= Bernoulli power substitution
{c}
{title2=$z=y^{1-n},\quad n\ne0,1$}
For arbitrary real $n\ne0,1$, 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> gives
$$
z'= (1-n)y^{-n}y',\qquad y=z^{1/(1-n)},\quad z>0.
$$
Multiplying the <Bernoulli differential equation> by $(1-n)y^{-n}$ yields the <first-order linear differential equation>
$$
z'+(1-n)P(x)z=(1-n)Q(x).
$$
This 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> are
$$
\mu(x)=\exp\!\left((1-n)\int_{x_0}^xP(s)\,ds\right),\qquad z(x)=\frac{y_0^{1-n}+(1-n)\int_{x_0}^x\mu(s)Q(s)\,ds}{\mu(x)}.
$$
The <product rule> verifies $(\mu z)'=(1-n)\mu Q$. 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.<Notes on Diffy Qs: Differential Equations for Engineers by Jiří Lebl/1.4 Linear equations and the integrating factor>[Notes on Diffy Qs, §1.4 Linear equations and the integrating factor: product derivative, formula (1.4), and Remark 1.4.1 on existence throughout a coefficient-continuity interval]{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, $y\equiv0$ 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 $z\downarrow0$, the inverse sends $y\to\infty$ for $n>1$, but sends $y\downarrow0$ 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.
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