Up until now, we simply rearranged and integrated both sides, so
$$y'=y(1-y)$$$$\frac{dy}{dx}=y(1-y)$$$$\frac{dy}{y(1-y)}=dx$$$$\int\frac{dy}{y(1-y)}=\int dx$$
With partial fraction decomposition one gets
$$\int\frac{dy}{y} +\int\frac{dy}{(1-y)}=\int dx$$
$$\log|y| + C_1 + \log|1-y|+C_2=x+C_3$$
$$\log|y| + \log|1-y|=x+C_4$$
$$|y||1-y|=C_5e^x$$
How do I continue from here?
$\endgroup$ 6Edit: It should be
$$\int\frac{dy}{(1-y)}=-\log|1-y|+C$$
2 Answers
$\begingroup$You have $|y||1-y|=C_5e^x$, and $C_5$ must be positive because it's $e$ raised to a power. This implies $y(1-y)=C_6 e^x$, where $C_6$ need not be positive.
However, when you divided by $y(1-y)$, you tacitly assumed $y\ne0$ and $1-y\ne 0$. That means you need to check separately to see if there are solutions consistent with those.
However: You neglected the chain rule and should have instead $$ \frac y{1-y} = C_6 e^x. $$ That implies: $$ y = (1-y)C_6e^x $$ $$ y = C_6e^x - yC_6 e^x $$ $$ y+yC_6 e^x = C_6 e^x $$ $$ y(1+C_6 e^x) = C_6e^x $$ $$ y = \text{etc.}\cdots $$
$\endgroup$ 4 $\begingroup$A hint: notice that $y(x) = 1$ and $y(x) = 0$ are constant solutions to the DE. Existence and uniqueness then implies that no solution can ever cross these values.
Using this, you could simplify all those absolute values, considering each case... Do you think you can take it from here?
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