## Calculus with Applications (10th Edition)

$$y = k{e^{{x^2}}}$$
\eqalign{ & \frac{{dy}}{{dx}} = 2xy \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = 2xdx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{{dy}}{y}} = \int {2xdx} \cr & {\text{integrating we obtain}} \cr & \ln \left| y \right| = {x^2} + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{{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^2}}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y = {e^C}{e^{{x^2}}} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & \boxed{y = k{e^{{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^2}}}} \right] \cr & \frac{{dy}}{{dx}} = 2kx{e^{{x^2}}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = 2kx{e^{{x^2}}}{\text{ and }}y = k{e^{{x^2}}}{\text{ into }}\frac{{dy}}{{dx}} = 2xy \cr & 2kx{e^{{x^2}}} = 2x\left( {k{e^{{x^2}}}} \right) \cr & 2kx{e^{{x^2}}} = 2kx{e^{{x^2}}} \cr & {\text{The general solution is verified}} \cr}