Exists $t^*\in \mathbb{R}$ such that $y(t^*)=-1$?. $y'=y^2-13y+77, y(0) = \frac{13}{2}$
Exists $t^*\in \mathbb{R}$ such that $y(t^*)=-1$?.
How to prove without solving the ode? Any hint?
Using previous post:
We have: $y^2-13y+77=0 \Leftrightarrow y=\dfrac{13\pm \sqrt{139}i}{2}$
In $\mathbb{R}: \Delta(y^2-13y+77)=-139<0$, ie $y(t)>0$. So, how can that $t^*$ exist?
From wolfram, exists that $t^*$.
Sol. ode from wolfram:
 A: First, as $y^2-13y+77$ is of class $C^1$ on $y$, we have existence and uniqueness of the solution.
Note that $y'(t) = y^2-13y+77 = (y-\frac{13}{2})^2+\frac{139}{4}\ge \frac{139}{4}$, for all $t\in\mathbb{R}$. If there no exists a $t^*$ with $y(t^*)=-1$ then, by continuity of the solution, $y(t)>-1$, for all $t\in\mathbb{R}$.
By the Mean Value Theorem, for every $t>0$, there exists a $c\in(-1,\frac{13}{2})$, such that
$$ y'(c) = \frac{y(t)-\frac{13}{2}}{t-0}$$
As $y(t)\in (-1,\frac{13}{2})$, you can choose $t_0$ large enough such that $\frac{y(t)-\frac{13}{2}}{t-0}<1$. So there exists a $c$ with $y'(c)<1$, but it contradicts $y'(t)\ge\frac{139}{4}$.
So there exists a $t^*$ such that $y(t^*)=-1$.
A: In this case the situation is more involved (as compared with the other post).
The r.h.s. is always strictly positive, hence the solution is strictly increasing.
By separation of variables, its implicit form is
$$
F(y) = t, \qquad
F(y) := \int_{13/2}^y \frac{1}{s^2-13s+77}\, ds.
$$
($F$ can be explicitly computed, but there is no need to do the computation.)
By studying the convergence of the integral, there exist (and are finite) the limits
$$
T_1 := \lim_{y\to + \infty} F(y),
\qquad
T_0 := \lim_{y\to - \infty} F(y).
$$
Hence, your solution is defined for $t\in (T_0, T_1)$; moreover, it can be proved that
$$
\lim_{t\to T_0+} y(t) = -\infty,
\qquad
\lim_{t\to T_1-} y(t) = +\infty,
$$
so that the image of $y$ is $\mathbb{R}$.
A: One could also classify this equation as Riccati and set $y=-\frac{u'}{u}$. Then
$$
u''+13u'+77u=0.
$$
As the characteristic roots are complex, the solution $u$ oscillates with periodic root locations, which are poles for $y$. In consequence, the domain of $y$ is the interval between two such poles. As $y'$ is always positive, $y$ thus monotonously increasing, the range of $y$ is $\Bbb R$, including the value $-1$.
