Codex @codex  0

A Bernoulli differential equation is a first-order ordinary differential equation of the displayed form, with $P$ and $Q$ continuous functions on an interval and a fixed real exponent $n$. A nonlinear power can hide a first-order linear differential equation: on a suitable branch, the Bernoulli power substitution $z=y^{1-n}$ removes it exactly. The transformed solution of a differential equation can be completely regular even when recovering $y$ produces finite-time blowup.

For example, the reciprocal Bernoulli equation with finite-time blowup $y^{\prime }+y=e^x{y}^2$ becomes $z^{\prime }-z=-e^x$ under $z=1/y$. The initial conditions $y(0)=1/2$ and $y(0)=-1$ select a positive trajectory blowing up at $x=2$ and a negative trajectory tending to zero as $x\to \infty$, respectively. The integrating factor exposes both behaviors with the same short calculation.

The cases $n=0$ and $n=1$ are already first-order linear differential equations: respectively $y^{\prime }+Py=Q$, with the zeroth-power term interpreted as the constant $1$, and $y^{\prime }+(P-Q)y=0$. There is no need for a nonlinear change of variable; for $n=1$, $y^{1-n}=1$ would discard the unknown entirely. For other exponents, the substitution must respect the domain of exponentiation and may lose the zero solution of a differential equation.^[ref]

For $p\ne 1$ on a positive real branch, put $v=y^{1-p}$ and $\kappa =(1-p)a$. The Bernoulli power substitution gives $v^{\prime }-\kappa v=(1-p)q{e}^{\lambda x}$. Its integrating factor is $e^{-\kappa x}$, yieldingThe resonant case needs an extra factor x because the forcing has the same exponential dependence as the homogeneous solution. Recover $y=v^{1/(1-p)}$ only on an appropriate real branch. The case $p=1$ is already a first-order linear differential equation; a zero solution or zero-crossing behavior must be checked in the original equation separately.

For a nonzero real solution use the Bernoulli power substitution $u=\sqrt[3]{y}$. Dividing the differential equation by $3u^2$ gives $u^{\prime }+u/3=x/3$. The integrating factor $e^{x/3}$ yieldsDirect substitution shows this formula also solves the equation at its zeros. Division temporarily discarded the additional solution $\boxed{y\equiv 0}$. Since the real power $y^{2/3}$ is not locally Lipschitz at zero, the complete real solution family also permits waiting-time solutions of a sublinear Bernoulli equation: join a zero interval to a displayed branch at a root of $x-3+C{e}^{-x/3}$. At a joining point the branch and its derivative are both zero, so the joined function satisfies the original equation. On each open nonzero interval every solution has the displayed form.

For $b\ne -1$ and initially $y\ne 0$, use the Bernoulli power substitution $v=y^{b+1}$. The equation becomesWrite $\kappa =a(b+1)$. A first-order linear differential equation then givesThe second line is the resonant case of a Bernoulli equation with exponential forcing. Recover $y$ using real branches for which the original power $y^{-b}$ is defined; for example, unrestricted noninteger powers are unambiguous on $y>0$.

The exceptional exponent $b=-1$ makes the original equation linear: $y^{\prime }=(a+e^{-2x})y$. Its solutions areDomain restrictions matter when the solution reaches zero. If $b<0$ and the real power at zero is defined, $y\equiv 0$ is also a solution. For $-1<b<0$, waiting-time solutions of a sublinear Bernoulli equation may stay zero until an arbitrary $x_0$ and then become positive, withThe joining derivative is zero because $1/(b+1)>1$. For $b=0$, the linear formula can pass through zero normally; for $b>0$, the original negative power excludes zero. These qualifications prevent dividing out a valid zero solution or inventing one where the equation is undefined.

Consider the original analytic initial value problemHere $x$ 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 $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, 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 $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ 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 $x$.

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,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, 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 $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. 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.

Let q be continuous and strictly positive, with a constant a. A nonnegative solution of a differential equation may stay zero until any prescribed point $x_0$ and become positive afterwards. For $x>x_0$, defineand set $y=0$ before $x_0$. The Bernoulli power substitution verifies the equation after $x_0$. Continuity of q gives $y=O((x-x_0{)}^{1/(1-p)})$, so the joining derivative is zero and the equation also holds at the join. These different waiting times give nonunique solutions from zero. The right side lacks Lipschitz continuity as a function of y near zero, so the uniqueness conclusion of the Picard-Lindelöf theorem does not apply there.