finding leading behavior of a differential equation we have 
$$ y'' = \sqrt{x} y $$
I want to find the leading asymptotic behavior as $x \to \infty$
I tried substituting $y = e^{S(x)}$ and then after obtaining $S(x) \sim \pm \frac{4}{3} x^{3/4} $ and doing the correction $S = \pm \frac{4}{3} x^{3/4} + C(x) $ where $C << x^{3/4}$, I end up obtaining that 
$$ y \sim \pm \frac{1}{x^{1/4}} e^{1/3 x^{3/2} } $$
How do I check that this is indeed the leading asymptotic behavior as $x \to \infty$?

That the correction you obtain cancelled the dominant term is a sign that you did not obtain the correct asymptotic expression.
If you substitute $y=\exp(S)$ then you obtain the equation
$$S'' + S'^2 =\sqrt{x}.$$
Now you have to test the different cases, in order to see if they are consistent:
a) $ S'^2 \gg S''$ and thus $S'^2 \sim\sqrt{x}$ with the solution $$S= \frac{4}{5} x^{5/4}.$$
In this case, we have $S'^2 =\sqrt{x} \gg S''= O(x^{-3/4})$, so this case is consistent.
b) $S'' \gg S'^2$ and thus $S''\sim \sqrt{x}$ with the solution $$S= \frac{4}{15} x^{5/2}.$$
In this case, we have $S''=\sqrt{x} \not\gg S'^2= O(x^3)$.
So a) is correct. Let us find the correction: we set $S= (4/5) x^{5/4}+C$
and with $x^{1/4} C' \gg C'^2 , C''$, we obtain the result
$$ C(x) = \ln c-\frac{\ln x}{8}.$$
As a result, the asymptotic expansion is given by
$$y\sim \frac{c}{x^{1/8}} \exp\left[(4/5) x^{5/4}\right].$$
