Can a solution for x(t) be found from the following real valued differential equation
$$a\frac{d}{dt}\!(x(t))^2 + x(t) +b\frac{d}{dt}\!(y(t)) = 0$$
in terms of only y(t), it's integrals or derivatives and a, b ?
I tried using LaPlace operators to solve, but no luck.
$$\dfrac{2a}{b}\int x(t){dx(t)}+\dfrac{1}{b}\int x(t)dt=y(t)+C\\ \implies y(t)+C=\dfrac{a\times (x(t))^2+\int x(t)dt}{b}$$