Why these solutions from this ODE are global? Let consider the Cauchy's Problem : $y'(t) = y(t)(1-y(t))(1-K(t)y(t)), \ t\in J$ and  $y(0)=\frac{1}{4}$.
First if $K(t)=K\in \mathbb{R}$, I have to prove that the maximal of this problem is maximal.
Applying the Cauchy-Lipschitz's theorem, we know that there exists only one maximal solution in an open interval $I\subset J$ of the Cauchy's Problem. Here $f(t,y(t))=y(t)(1-y(t))(1-Ky(t))$ and $f\in \mathcal{C}^1$. Moreover the solution can be written as follow :
$y(t)=\frac{1}{4}+\int_{0}^{t}f(s,y(s)) \mathrm{d}s$.
Then I don't really know how to proceed. If $f$ is bounded on $I$, I know that the maximal solution is global but here we do not have lots of informations on $f$. If the interval was compact, I could have done it by contradiction using the "blow-up in finite time" theorem.
Second, same question if $K(t)=\frac{\sin(t)}{1+t}$.
Thanks in advance !
 A: If $K(t) = K\in\mathbb{R}$ is a constant, then this equation is autonomous: $f$ does not depend explicitly on $t$. In this case, it is enough to show that a solution $y(t)$ with $y(0) = \frac{1}{4}$ remains bounded in time on its maximal interval of existence. For if there exist constants $M\leq y(t) \leq N$ for all $t$, then you can just show that $f = f(y)$ has a Lipschitz constant $L$ on the set $\{y:M\leq y\leq N\}$. Since $y(t)$ never leaves this range, this Lipschitz constant is effectively global: you can extend your solution by a time $\sim \frac{1}{L}$ infinitely many times.
To see why $y(t)$ is bounded by constants, use the Picard-Lindelof theorem again: note that this equation has constant solutions $y(t) = 0$, $y(t) = 1$, and $y(t) = K$. Therefore a solution satisfying $y(0) = \frac{1}{4}$ obeys the global bounds $0 < y(t) < 1$. For otherwise, there would be some time $t_0$ for which either $y(t_0) = 0$ or $y(t_0) = 1$. But any solution satisfying $y(t_0) = 0$ for some $t_0$ must in fact be uniformly $0$, since we already know $0$ is a solution and you can show that solutions are locally unique. Since we know $y(0) = \frac{1}{4}$ it cannot be the $0$ solution. Similarly $y$ cannot be the constant solution $y(t) = 1$. Therefore the only possibility is that $y(t)$ remains bounded between $0$ and $1$.
