Bernoulli power substitution ($z=y^{1-n},n\ne 0,1$)

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.