## Calculus with Applications (10th Edition)

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