## University Calculus: Early Transcendentals (3rd Edition)

$2 \sqrt y -\dfrac{x^3}{3} =C$
We have: $\dfrac{dy}{dx}=x^2 \sqrt y$ or, $\dfrac{dy}{\sqrt y}=x^2 dx$ We have to integrate the above expression: $\int \dfrac{dy}{\sqrt y}= \int x^2 dx$ or, $2 (\int \dfrac{dy}{2\sqrt y})= \int x^2 dx$ This implies: $2 \sqrt y=\dfrac{x^3}{3}+C$ Hence, $2 \sqrt y -\dfrac{x^3}{3} =C$