## Calculus 10th Edition

$$y=\frac{1}{4}(100x^2+20Cx+C^2)$$
$$\frac{dy}{dx}=10\sqrt y$$ Rearranging gives, $$\frac{dy}{\sqrt y}=10dx$$ Integrating both sides. $$\int y^{-1/2}dy=\int 10dx$$ $$\frac{y^{1/2}}{1/2}=10x+c$$ Or, $$y=\frac{1}{4}(100x^2+20Cx+C^2)$$