Answer
$c=2 \tan \sqrt y-x$
Work Step by Step
Given: $\dfrac{dy}{dx}=\sqrt y \cos^2 \sqrt y$
$\dfrac{dy}{dx}=\dfrac{3x^2}{e^y}$ or, $e^y dy=3x^2 dx$
Re-arrange the given equation and integrate as follows:.
Then $\int dx= \int \dfrac{2}{2 \sqrt y \cos^2 \sqrt y} dy$ ....(1)
Let us take the help of $u$ substitution.
plug $u=\sqrt y \implies \dfrac{dy}{2 \sqrt y}=du$
Equation (1) becomes: $ \int \dfrac{2}{ \cos^2 u} du=x+c$
$\implies 2 \tan u=x+c$
Hence, $ 2 \tan \sqrt y=x+c$
$c=2 \tan \sqrt y-x$
