#### Answer

$$y = \frac{{k{x^2} - 1}}{2}$$

#### Work Step by Step

$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{2y + 1}}{x} \cr
& {\text{Separating variables leads to}} \cr
& \frac{{dy}}{{2y + 1}} = \frac{1}{x}dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {\frac{{dy}}{{2y + 1}}} = \int {\frac{1}{x}dx} \cr
& {\text{integrating we obtain}} \cr
& \frac{1}{2}\ln \left| {2y + 1} \right| = \ln \left| x \right| + C \cr
& {\text{multiplying by 2}} \cr
& \ln \left| {2y + 1} \right| = 2\ln \left| x \right| + C \cr
& {\text{solving the equation for }}y \cr
& {e^{\ln \left| {2y + 1} \right|}} = {e^{2\ln \left| x \right| + C}} \cr
& {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5 in the section 10}}{\text{.1}}} \right) \cr
& {e^{\ln \left| {2y + 1} \right|}} = {e^C}{e^{2\ln \left| x \right|}} \cr
& {\text{replace the constant }}{e^C}{\text{ with }}k \cr
& {e^{\ln \left| {2y + 1} \right|}} = k{e^{2\ln \left| x \right|}} \cr
& {\text{simplify by using the logarithmic properties}} \cr
& 2y + 1 = k{x^2} \cr
& y = \frac{{k{x^2} - 1}}{2} \cr} $$