Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?

Multiply through by $3/2$ and rename $\frac32 C = c$
$$ y^{3/2} = c - x^{3/2} $$
The initial condition $y(1)=9$ gives $c = 28$. Solving for $y$ gives
$$ y^{3/2} = 28 - x^{3/2} \implies y = (28-x^{3/2})^{2/3} $$
