This ode is called reduced Riccati. It has direct solution because of this. i.e. its solution can be written directly without the need to solve associated Airy ode. (you can ofcourse still do that, but why?).
For the general Riccati, there is no such direct solution. Need to solve the associated second order Airy ode directly.
The reduced Riccati has the form
$$ y' = a x^n + b y^2 $$ There are 2 subcases. When $n=-2$ and when it is not.
Comparing the given ode to the above shows that \begin{align*} a &= 2\\ b &= -1\\ n &= 1 \end{align*} Since $n\neq -2$ then the solution of the reduced Riccati ode is given by (reference 1,reference 2, reference 3) \begin{align*} w & =\sqrt{x}\left\{ \begin{array}[c]{cc} c_{1}\operatorname{BesselJ}\left( \frac{1}{2k},\frac{1}{k}\sqrt{ab} x^{k}\right) +c_{2}\operatorname{BesselY}\left( \frac{1}{2k},\frac{1}% {k}\sqrt{ab}x^{k}\right) & ab>0\\ c_{1}\operatorname{BesselI}\left( \frac{1}{2k},\frac{1}{k}\sqrt{-ab}% x^{k}\right) +c_{2}\operatorname{BesselK}\left( \frac{1}{2k},\frac{1}% {k}\sqrt{-ab}x^{k}\right) & ab<0 \end{array} \right. \tag{1}\\ y & =-\frac{1}{b}\frac{w^{\prime}}{w}\nonumber\\ k &=1+\frac{n}{2}\nonumber \end{align*} Since $ab<0$ then EQ(1) gives \begin{align*} k &= {\frac{3}{2}}\\ w &= \sqrt{x}\, \left(c_1 \operatorname{BesselI}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)+c_2 \operatorname{BesselK}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)\right) \end{align*} Therefore the solution becomes \begin{align*} y & =-\frac{1}{b}\frac{w^{\prime}}{w} \end{align*} Substituting the value of $b,w$ found above and simplifying gives \begin{align*} y &= \frac{\sqrt{2}\, \sqrt{x}\, \left(\operatorname{BesselI}\left(-\frac{2}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right) c_1 -\operatorname{BesselK}\left(\frac{2}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right) c_2 \right)}{c_1 \operatorname{BesselI}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)+c_2 \operatorname{BesselK}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)} \end{align*} Letting $c_2 = 1$ the above becomes \begin{align*} y = \frac{\sqrt{2}\, \sqrt{x}\, \left(\operatorname{BesselI}\left(-\frac{2}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right) c_1 -\operatorname{BesselK}\left(\frac{2}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)\right)}{c_1 \operatorname{BesselI}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)+\operatorname{BesselK}\left(\frac{1}{3}, \frac{2 \sqrt{2}\, x^{{3}/{2}}}{3}\right)} \end{align*}
