= Reciprocal Bernoulli equation with finite-time blowup
{title2=$y(x)=e^{-x}/(C-x)$}
Consider the original analytic <initial value problem>
$$
y'+y=e^xy^2,\qquad y(0)=a.
$$
Here $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact <solutions of a differential equation>, with no measured data or empirical parameter fit. For $a\ne0$, the <Bernoulli power substitution> $z=1/y$ gives
$$
z'-z=-e^x,\qquad (e^{-x}z)'=-1,\qquad z(x)=e^x(C-x),\qquad y(x)=\frac{e^{-x}}{C-x},\qquad C=\frac1a.
$$
The <integrating factor> is $e^{-x}$. Direct <differentiation> gives $y'=-y+e^xy^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear <solution of a differential equation> uses only the <interval> containing $0$ where $z\ne0$:
$$
\begin{array}{c|c|c}
\text{initial value}&\text{maximal interval}&\text{forward behavior}\\
a>0&(-\infty,1/a)&y\to+\infty\text{ as }x\uparrow1/a\\
a<0&(1/a,\infty)&y\to0\text{ from below as }x\to\infty\\
a=0&\mathbb R&y\equiv0.
\end{array}
$$
For $a<0$, $y'>0$, so its absolute value decreases toward zero. The lost zero <solution of a differential equation> is recovered by substitution into the original equation. Since $f(x,y)=-y+e^xy^2$ is <locally Lipschitz> in $y$ everywhere, the <Picard-Lindelöf theorem> ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.
\Image[/bernoulli-reciprocal-trajectories.png]
{title=Two Bernoulli trajectories and their exact linearized coordinates}
{description=The <initial conditions> are $y(0)=1/2$ in orange and $y(0)=-1$ in blue. Left: the positive trajectory reaches <finite-time blowup> at $x=2$, while the negative trajectory approaches zero. The vertical axis clips the positive curve at $4$; this is a display limit. Right: the corresponding $z=1/y$ trajectories satisfy $z'-z=-e^x$. The dashed orange continuation past $z=0$ solves the <first-order linear differential equation> but cannot extend the original positive <initial value problem> across blowup.}
For $a=1/2$, the exact <solution of a differential equation> is $y=e^{-x}/(2-x)$ on $(-\infty,2)$. It first decreases, reaches a <local minimum> $y(1)=e^{-1}$, then increases without bound. At the endpoint,
$$
\lim_{x\uparrow2}(2-x)y(x)=e^{-2}.
$$
The <finite-time blowup> comes from a simple zero of $z$ at $2$, where $z'(2)=-e^2$, rather than from a singular coefficient in the <first-order linear differential equation>. The same algebraic expression for $y$ is negative on $(2,\infty)$, but that is a separate <solution of a differential equation> on a disjoint <interval>; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty)$, giving a distinct decaying branch whose finite blowup is instead in the past.
The <direction field> also explains the positive turning point. Its <nullclines> are $y=0$ and $y=e^{-x}$, because
$$
y'=y(e^xy-1).
$$
Above the positive <nullcline>, $y'>0$; between it and zero, $y'<0$; below zero, $y'>0$. The moving <nullcline> $y=e^{-x}$ is not a <solution of a differential equation>: its <derivative> is $-e^{-x}$, whereas the prescribed <slope> there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its <local minimum>.
\Image[/bernoulli-slope-field-and-branches.png]
{title=The Bernoulli direction field and the disconnected inverse branches at blowup}
{description=Left: the <direction field>, its two <nullclines>, and the exact $a=1/2$ and $a=-1$ trajectories. Right: $y=e^{-x}/(2-x)$ has a positive branch on $x<2$ and a negative branch on $x>2$; the excluded point separates them. The zero <solution of a differential equation> belongs to the original equation but has no finite reciprocal coordinate. The branches continue beyond the plotted vertical limits.}
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