## Calculus with Applications (10th Edition)

$$y = kx$$
\eqalign{ & \frac{{dy}}{{dx}} = \frac{y}{x},{\text{ with }}x > 0 \cr & {\text{Separating variables leads to}} \cr & \frac{1}{y}dy = \frac{1}{x}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{y}dy} = \int {\frac{1}{x}dx} \cr & {\text{integrating we obtain}} \cr & \ln \left| y \right| = \ln \left| x \right| + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{\ln \left| x \right| + 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^{\ln \left| x \right|}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y = {e^C}x \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & y = kx \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {kx} \right] \cr & \frac{{dy}}{{dx}} = k \cr & {\text{substitute }}\frac{{dy}}{{dx}} = k{\text{ and }}y = kx{\text{ into }}\frac{{dy}}{{dx}} = \frac{y}{x} \cr & k = \frac{{kx}}{x} \cr & k = k \cr & {\text{The general solution is verified}} \cr}