Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 13

$${y^2} = k{e^x} - 6$$

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{y^2} + 6}}{{2y}} \cr & {\text{Separating variables leads to}} \cr & \frac{{2y}}{{{y^2} + 6}}dy = dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{{2y}}{{{y^2} + 6}}dy} = \int {dx} \cr & {\text{integrating by using the logarithmic rule and the power rule we obtain}} \cr & \ln \left( {{y^2} + 6} \right) = x + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left( {{y^2} + 6} \right)}} = {e^{x + C}} \cr & {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5}}} \right) \cr & {e^{\ln \left( {{y^2} + 6} \right)}} = {e^C}{e^x} \cr & {\text{simplify by using the logarithmic properties}} \cr & {y^2} + 6 = {e^C}{e^x} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & {y^2} + 6 = k{e^x} \cr & {y^2} = k{e^x} - 6 \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{d}{{dx}}\left[ {{y^2}} \right] = \frac{d}{{dx}}\left[ {k{e^x} - 6} \right] \cr & 2y\frac{{dy}}{{dx}} = k{e^x} \cr & \frac{{dy}}{{dx}} = \frac{{k{e^x}}}{{2y}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = \frac{{k{e^x}}}{{2y}}{\text{ into }}\frac{{dy}}{{dx}} = \frac{{{y^2} + 6}}{{2y}} \cr & \frac{{k{e^x}}}{{2y}} = \frac{{{y^2} + 6}}{{2y}} \cr & {\text{where }}{y^2} + 6 = k{e^x} \cr & \frac{{k{e^x}}}{{2y}} = \frac{{k{e^x}}}{{2y}} \cr & {\text{The general solution is verified}} \cr} $$