How do solve $\displaystyle \displaystyle\frac{dy}{dx}+y^2+2x=0$ ?
Why? Click on this: solve dy/dx + y^2 + 2x = 0 - Wolfram|Alpha
Your method is wrong and your solution is wrong (did you even bother to check the solution by substituting it into the original DE?)
Although $\displaystyle y = \frac{1}{2x + a}$ satisfies $\displaystyle \frac{dy}{dx} + 2y^2 = 0$, it is also required to satisfy $\displaystyle \frac{dy}{dx} + 4 = 0$ and it doesn't. Similarly, the other part of your 'solution' has to satisfy both $\displaystyle \frac{dy}{dx} + 2y^2 = 0$ and $\displaystyle \frac{dy}{dx} + 4 = 0$ ....
My opinion is : $\displaystyle \begin{bmatrix}x \\y \end{bmatrix} \rightarrow \begin{bmatrix}x(a,b) \\ y(a,b) \end{bmatrix} $
If equation#1 's solution is $\displaystyle E_1 $ and equation#2 's is $\displaystyle E_2 $ , then $\displaystyle E_1\cap E_2 $ must fit both equations, is it right ?
And F(x, a, b) = 0 , it may simply solution