Bernoulli differential equation

In mathematics, an ordinary differential equation of the form:

y'+P(x)y=Q(x)y^{n}\,

is called a Bernoulli differential equation where $n$ is any real number and $n\neq 0$ and $n\neq 1$ .^[1] It is named after Jacob Bernoulli who discussed it in 1695. Bernoulli equations are special because they are nonlinear differential equations with known exact solutions. A famous special case of the Bernoulli equation is the logistic differential equation.

Transformation to a linear differential equation

Note that for $n=0$ and $n=1$ , the Bernoulli equation is linear. For $n\neq 0$ and $n\neq 1$ , the substitution $u=y^{1-n}$ reduces any Bernoulli equation to a linear differential equation. For example:

Let's consider the following differential equation: $x{\frac {dy}{dx}}+y=x^{2}y^{2}$

Rewriting it in the Bernoulli form (with $n=2$ ): ${\frac {dy}{dx}}+{\frac {1}{x}}y=xy^{2}$

Now, substituting $u=y^{-1}$ we get: ${\frac {du}{dx}}-{\frac {1}{x}}u=-x$ , which is a linear differential equation.

Solution

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

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

be a solution of the linear differential equation

z'(x)=(1-\alpha )P(x)z(x)+(1-\alpha )Q(x).

Then we have that $y(x):=[z(x)]^{{{\frac {1}{1-\alpha }}}}$ is a solution of

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

And for every such differential equation, for all $\alpha >0$ we have $y\equiv 0$ as solution for $y_{0}=0$ .

Example

Consider the Bernoulli equation (more specifically Riccati's equation).^[2]

y'-{\frac {2y}{x}}=-x^{2}y^{2}

We first notice that $y=0$ is a solution. Division by $y^{2}$ yields

y'y^{{-2}}-{\frac {2}{x}}y^{{-1}}=-x^{2}

Changing variables gives the equations

w={\frac {1}{y}}

w'={\frac {-y'}{y^{2}}}.

-w'-{\frac {2}{x}}w=-x^{2}

w'+{\frac {2}{x}}w=x^{2}

which can be solved using the integrating factor

M(x)=e^{2\int {\frac {1}{x}}\,dx}=e^{2\ln x}=x^{2}.

Multiplying by $M(x)$ ,

w'x^{2}+2xw=x^{4},\,

Note that left side is the derivative of $wx^{2}$ . Integrating both sides, with respect to $x$ , results in the equations

\int w'x^{2}+2xw\,dx=\int x^{4}\,dx

wx^{2}={\frac {1}{5}}x^{5}+C

{\frac {1}{y}}x^{2}={\frac {1}{5}}x^{5}+C

The solution for $y$ is

y={\frac {x^{2}}{{\frac {1}{5}}x^{5}+C}}

.

References

Bernoulli, Jacob (1695), "Explicationes, Annotationes & Additiones ad ea, quae in Actis sup. de Curva Elastica, Isochrona Paracentrica, & Velaria, hinc inde memorata, & paratim controversa legundur; ubi de Linea mediarum directionum, alliisque novis", Acta Eruditorum . Cited in Hairer, Nørsett & Wanner (1993).
Hairer, Ernst; Nørsett, Syvert Paul; Wanner, Gerhard (1993), Solving ordinary differential equations I: Nonstiff problems, Berlin, New York: Springer-Verlag, ISBN 978-3-540-56670-0 .

↑ Weisstein, Eric W. "Bernoulli Differential Equation." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/BernoulliDifferentialEquation.html
↑ y'-2*y/x=-x^2*y^2, Wolfram Alpha, 01-06-2013

External links

"Bernoulli equation". PlanetMath.
"Differential equation". PlanetMath.
"Index of differential equations". PlanetMath.

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[1] Weisstein, Eric W. "Bernoulli Differential Equation." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/BernoulliDifferentialEquation.html

[2] y'-2*y/x=-x^2*y^2, Wolfram Alpha, 01-06-2013