$dX_t=\alpha X_t \,dt + \sqrt{X_t} \,dW_t, $ with $X_0=x_0,\,\alpha,\sigma>0.$ Compute $E[X_t] $ and $E[Y]$ for $Y=\lim_{t\to\infty}e^{-\alpha t}X_t$ First of all i suppose that there is no easy analytical solution to the SDE? I got the hint to use Gronwall to show that there is a constant $C$ , s.t. $E[X_{t\land\tau_n}]\le C$ for all $n\in \mathbb{N}$, with $\tau_n=\inf(t\ge0: |X_t|\ge n)\land n.$

It is unclear to me how the hint is helping. If I could show it, I still would only have an upper bound for $E[X_t]$. 

I think Y will have expectation $x_0$, since the drift gets eliminated but I don`t know how to prove the existence of Y and my guess.

Edit: $\alpha$ and $ \sigma$ are constants
Assume for the moment being that we already know that $$\sup_{t \in [0,T]} \mathbb{E}(|X_t|) \leq C &lt; \infty \tag{1}$$ for any $T&gt;0$. Follow the following steps to solve the problem:


Recall the following statement: If $f$ is a progressively mapping such that $\mathbb{E}\left( \int_0^T f(s)^2 \, ds \right)&lt;\infty$ for any $T&gt;0$, then $$M_t := \int_0^t f(s) \, dW_s$$ is a martingale; in particular $$\mathbb{E}(M_t)=\mathbb{E}(M_0)=0.$$
Use Step 1 and Fubini's theorem to show that $$\mathbb{E}(X_t)-x_0 =\alpha \int_0^t \mathbb{E}(X_s) \, ds. \tag{2}$$
By $(2)$, the mapping $u(t) := \mathbb{E}(X_t)$ solves the ordinary differential equation $$\frac{d}{dt} u(t) = \alpha u(t), \qquad u(0)=x_0.$$ Solve it.
Apply Itô's formula to show that $$e^{-\alpha t} X_t - x_0 = \int_0^t e^{-\alpha s} \sqrt{X_s} \, W_s.$$
Conclude that $$Y=\lim_{t \to \infty} e^{-\alpha t} X_t$$ almost surely and in $L^1$. (Hint: Show that $e^{-\alpha t}X_t$ is an $L^2$-bounded martingale.)
Because of the $L^1$-convergence, it follows from Step 4 that $$\mathbb{E}(Y) = \lim_{t \to \infty} \mathbb{E}(e^{-\alpha t} X_t) = x_0.$$


It remains to prove $(1)$:


Show that the process $$M_t^{(n)} := \int_0^{t \wedge \tau_n} \sqrt{X_s} \, dW_s$$ is a martingale for any $n \in \mathbb{N}$; here $$\tau_n := \inf\{t \geq 0; |X_t| \geq n\} \wedge n.$$ In particular, $M_t^{(n)}$ has expectation zero.
Use Itô's formula and step 1 to show that $$\mathbb{E}(X_{t \wedge \tau_n})-x_0 = \alpha \mathbb{E}\left( \int_0^{t \wedge \tau_n} X_s \, ds \right) \leq \alpha \int_0^t \mathbb{E}(X_{s \wedge \tau_n}) \, ds.$$
Apply the Gronwall lemma to prove that there exists a constant $C=C(T,x_0)&gt;0$ such that $$\sup_{t \leq T} \mathbb{E}(X_{t \wedge \tau_n}) \leq C \qquad \text{for all $n \geq 1$}.$$ Use Fatous lemma to conclude that $$\sup_{t \leq T} \mathbb{E} \left(  X_t \right) \leq C.$$
You did not specify what kind of function $\alpha$ is. Assume it is a constant. Take expectation of the SDE, as $\mathbf E[dW_t]=0$, 
$$dx = \alpha xdt \Longleftrightarrow x(t)=x(0)e^{\alpha t}\Longleftrightarrow y(t)=e^{-\alpha t}x(t)=x(0),$$
where $x(t):=\mathbf E[X_t]$ and $y(t):=\mathbf E[e^{-\alpha t}X_t]$.



I will add the existence of $\mathbf E[X_t]$ later, but soon.