= Solution
On an interval where $y>0$, the <Bernoulli differential equation> is transformed by $u=y^{1-p}$:
$$
u'=(1-p)y^{-p}y'=(1-p)(f_1u+f_2).
$$
For $f_1=1$, $f_2=x$, put $s=1-p\ne0$. An <integrating factor> $e^{-sx}$ solves $u'-su=sx$, giving $u=Ce^{sx}-x-s^{-1}$. Thus the positive branches are
$$
\boxed{y(x)=\left(Ce^{(1-p)x}-x-\frac1{1-p}\right)^{1/(1-p)},}
$$
on intervals where the expression in parentheses is positive. The transformation omits the separate solution $y\equiv0$. For $0<p<1$, the right-hand side need not be locally <Lipschitz continuous> at zero, so nonnegative solutions can also join zero and positive branches where the original equation permits; the formula describes every strictly positive segment.
For the autonomous case, $y'=y-\alpha^2y^p$. Its two nonnegative <equilibrium points> are
$$
\boxed{y=0,\qquad y_*=|\alpha|^{2/(1-p)}.}
$$
At the positive <equilibrium point>, $f'(y_*)=1-p$. If $p>1$, $f(y)>0$ below $y_*$ and $f(y)<0$ above it; hence \b[$0$ is unstable and $y_*$ is asymptotically stable]. Every strictly positive solution tends to $y_*$ forward in time.
If $0<p<1$, those signs reverse: \b[$0$ is stable from the nonnegative side and $y_*$ is unstable]. The stability at zero follows from the phase-line signs, not a finite derivative there. For initial value $0<y_0<y_*$, the transformed equation is $u'=s(u-\alpha^2)$ with $s>0$. The solution reaches zero at
$$
T=\frac1{1-p}\log\frac{\alpha^2}{\alpha^2-y_0^{1-p}},
$$
and remains zero thereafter as a nonnegative forward solution. This is <finite-time extinction in a sublinear Bernoulli equation>. Initial values above $y_*$ increase without bound; the sublinear negative term prevents neither that growth nor the instability of $y_*$.
Back to article page
