## Calculus with Applications (10th Edition)

$$y = {e^{\frac{{1 - x}}{x}}}$$
\eqalign{ & {x^2}\frac{{dy}}{{dx}} = y;\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = - 1 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = \frac{{dx}}{{{x^2}}} \cr & \frac{{dy}}{y} = {x^{ - 2}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{y}dy} = \int {{x^{ - 2}}} dx \cr & {\text{integrate by using the logarithmic rule and the power rule for integration}} \cr & \ln \left| y \right| = \frac{{{x^{ - 1}}}}{{ - 1}} + C \cr & \ln \left| y \right| = \frac{1}{x} + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = - 1 \cr & \,y\left( 1 \right) = - 1{\text{ implies that }}y = - 1{\text{ when }}x = 1 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \ln \left| { - 1} \right| = \frac{1}{1} + C \cr & 0 = 1 + C \cr & C = - 1 \cr & {\text{substitute }}C = - 4{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & \ln \left| y \right| = \frac{1}{x} - 1 \cr & \ln \left| y \right| = \frac{{1 - x}}{x} \cr & {\text{Solve the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{\frac{{1 - x}}{x}}} \cr & y = {e^{\frac{{1 - x}}{x}}} \cr}