Bernoulli's differential equation

The Bernoulli equation (after Jacob Bernoulli ) is a non-linear ordinary differential equation of the first order of the mold

${\displaystyle y'(x)=f(x)y(x)+g(x)y^{\alpha }(x),\ \alpha \notin \lbrace 0,1\rbrace .}$

Through the transformation

${\displaystyle \ z(x):=(y(x))^{1-\alpha }}$

one can apply it to the linear differential equation

${\displaystyle z'(x)=(1-\alpha ){\bigl (}f(x)z(x)+g(x){\bigr )}}$

lead back.

The equation is not to be confused with the Bernoulli equation of fluid mechanics .

Theorem on the transformation of Bernoulli's differential equation

Be and ${\displaystyle x_{0}\in (a,b)}$

${\displaystyle \left\{{\begin{array}{ll}z:(a,b)\rightarrow (0,\infty )\ ,&{\textrm {falls}}\ \alpha \in \mathbb {R} \setminus \{1,2\},\\z:(a,b)\rightarrow \mathbb {R} \setminus \{0\}\ ,&{\textrm {falls}}\ \alpha =2,\\\end{array}}\right.}$

a solution to the linear differential equation

${\displaystyle z'(x)=(1-\alpha )f(x)z(x)+(1-\alpha )g(x).}$

Then

${\displaystyle y(x):=[z(x)]^{\frac {1}{1-\alpha }}}$

the solution of Bernoulli's differential equation

${\displaystyle y'(x)=f(x)y(x)+g(x)y^{\alpha }(x)\ ,\ y(x_{0})=y_{0}:=[z(x_{0})]^{\frac {1}{1-\alpha }}.}$

Furthermore, Bernoulli's differential equation has for each trivially as a solution for . ${\displaystyle \alpha >0}$ ${\displaystyle y\equiv 0}$ ${\displaystyle y_{0}=0}$

proof

It applies

${\displaystyle {\begin{array}{lll}y'(x)&=&{\frac {1}{1-\alpha }}z(x)^{{\frac {1}{1-\alpha }}-1}z'(x)\\&=&{\frac {1}{1-\alpha }}z(x)^{{\frac {1}{1-\alpha }}-1}{\bigl (}(1-\alpha )f(x)z(x)+(1-\alpha )g(x){\bigr )}\\&=&f(x)z(x)^{\frac {1}{1-\alpha }}+g(x)z(x)^{\frac {\alpha }{1-\alpha }}\\&=&f(x)y(x)+g(x)y^{\alpha }(x)\ ,\end{array}}}$

while the initial value is trivially fulfilled.

${\displaystyle \Box }$

Example: Logistic differential equation

The logistic differential equation

${\displaystyle y'(x)=ay(x)-by^{2}(x),\ y(0)=y_{0}>0,\ a,b>0}$

is a Bernoulli differential equation with . So you solve ${\displaystyle \alpha =2}$

${\displaystyle z'(x)=-az(x)+b\ ,\ z(0)={\frac {1}{y_{0}}},}$

surrendered

${\displaystyle z(x)={\frac {b}{a}}+\left({\frac {1}{y_{0}}}-{\frac {b}{a}}\right)e^{-ax}.}$

As for everyone with ${\displaystyle z(x)>0}$ ${\displaystyle x>x^{-}}$

${\displaystyle x^{-}:=\left\{{\begin{array}{ll}-\infty \ ,&{\textrm {falls}}\ a\geq by_{0},\\{\frac {1}{a}}\ln(1-{\frac {a}{by_{0}}})\ ,&{\textrm {falls}}\ a<by_{0},\\\end{array}}\right.}$

is

${\displaystyle y(x):={\frac {1}{z(x)}}={\frac {1}{{\frac {b}{a}}+\left({\frac {1}{y_{0}}}-{\frac {b}{a}}\right)e^{-ax}}}}$

the solution of the above equation . ${\displaystyle (x^{-},\infty )}$

literature

• Harro Heuser: Ordinary differential equations. Teubner, Stuttgart; Leipzig; Wiesbaden 2004, ISBN 3-519-32227-7