## Calculus with Applications (10th Edition)

$$y = k{e^{ - \frac{1}{x}}}$$
\eqalign{ & \frac{{dy}}{{dx}} = \frac{y}{{{x^2}}} \cr & {\text{Separating variables leads to}} \cr & \frac{1}{y}dy = \frac{1}{{{x^2}}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{y}dy} = \int {\frac{1}{{{x^2}}}dx} \cr & \int {\frac{1}{y}dy} = \int {{x^{ - 2}}dx} \cr & {\text{integrating by using the logarithmic rule and the power rule we obtain}} \cr & \ln \left| y \right| = \frac{{{x^{ - 1}}}}{{ - 1}} + C \cr & \ln \left| y \right| = - \frac{1}{x} + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{ - \frac{1}{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 \right|}} = {e^C}{e^{ - \frac{1}{x}}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y = {e^C}{e^{ - \frac{1}{x}}} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & y = k{e^{ - \frac{1}{x}}} \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {k{e^{ - \frac{1}{x}}}} \right] \cr & \frac{{dy}}{{dx}} = \frac{{k{e^{ - 1/x}}}}{{{x^2}}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = \frac{{k{e^{ - 1/x}}}}{{{x^2}}}{\text{ and }}y = k{e^{ - \frac{1}{x}}}{\text{ into }}\frac{{dy}}{{dx}} = \frac{y}{{{x^2}}} \cr & \frac{{k{e^{ - 1/x}}}}{{{x^2}}} = \frac{{k{e^{ - \frac{1}{x}}}}}{{{x^2}}} \cr & {\text{The general solution is verified}} \cr}