## Calculus with Applications (10th Edition)

$$y = - \frac{9}{{6{x^{3/2}} - 49}}$$
\eqalign{ & \frac{{dy}}{{dx}} = {x^{1/2}}{y^2};\,\,\,\,\,\,\,\,\,\,\,y\left( 4 \right) = 9 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{{{y^2}}} = {x^{1/2}}dx \cr & {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr & {y^{ - 2}}dy = {x^{1/2}}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{y^{ - 2}}dydy} = \int {{x^{1/2}}} dx \cr & {\text{integrate by using the power rule for integration}} \cr & \frac{{{y^{ - 1}}}}{{ - 1}} = \frac{{{x^{3/2}}}}{{3/2}} + C \cr & - \frac{1}{y} = \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( 4 \right) = 9 \cr & \,\,y\left( 4 \right) = 9{\text{ implies that }}y = 9{\text{ when }}x = 4 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & - \frac{1}{9} = \frac{{2{{\left( 4 \right)}^{3/2}}}}{3} + C \cr & - \frac{1}{9} = \frac{{16}}{3} + C \cr & C = - \frac{{49}}{9} \cr & {\text{substitute }}C = - \frac{{49}}{9}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & - \frac{1}{y} = \frac{{2{x^{3/2}}}}{3} - \frac{{49}}{9} \cr & {\text{Solve the equation for }}y \cr & - \frac{1}{y} = \frac{{6{x^{3/2}} - 49}}{9} \cr & y = - \frac{9}{{6{x^{3/2}} - 49}} \cr}