How to solve $0=\dot{y}-\dot{y}^2-y-t^2$? I was messing around with some differential equations(long story involving a game of snake, a game of minesweeper, chemical bonds, a basketball, and a friend) when I made a mistake and got the equation $0=\dot{y}-\dot{y}^2-y-t^2$. However, this seemed like a much more interesting equation than what I should've got but I couldn't figure out how to solve this which brought me here. Could someone give me a hint or point me towards some reading or techniques that could help me solve it?  Thanks.
I apologize if I violated any customs of MSE or rules, I'm not familiar.
 A: Not a general solution (and the wolframalpha link suggests that the general solution is horrible...).
But an Ansatz of $y = at^2 + bt + c$ actually gives two complex valued particular solutions straightforwardly. Let $r$ be a root of $4z^2+z+1$, i.e.
$$
r=\frac{-1\pm i\sqrt{15}}{8}.
$$
Then
$$
y(t) = r\cdot t^2 + \frac{r+1}{2}\cdot t + \frac{r+5}{16}
$$
is a solution for either choice of $r$.

You could also go for power series. If $y=\sum_n a_n t^n$, then $\dot y=\sum_n (n+1)a_{n+1} t^n$. Using this, we have
$$
t^2 
= - y + y' - (y')^2 
= \sum_n  \left( - a_n + (n+1)a_{n+1} - \sum_{j+k=n}(j+1)(k+1)a_{j+1}a_{k+1}\right)t^n 
= : \sum_n b_n t^n.
$$
So
$$
b_n = \begin{cases}0 & n\ne 2 \\ 1 & n=2\end{cases}
$$
lets you solve recursively for $a_n$. In fact, $a_0 = a_1(1-a_1)$ and
$$
a_{n+1} = \frac{\delta_{n,2} + a_n + \sum_{k=1}^{n-1}(k+1)(n-k+1)a_{k+1}a_{n-k+1}}{(n+1)(1-2a_1)}
, \quad n\ge1.
$$
In particular, there should be exactly two solutions for each initial condition $y(0) \ne \frac14$.
Here are some numeric real solutions from maple. Note the parabola $y=t^2+\frac14$. Beyond that, $y'$ would have to be complex.

