Solution to Bernoulli's Equation
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Theorem
- $(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 18$: Basic Differential Equations and Solutions: $18.3$: Bernoulli's equation
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $3$