## Calculus with Applications (10th Edition)

$$y = 4{e^{\frac{{2{x^{3/2}}}}{3} - \frac{2}{3}}}$$
\eqalign{ & \sqrt x \frac{{dy}}{{dx}} = xy;\,\,\,\,\,\,\,\,\,y\left( 1 \right) = 4 \cr & {\text{Separating variables leads to}} \cr & \frac{1}{y}dy = \frac{x}{{\sqrt x }}dx \cr & \frac{1}{y}dy = {x^{1/2}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{1}{y}dy} = \int {{x^{1/2}}dx} \cr & {\text{integrating by using the power rule}} \cr & \ln \left| y \right| = \frac{{{x^{3/2}}}}{{3/2}} + C \cr & \ln \left| y \right| = \frac{{2{x^{3/2}}}}{3} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 1 \right) = 4 \cr & y\left( 1 \right) = 4{\text{ implies that }}y = 4{\text{ when }}x = 1 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \ln \left| 4 \right| = \frac{{2{{\left( 1 \right)}^{3/2}}}}{3} + C \cr & C = \ln 4 - \frac{2}{3} \cr & {\text{substitute }}C = \ln 4 - \frac{2}{3}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & \ln \left| y \right| = \frac{{2{x^{3/2}}}}{3} + \ln 4 - \frac{2}{3} \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{\frac{{2{x^{3/2}}}}{3} - \frac{2}{3}}}{e^{\ln 4}} \cr & y = 4{e^{\frac{{2{x^{3/2}}}}{3} - \frac{2}{3}}} \cr}