Calculus 10th Edition

$8y^3=3x^4+C$
$\frac{dy}{dx}=\frac{x^3}{2y^2}$ Therefore, $2y^2dy=x^3dx$ $\int 2y^2dy=\int x^3dx$ Or, $\frac{2y^3}{3}=\frac{x^4}{4}+C'$ Multiplying both sides by 12 and taking $C=12C'$. $$8y^3=3x^4+C$$