simple way to solve differential equation $y''=3\sqrt y, y|_{x=0}=1, y'|_{x=0}=2$

Question

$$y''=3\sqrt y, y|_{x=0}=1, y'|_{x=0}=2$$ On finding the differential equation solution, let $y'=p$, so $ p'=3 \sqrt y $
$p'= \frac{dp}{dx}=\frac{dp}{dy} \frac{dy}{dx} =p \frac{dp}{dy}$, then $ p dp = 3 \sqrt y dy$.
Solve $$\int p dp = \int 3 \sqrt y dy$$ I have $$\frac12p^2=2y^{\frac32} +\frac12c,\ p= \sqrt{4y^{\frac32}+c}$$ $$\int (4y^{\frac32}+c)^{-\frac12} dy = \int dx$$ It's impossible for me to do this integration. I guess there is a simpler approach to use the initial values, but apparently, it can't be used till now.

Answer 1

$$p(0)=y'(0)=2=\sqrt{4+c}$$

So $c=0$