# Power series solution to $y' = y(1-y)$

Find the first five terms of the power series solution to the differential equation:

$$y' = y(1-y)$$

Letting $y = a_0+a_1x+a_2x^2+a_3x^3+...$

It's evident that: $$y' = \frac{dy}{dx} = a_1+2a_2x+3a_3x^2+4a_4x^3+...$$

And from the differential equation: $$y' = y(1-y) = (a_0+a_1x+a_2x^2+a_3x^3+...)(1-a_0-a_1x-a_2x^2-a_3x^3-...)$$

However, I don't know how to handle the product of these two infinite series. The Cauchy product comes to mind, but I'm not sure how helpful it'll be here.

Is it possible to solve this differential equation using power series alone?

You need to collect coefficients on the RHS and equate them to corresponding coefficients on the LHS. So for instance, the coefficient of $x^0$ on $y'$ is $a_1$, and the coefficient of $x^0$ in $y(1-y)$ is $a_0(1-a_0)$, so that $a_1=a_0(1-a_0)$.
More generally, the coefficient of $x^n$ in $y'$ is $(n+1)a_{n+1}$. We can write the RHS as $y-y^2$, so the coefficient of $x^n$ in the RHS is $a_n -c_n$ where
$$c_n=\sum\limits_{k=0}^na_{k}a_{n-k}$$