2} - 2x} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{{dy}}{y}} = \int {\left( {3{x^2} - 2x} \right)} dx \cr & {\text{integrating by using the power rule we obtain}} \cr & \ln \left| y \right| = 3\left( {\frac{{{x^3}}}{3}} \right) - 2\left( {\frac{{{x^2}}}{2}} \right) + C \cr & \ln \left| y \right| = {x^3} - {x^2} + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{{x^3} - {x^2} + C}} \cr & {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5}}} \right) \cr & {e^{\ln \left| y \right|}} = {e^C}{e^{{x^3} - {x^2}}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y = {e^C}{e^{{x^3} - {x^2}}} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & y = k{e^{{x^3} - {x^2}}} \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {k{e^{{x^3} - {x^2}}}} \right] \cr & \frac{{dy}}{{dx}} = k\left( {3{x^2} - 2x} \right){e^{{x^3} - {x^2}}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = k\left( {3{x^2} - 2x} \right){e^{{x^3} - {x^2}}}{\text{ and }}y = k{e^{{x^3} - {x^2}}}{\text{ into }}\frac{{dy}}{{dx}} = \left( {3{x^2} - 2x} \right)y \cr & k\left( {3{x^2} - 2x} \right){e^{{x^3} - {x^2}}} = \left( {3{x^2} - 2x} \right)\left( {k{e^{{x^3} - {x^2}}}} \right) \cr & k\left( {3{x^2} - 2x} \right){e^{{x^3} - {x^2}}} = k\left( {3{x^2} - 2x} \right){e^{{x^3} - {x^2}}} \cr & {\text{The general solution is verified}} \cr} $$