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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 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 the positive half-line, divide by the stochastic exponential $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$. The Itô product rule reduces the equation to the random Bernoulli differential equation $d(X/F)/dt=F^{\delta -1}(X/F{)}^{\delta }$. For $\delta \ne 1$ this gives $X_t=F_t[x_0^{1-\delta }+(1-\delta ){\int }_0^t{F}_s^{\delta -1}ds{]}^{1/(1-\delta )}$, until the bracket vanishes; for $\delta =1$, $X_t=x_0{F}_t{e}^t$.

For the Bernoulli differential equation, work on a branch where $y\ne 0$ and the required powers are defined. Multiplication by $(1-m)y^{-m}$ makes the chain rule useful:Thus $\alpha =1-m$, $h=(1-m)f$ and $k=(1-m)g$. The case $m=1$ is already linear and must be treated separately, because the proposed substitution would be constant. Solutions lost by dividing by a power of $y$, particularly $y=0$ when it solves the original equation, must also be checked separately.

The real logarithm requires $x>0$. The identically zero function is a solution and must be retained before dividing by $y$. For a nonzero solution, divide by $x$ and make the Bernoulli differential equation substitution $u=1/y$. This givesIntegration yields $u/x=-x\ln x+x+C$. Consequentlywith each nonzero branch restricted to an interval where its denominator is nonzero. The right-hand side of the first-order equation is locally Lipschitz in $y$ for $x>0$, so a nonzero solution cannot cross the zero solution at a finite positive $x$.

For $n\ne 1$ and a nonzero solution, set $v=y^{1-n}$. Since $v^{\prime }=(1-n)y^{-n}{y}^{\prime }$, multiplying the Bernoulli differential equation by $(1-n)y^{-n}$ gives the linear ordinary differential equationAn integrating factor $\mu =\exp ((1-n)\int p(x)dx)$ gives $v={\mu }^{-1}(C+(1-n)\int \mu qdx)$. Recover $y$ from $y^{1-n}=v$ on an appropriate real or complex branch. This substitution can omit the zero solution, which must be checked in the original equation. For $n=1$, no substitution is needed: $y^{\prime }+(p-q)y=0$ is already a linear ordinary differential equation.

The zero function is a solution on any interval avoiding $x=0$. For a nonzero solution, the Bernoulli differential equation substitution $u=1/y$ gives $u^{\prime }=-y^{\prime }/y^2$ andAn integrating factor is $1/x$ on either side of zero. ThusThe full solution set isUse connected intervals with $x\ne 0$ and, in the nonzero family, $Cx-x^2\ne 0$. The separate zero solution would be lost by dividing by $y^2$.

Set $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$, a positive stochastic exponential with $d{F}_t=\alpha F_{t}d{B}_t$. The Itô formula givesFor $Y=X/F$, the Itô product rule and $d[X,F^{-1}{]}_t=-{\alpha }^2{X}_t{F}_t^{-1}dt$ cancel both the Brownian terms and the extra drift. ConsequentlyThis is an ordinary Bernoulli differential equation with continuous random coefficients. For $\delta \ne 1$, separate variables to obtainTherefore the Bernoulli stochastic differential equation has the explicit maximal positive solutionvalid while its bracket remains strictly positive. When $\delta =1$, the reduced equation is $d{Y}_t=Y_{t}dt$, andSubstituting these expressions back through the product rule verifies the original stochastic differential equation, not just the reduced equation.

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 $y_{\ast }$.

Multiplyingby $y^{-n}$ and setting $z=y^{1-n}$ givesTherefore the Bernoulli differential equation becomes

For the stated equation,Set $z=\sqrt{x}$. ThenThe integrating factor is $e^{t^3}$, soHenceThe condition $x(1)=4$ gives $C=2e-\frac{1}{2}$. Consequently

The Bernoulli differential equation becomes linear after $z=1/y$: multiplying by $-y^{-2}$ gives $z^{\prime }-2z=-x^2$. A quadratic particular solution is $z_p=x^2/2+x/2+1/4$, while the homogeneous solution is $C{e}^{2x}$. The initial condition gives $z(0)=1$ and $C=3/4$. HenceThe denominator is positive for all real $x$, since $2x^2+2x+1=2(x+1/2{)}^2+1/2$. Thus this initial value problem has the displayed solution on the entire real line. The separate solution $y=0$ does not satisfy its initial condition.

For $y\ne 0$, the Bernoulli differential equation becomes linear under $u=1/y$. Multiplying the equation by $-y^{-2}$ givesOn an interval with $x\ne 0$, the integrating factor is $x^{-2}$, henceTherefore the complete set of solution branches isThe original equation requires $x\ne 0$; each nonzero branch additionally requires $C+\sin x\ne 0$. The zero solution is restored explicitly because the reciprocal substitution excludes it.

This is a Bernoulli differential equation with exponent four. On the nonzero branch selected by $y(0)=1$, put $v=y^{-3}$. Multiplying the original equation by $-3y^{-4}$ givesIts integrating factor is $e^{-x}$, so $(e^{-x}v{)}^{\prime }=-3$ and $e^{-x}v=C-3x$. The initial condition sets $C=1$. ThusThis is the maximal real solution interval containing zero: the solution diverges as $x\to (1/3{)}^{-}$. The real negative branch of the same algebraic formula for $x>1/3$ is disconnected from this initial-value solution. The original equation also admits $y=0$, but it does not satisfy the prescribed initial condition.

First retain the equilibrium solution $\boxed{y\equiv 0}$. Any nonzero real solution has constant sign on its interval of existence, since the polynomial right-hand side is locally Lipschitz and Picard-Lindelöf theorem prevents crossing the zero solution.

For a nonzero solution, the Bernoulli differential equation substitution $u=y^{-4}$ givesThis linear ordinary differential equation has solution $u+2=C{e}^{-8x}$. Therefore every nonzero real solution isEither sign may be chosen and remains fixed. The restriction ensures $y^{-4}>0$; nonpositive $C$ gives no nonzero real solution. These are maximal real intervals, and their restrictions are also solutions. As the right endpoint is approached, $|y|$ diverges in finite-time blowup. The substitution would have lost $y=0$ if it had not been checked separately.

For a nonzero solution on a branch where the real power is differentiable, set $v=u^{1-\alpha }$. Multiplying the Bernoulli differential equation by $(1-\alpha )u^{-\alpha }$ gives the linear equationIt can be solved by an integrating factor. For the specified cubic equation take $\alpha =3$, $f=g=x$, so $v=u^{-2}$ satisfies $v^{\prime }-2xv=-2x$. Its integrating factor is $e^{-x^2}$, giving $(v{e}^{-x^2}{)}^{\prime }=-2x{e}^{-x^2}$ and hence $v=1+C{e}^{x^2}$. Thus all nonzero real solutions areon connected intervals where the denominator is positive, with a fixed sign on each interval. Also $u\equiv 0$ solves the original equation and must be restored because the substitution excluded it. The choices $C=0$ give the constant solutions $u=\pm 1$. Smoothness of the original right-hand side gives uniqueness, so a nonzero solution cannot cross through the zero solution at a finite regular point.