## Calculus with Applications (10th Edition)

$${y^2} = \frac{{{x^4}}}{2} + 25$$
\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{x^3}}}{y};\,\,\,\,\,\,\,y\left( 0 \right) = 5 \cr & {\text{Separating variables leads to}} \cr & ydy = {x^3}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {ydy} = \int {{x^3}dx} \cr & {\text{integrate by using the power rule}} \cr & \frac{{{y^2}}}{2} = \frac{{{x^4}}}{4} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 5 \cr & y\left( 0 \right) = 5{\text{ implies that }}y = 5{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & \frac{{{{\left( 5 \right)}^2}}}{2} = \frac{{{{\left( 0 \right)}^4}}}{4} + C \cr & C = \frac{{25}}{2} \cr & \cr & {\text{substitute }}C = \frac{{25}}{2}{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & \frac{{{y^2}}}{2} = \frac{{{x^4}}}{4} + \frac{{25}}{2} \cr & {\text{multiply both sides by 2}} \cr & {y^2} = \frac{{{x^4}}}{2} + 25 \cr}