Find a IVP associated with $y' = y - y^{2}$

Question

The function $g(x, c) = \frac{1}{1-c e^{-x}} $ is a family of solutions (of one parameter) of the first order DE $y'(x) = y(x) - y^{2}(x)$. Find a IVP associated with this differential equation and find the solution corresponding to the initial condition $y(0) = \frac{-1}{3}$

I did the next:

The IVP associated with $y' = y- y^{2}$ is: $y' - y + y^{2} = 0$ and $y(0) = \frac{-1}{3}$ this is defined in $(- \infty,1) $

I'm not sure if that's right. It's seems incomplete to me.

Answer 1

Hint

Make the problem simpler switching variables $$y' = y- y^{2}\implies x'=\frac1 {y(1-y)}=\frac{1}{y}-\frac{1}{y-1}$$

Integrate both sides and solve for $y$. Now, apply the condition.