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Bernoulli Differential Equation

{dy \over dx} + p(x)y = q(x)y^n.
\end{displaymath} (1)

Let $v \equiv y^{1-n}$ for $ n\not = 1$, then
{dv\over dx} = (1-n)y^{-n}{dy\over dx}.
\end{displaymath} (2)

Rewriting (1) gives
y^{-n}{dy \over dx}= q(x)-p(x)y^{1-n} = q(x)-vp(x).
\end{displaymath} (3)

Plugging (3) into (2),
{dv\over dx} = (1-n)[q(x)-vp(x)].
\end{displaymath} (4)

Now, this is a linear First-Order Ordinary Differential Equation of the form
{dv\over dx}+vP(x)=Q(x),
\end{displaymath} (5)

where $P(x)\equiv (1-n)p(x)$ and $Q(x)\equiv (1-n)q(x)$. It can therefore be solved analytically using an Integrating Factor
$\displaystyle v$ $\textstyle =$ $\displaystyle {{\int e^{\int P(x)\,dx} Q(x)\,dx + C}\over e^{\int P(x)\,dx}}$  
  $\textstyle =$ $\displaystyle {{(1-n)\int e^{(1-n)\int p(x)\,dx} q(x)\,dx + C}\over e^{(1-n)\int p(x)\,dx}},$ (6)

where $C$ is a constant of integration. If $n=1$, then equation (1) becomes
{dy \over dx}=y(q-p)
\end{displaymath} (7)

{dy\over y}=(q-p)\,dx
\end{displaymath} (8)

y=C_2 e^{\int [q(x)-p(x)]\,dx}.
\end{displaymath} (9)

The general solution is then, with $C_1$ and $C_2$ constants,

y = \cases{
\left[{{{(1-n)\int e^{(1-n)\int p(x)\,dx} q(x)\...
...or $n\not=1$\cr
C_2 e^{\int [q(x)-p(x)]\,dx} & for $n=1$.\cr}
\end{displaymath} (10)

© 1996-9 Eric W. Weisstein