6. Consider
\frac{d}{dt}y(t) = ay - by^2, \quad y(0) = y_0 \neq 0,
(0.1)
with $b > 0$.
(a) Solve the ODE. Show that if $y_0 > 0$, the solution exists globally, or $y(t)$ does not become
infinity for finite $t$. There exists some $y_0 < 0$, such that the solution becomes infinite at some
fintie time $t$.
Now, for $y_0 > 0$, we use another method to prove that the solution exists globally. For
simplicity, we consider $a = 1, b = 1$. Then (0.1) reduces to
\frac{d}{dt}y(t) = y - y^2
(b) Show that $\frac{d}{dt}y(t) \leq y$.
(c) By solving the above differential inequality, show that $y(t) \leq e^t y_0$. We will assume the
property that $y(t) > 0$.
(d) Suppose that the solution exists on $[0, T]$ for some $T$. Now, apply the existence and
uniqueness theorem to the new initial data $(T, y(T))$. By choosing suitable $a$ and $b = e^T y_0$ in
the existence and uniqueness theorem, show that the solution $y(t)$ exists on a new time interval
$[T, T + \alpha]$ with
$\alpha \geq Ce^{-T}$,
where the constant $C$ can depend on $y_0$, but it is independent on $T$.
Hint: Use $y(T) \in [0, e^T y_0]$ from the result (c).
(e) For $T_n = \varepsilon \log n$ with $0 < \varepsilon < \min(C, 1)$, show that $T_{n+1} \leq T_n + Ce^{-T_n}$. Use result (d)
to show that if the solution exists on $[0, T_n]$, then it also exists on $[0, T_{n+1}]$.
Finally, since $T_n \to \infty$ as $n \to \infty$, by taking $n \to \infty$, we obtain the existence of the solution
$y(t)$ for all $t > 0$.
