The homogeneous equation $$y'=-y$$ has the solution $$y_h(t)=ce^{-t}\;\;\;\;\;(c,t\in\mathbb{R})$$ In order to find a particular solution we can take the approach $$y_p(t)\stackrel{!}{=}c(t)e^{-t}$$ Since $$y_p'(t)=c'(t)e^{-t}-c(t)e^{-t}\tag{1}$$ insertion into the differential equation yields $$c'(t)e^{-t}=\sqrt{t}$$ So, we've got
\begin{equation} \begin{split} c(t) &= \int\sqrt{t}e^tdt \\ &= 2\int\theta^2e^{\theta^2}d\theta\;\;\;\;\left(\theta=\sqrt{t},\frac{d\theta}{dt}=\frac{1}{2\sqrt{t}}\right) \end{split} \end{equation}$\tag{2}$
However, the integral is related to the error function. So, I can't find a closed form for it. The solution to the initial value problem is given by $(1)$ and from $(2)$ we've got an idea how $c$ looks like.
How can we show $$\lim_{t\to\infty}\frac{y(t)}{\sqrt{t}}=1$$
You do not need an explicit formula for the integral to compute the limit. According to your calculations, the solution is $$ y(t)=y_0\,e^{-t}+e^{-t}\int_0^t\sqrt s\,e^s\,ds. $$ Then $$\begin{align} \lim_{t\to\infty}\frac{y(t)}{\sqrt t}&=y_0\lim_{t\to\infty}\frac{e^{-t}}{\sqrt t}+\lim_{t\to\infty}\frac{e^{-t}}{\sqrt t}\int_0^t\sqrt s\,e^s\,ds\\ &=\lim_{t\to\infty}\frac{\int_0^t\sqrt s\,e^s\,ds}{e^t\sqrt t}\\ &=\lim_{t\to\infty}\frac{\sqrt t\,e^t}{e^t\sqrt t+e^t/(2\,\sqrt t)}\\ &=\lim_{t\to\infty}\frac{1}{1+1/(2\,t)}\\ &=1. \end{align}$$ Use has been made of L'Hôpital's rule.