Normal form of differential equation $y'=y-y^2$

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I've been told that I can transform any equation of form $y'=ay+g(y)$, where $g(y)$ is polynomial, to form $y'=y$ (Poincaré-Dulac theorem if I'm not mistaken).

So I want to find the normal form of differential equation $y'=y-y^2$. What I've done:

Let $y=\tilde{y}+a\tilde{y}^2$.

LHS: $\tilde{y}'+2a\tilde{y}'\tilde{y}=\tilde{y}'(1+2a\tilde{y})$

RHS: $\tilde{y}+a\tilde{y}^2-\tilde{y}^2-2a\tilde{y}^3-a^2\tilde{y}=(1+2a\tilde{y})\tilde{y}-(1+a)\tilde{y}^2-2a\tilde{y}^3-a^2\tilde{y}^4$

So $$\tilde{y}'(1+2a\tilde{y})=(1+2a\tilde{y})\tilde{y}-(1+a)\tilde{y}^2-2a\tilde{y}^3-a^2\tilde{y}^4$$

We can divide both sides by $(1+2a\tilde{y}) $:

$$ \tilde{y}'=\tilde{y}-\frac{1+a}{1+2a\tilde{y}}\tilde{y}^2-\frac{2a}{1+2a\tilde{y}}\tilde{y}^3-\frac{a^2}{1+2a\tilde{y}}\tilde{y}^4$$

So if $a=-1$ we have:

$$\tilde{y}'=\tilde{y}+\frac{2}{1-2\tilde{y}}\tilde{y}^3-\frac{1}{1-2\tilde{y}}\tilde{y}^4$$

Then we can expand fractions in Taylor series and use the same algorithm to "kill" the third degree etc.

The question is: Can I obtain the explicit form of the change of variables for any order? Does this series converge?

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To get the normal form of $y'=y-y^2$, consider $$ z=\frac{y}{1-y}, $$ then $$ z'=\frac{y'}{(1-y)^2}=\frac{y}{1-y}=z. $$