Solution to Bernoulli's Equation

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Theorem

Bernoulli's equation:

$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$

where:

$n \ne 0, n \ne 1$

has the general solution:

$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$

where:

$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$


Proof

Make the substitution:

$z = y^{1 - n}$

in $(1)$.


Then we have:

\(\ds \frac {\d z} {\d y}\) \(=\) \(\ds \paren {1 - n} y^{-n}\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n}\) \(=\) \(\ds \map Q x y^n \paren {1 - n} y^{-n}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x} + \paren {1 - n} \map P x y^{1 - n}\) \(=\) \(\ds \paren {1 - n} \map Q x\) Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x} + \paren {1 - n} \map P x z\) \(=\) \(\ds \paren {1 - n} \map Q x\)

This is now a linear first order ordinary differential equation in $z$.


It has an integrating factor:

\(\ds \map \mu x\) \(=\) \(\ds e^{\int \paren {1 - n} \map P x \rd x}\)
\(\ds \) \(=\) \(\ds e^{\paren {1 - n} \int \map P x \rd x}\)


and this can be used to obtain:

$\ds \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$

Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof.

$\blacksquare$


Also see

When $n = 0$ or $n = 1$ the equation is linear, and Solution to Linear First Order Ordinary Differential Equation can be used.


Sources