University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{2}{3}y^{3/2}-\sqrt {2x}=c$
Given: $\sqrt {2yx}\dfrac{dy}{dx}=1$ Re-arrange the given equation and inetgrate as follows:. Then $\int \dfrac{dx}{\sqrt {2x}}=\int \sqrt y dy$ ....(1) As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$ Equation (1) becomes: $\sqrt 2 \int \dfrac{dx}{2 \sqrt {x}}=\int y^{1/2} dy \implies \dfrac{y^{1/2+1}}{1/2+1}=\sqrt {2x}+c$ Hence, $\dfrac{2}{3}y^{3/2}-\sqrt {2x}=c$