Calculus with Applications (10th Edition)

$${e^{ - {y^2}}} = - 2x + C$$
\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{e^{{y^2}}}}}{y} \cr & {\text{Separating variables leads to}} \cr & \frac{y}{{{e^{{y^2}}}}}dy = dx \cr & {\text{use }}\frac{1}{{{a^n}}} = {a^{ - n}} \cr & y{e^{ - {y^2}}}dy = dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {y{e^{ - {y^2}}}dy} = \int {dx} \cr & {\text{rewrite the integral on the left side}} \cr & - \frac{1}{2}\int {{e^{ - {y^2}}}\left( { - 2y} \right)dy} = \int {dx} \cr & {\text{integrating by using }}\int {{e^u}du} = {e^u} + C{\text{ and the power rule }} \cr & - \frac{1}{2}{e^{ - {y^2}}} = x + C \cr & {\text{multiply both sides of the equation by }} - 2 \cr & {e^{ - {y^2}}} = - 2x + C \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{d}{{dx}}\left[ {{e^{ - {y^2}}}} \right] = \frac{d}{{dx}}\left[ { - 2x + C} \right] \cr & {e^{ - {y^2}}}\left( { - 2y} \right)\frac{{dy}}{{dx}} = - 2 \cr & y{e^{ - {y^2}}}\frac{{dy}}{{dx}} = 1 \cr & \frac{{dy}}{{dx}} = \frac{1}{{y{e^{ - {y^2}}}}} \cr & or \cr & \frac{{dy}}{{dx}} = \frac{{{e^{{y^2}}}}}{y}.{\text{ then}} \cr & {\text{The general solution is verified}} \cr}