How is this ODE solution correct? I just want to solve $\frac{dy}{dt} = 2y^2.$ Should be easy right? The answer should be $y=\frac{-1}{2t_0+2t}.$ Somehow this doesn't work when I'm trying to solve my problem, somehow what works is $y=\frac{y_0}{1-2y_0t}$ and I have no idea why. The only possible explanation can be traced back to this differential equation, which as far as I know I did correctly, yet somehow it's not correct. Is there some magical way I can turn what I have into the second form?

Both of your answers solve the differential equation, it's just a  matter of checking initial values. In general the solutions look like $$
y(t) = \frac{-1}{c + 2t}
$$
The first answer you gave is the solution if you have initial value $y(0) = \frac1{t_0}$ and the second if $y(0) = y_0$. If for example you want to have $y(t_0) = y_0$, then you could take the equation $$
y_0 = \frac{-1}{c + 2t_0}
$$
and solve for $c$ to determine $y$ with that initial value.
