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