Solve $y'+2xy^2 = 1$ 
Solve $y'+2xy^2 = 1$

I've tried to test if the equation is exact or if it can be made exact. Nothing worked and then I understood that it's a non-linear equation. I've tried a lot to think of a substitution so that I can do separation of variables but that also didn't work. It's also not even a Bernoulli's equation.
This might be a simple problem for some but I can't get my head around it. Can I please get some hint? I'm not asking for a full solution so please don't close this question. I can't solve it and I need some help doing it on my own.

This is not a full solution, just some pointers which might help on the way towards a solution.
This is a Riccati equation which according to Wolfram Alpha has solutions in so called Airy and Bairy functions which are defined as solutions to the related family of differential equations.
$$\frac{d^2y}{dy^2} -xy = 0$$
These solutions can not be expressed in elementary functions, but are rather commonly expressed as trigonometric integrals $$Ai(x) = \frac 1 \pi \int_0^\infty \cos\left(\frac{t^3}3 +xt\right)dt$$
Moreover, we can easily see by ocular inspection that there can't be a polynomial solution to our question.
This is because the $y'$ shrinks in order, while $2xy^2$ grows (as compared to $y$).
If we slightly modify the differential equation to $$y' + y^2 = 1$$
This becomes a separable equation:
$$\frac{y'}{(1+y)(1-y)} = 1$$
which we can integrate
$$\int\frac{dy}{(1+y)(1-y)} = \int 1 dx$$
This gives us a hint that exponential functions shall be involved as the left hand sides gives us logarithms in $y$ (after some partial fraction decomposition).
How to handle the $2x$ factor, I don't know. Maybe we can do $\sqrt{2x}$ and a variable substitution of some kind.
