#### Answer

$$y = kx$$

#### Work Step by Step

$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{y}{x},{\text{ with }}x > 0 \cr
& {\text{Separating variables leads to}} \cr
& \frac{1}{y}dy = \frac{1}{x}dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {\frac{1}{y}dy} = \int {\frac{1}{x}dx} \cr
& {\text{integrating we obtain}} \cr
& \ln \left| y \right| = \ln \left| x \right| + C \cr
& {\text{solving the equation for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{\ln \left| x \right| + C}} \cr
& {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5}}} \right) \cr
& {e^{\ln \left| y \right|}} = {e^C}{e^{\ln \left| x \right|}} \cr
& {\text{simplify by using the logarithmic properties}} \cr
& y = {e^C}x \cr
& {\text{replace the constant }}{e^C}{\text{ with }}k \cr
& y = kx \cr
& \cr
& {\text{verifying that the solution satisfies the original differential equation}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {kx} \right] \cr
& \frac{{dy}}{{dx}} = k \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = k{\text{ and }}y = kx{\text{ into }}\frac{{dy}}{{dx}} = \frac{y}{x} \cr
& k = \frac{{kx}}{x} \cr
& k = k \cr
& {\text{The general solution is verified}} \cr} $$