#### Answer

$${y^2} = \frac{{{x^4}}}{2} + 25$$

#### Work Step by Step

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