## Calculus with Applications (10th Edition)

$$y = {e^{ - 2{x^{ - 1/2}}}}$$
\eqalign{ & {x^2}\frac{{dy}}{{dx}} - y\sqrt x = 0;\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = {e^{ - 2}} \cr & {\text{add }}y\sqrt x {\text{ to both sides of the equation}} \cr & {x^2}\frac{{dy}}{{dx}} = y\sqrt x \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = \frac{{\sqrt x }}{{{x^2}}} \cr & \frac{{dy}}{y} = {x^{ - 3/2}} \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{{dy}}{y}} = \int {{x^{ - 3/2}}} \cr & {\text{integrate}} \cr & \ln \left| y \right| = \frac{{{x^{ - 1/2}}}}{{ - 1/2}} + C \cr & \ln \left| y \right| = - 2{x^{ - 1/2}} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 1 \right) = {e^{ - 2}} \cr & y\left( 1 \right) = {e^{ - 2}}{\text{ implies that }}y = {e^{ - 2}}{\text{ when }}x = 1 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \ln \left| {{e^{ - 2}}} \right| = - 2{\left( 1 \right)^{ - 1/2}} + C \cr & - 2 = - 2 + C \cr & C = 0 \cr & {\text{substitute }}C = 0{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & \ln \left| y \right| = - 2{x^{ - 1/2}} + 0 \cr & {\text{Solve the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{ - 2{x^{ - 1/2}}}} \cr & y = {e^{ - 2{x^{ - 1/2}}}} \cr}