Bernoulli differential equation ($y^{\prime }+P(x)y=Q(x)y^n$)

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]

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.

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 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.

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.

For $y^{\prime }=y-A{y}^p$, $A>0$ and $0<p<1$, put $u=y^{1-p}$ while $y>0$. Then $u^{\prime }=(1-p)(u-A)$. If $0<y_0<A^{1/(1-p)}$, $u$ decreases to zero at the displayed finite time. The positive branch joins the zero solution with zero first derivative, giving a nonnegative forward solution. The equilibrium at zero is one-sided stable even though the vector field is not locally Lipschitz continuous there; the positive equilibrium is unstable.