>>10873379Well, one should obviously replace y' with z here to get z' = (z - 1)^2, at which point -1/(z - 1) = t + C. Since z(1) = -1, C = -1/2.
Now we have y' = z = 1/(1/2 - t) + 1. So then y = -log|1/2- t| + t + D. Now y(0)=0, so D = log(1/2). There's your solution, but it's fatally flawed to begin with. y blows up in finite time if you start from t = 0, before time t = 1, so the second initial condition doesn't make sense. But if you start from t > 1/2, you cannot extend back to 0 due to the blowup, so again the initial condition is nonsensical. The problem is ill-posed.
