Analytic function solution of a second order Riccati equation $\frac{1}{2}y''+\frac{x}{2}y'+y-\frac{1}{2}y^2=0$ Consider nonlinear ode
$$\frac{1}{2}y''+\frac{x}{2}y'+y-\frac{1}{2}y^2=0$$
$$y'(0)=0$$
$$0< y(x)\sim xe^{-\frac{x^2}{2}}$$
It is also known that $y$ is symmetric and we want the solution when $y$ is sufficiently close to $0$ so that it has thin tails on both sides. How can we find an analytic function solution to this problem?
 A: This isn't a full answer to the question, but a comment too long to be posted in the comments section. Hopping this will provide some clarification insofar there is no clue from where $y\sim xe^{-x^2/2}$ is coming. 
$$y''+xy'+2y-y^2=0\quad\text{with condition}\quad y'(0)=0 \tag 1$$
Since I am not quite sure of what is the second condition to determine $y(x)$, the figure below shows the map of $y(x)$ with the first condition only. In interest of readability, it is split into two graphs.

As far as I can understand, the question concerns the behavior at $x\to\infty$ only in the cases where $y(x)\to 0$.
In these cases, $2y-y^2\sim 2y$ and the differential equation is simplified to 
$$y''+xy'+2y\simeq 0\quad \text{in }\begin{cases}x\to\infty\\y\to 0\end{cases}\tag 2$$ 
Of course, the condition $y'(0)=0$ is no longer valid to the simplified ODE since the simplification is only for $x$ large.
The solution of the simplified equation (2) is : 
$$y=c_1xe^{-x^2/2}+c_2\left(\sqrt{\frac{\pi}{2}}xe^{-x^2/2}\text{Erfi}\left(\frac{x}{\sqrt{2}}\right) -1\right)$$
In $x\to\infty$, the asymptotic expansion leads to $\quad \sqrt{\frac{\pi}{2}}xe^{-x^2/2}\text{Erfi}\left(\frac{x}{\sqrt{2}}\right) -1 \sim \frac{1}{x^2}$
$$y\sim c_1xe^{-x^2/2}+c_2\frac{1}{x^2}$$
$$y\sim c_2\frac{1}{x^2}\quad\text{if}\quad c_2\neq 0$$
Coming back to the ODE (1), in the cases where $y_{(x\to\infty)}\to 0$ , in general $\quad y\sim c_2\frac{1}{x^2}$
To be in the case of $c_2=0$, it is necessary  to have a particular second condition. If so, $y\sim c_1xe^{-x^2/2}$ .
For example, with the first condition $y'(0)=0$ and the second condition $0<y(0)<2$ the function $y(x)$ tends to $0$ as $y(x)\sim c_2\frac{1}{x^2}$ in general.
But, for a particular value $y_0$ the function is likely to tends to $0$ as $y\sim c_1xe^{-x^2/2}$ .
So, if I well understand the question, the challenge is to analytically express this particular value $y_0$. I am afraid that it is a too big chalenge. Most likely, numerical calculus could give approximate of $y_0$. 
In addition :
The behavior when $x\to\infty$ described above from analytical calculus is in well agreement with numerical simulations. The next figure shown how $x^2y(x)$ varies for various initial point $y(0)$. 

We observe an horizontal asymptote which give the approximate of $c_1$ corresponding to each initial value $y(0)$. This confirms that
$$y(x)\sim \frac{c_1}{x^2}$$ 
We observe that it exists a particular value of the initial point $y(0)=y_0$ to which $c_1\simeq 0$. A rough numerical approach gives $y_0$ close to $1.37958$ . 
Of course, all this should be verified and more accurate result can be obtained. 
