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For arbitrary real $n\ne 0,1$, work first on an interval where $y>0$. Define exponentiation by $y^r=\exp (r\log y)$, using the real logarithm. The change of variable is differentiable and invertible there, and the chain rule givesMultiplying the Bernoulli differential equation by $(1-n)y^{-n}$ yields the first-order linear differential equationThis is an exact transformation of nonzero solutions of a differential equation on the chosen branch; it is valid far from any reference state and does not discard higher powers as a local linearization would.

Given the initial condition $y(x_0)=y_0>0$, the integrating factor and transformed solution of a differential equation areThe product rule verifies $(\mu z{)}^{\prime }=(1-n)\mu Q$. Recover $y=z^{1/(1-n)}$ only on the connected component of $\{x:z(x)>0\}$ containing $x_0$, inside the original coefficient interval. The first-order linear differential equation can have a longer existence interval than this inverse transformation allows.^[ref]

Negative values need a separately justified real branch. For integer $n$, the required powers are real and differentiable for $y<0$ too, but the inverse must retain the sign selected by the initial condition. For $n=3$, $z=y^{-2}>0$ represents either $y=+1/\sqrt{z}$ or $y=-1/\sqrt{z}$; both square roots are taken of positive $z$. For $n=2$, $z=1/y$ retains the sign directly. For a general noninteger exponent, use the positive branch unless a specific differentiable real negative branch has been established. Never evaluate a real fractional power of a negative value merely because the linear formula has become negative.

Division by $y^n$ excludes $y=0$. When $n>0$ and the original real power is defined at zero, $y\equiv 0$ is an additional solution of a differential equation and must be checked in the original equation. When $n<0$, the equation as written is undefined at zero. On a positive branch the right-hand side is locally Lipschitz in $y$, so the Picard-Lindelöf theorem gives local uniqueness. At zero, the same hypothesis needs a fresh check: it holds for the polynomial $n=2$ example, whereas for $0<n<1$ a nonzero power coefficient prevents local Lipschitz continuity there. If $z\downarrow 0$, the inverse sends $y\to \infty$ for $n>1$, but sends $y\downarrow 0$ for $0<n<1$. The latter endpoint can lead to finite-time extinction in a sublinear Bernoulli equation, where the original equation determines how to join to zero.

On an interval where $y>0$, the Bernoulli differential equation is transformed by $u=y^{1-p}$:For $f_1=1$, $f_2=x$, put $s=1-p\ne 0$. An integrating factor $e^{-sx}$ solves $u^{\prime }-su=sx$, giving $u=C{e}^{sx}-x-s^{-1}$. Thus the positive branches areon intervals where the expression in parentheses is positive. The transformation omits the separate solution $y\equiv 0$. 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, $y^{\prime }=y-{\alpha }^2{y}^p$. Its two nonnegative equilibrium points areAt the positive equilibrium point, $f^{\prime }(y_{\ast })=1-p$. If $p>1$, $f(y)>0$ below $y_{\ast }$ and $f(y)<0$ above it; hence $0$ is unstable and $y_{\ast }$ is asymptotically stable. Every strictly positive solution tends to $y_{\ast }$ forward in time.

If $0<p<1$, those signs reverse: $0$ is stable from the nonnegative side and $y_{\ast }$ is unstable. The stability at zero follows from the phase-line signs, not a finite derivative there. For initial value $0<y_0<y_{\ast }$, the transformed equation is $u^{\prime }=s(u-{\alpha }^2)$ with $s>0$. The solution reaches zero atand remains zero thereafter as a nonnegative forward solution. This is finite-time extinction in a sublinear Bernoulli equation. Initial values above $y_{\ast }$ increase without bound; the sublinear negative term prevents neither that growth nor the instability of $y_{\ast }$.