# Bernoulli differential equation

Template:No footnotes In mathematics, an ordinary differential equation of the form

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

is called a Bernoulli equation when n≠1, 0, which is named after Jacob Bernoulli, who discussed it in 1695 Template:Harv. Bernoulli equations are special because they are nonlinear differential equations with known exact solutions.

## Solution

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

be a solution of the linear differential equation

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

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

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

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

## Example

Consider the Bernoulli equation (more specifically Riccati's equation).[1]

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

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

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

Changing variables gives the equations

${\displaystyle w={\frac {1}{y}}}$
${\displaystyle w'={\frac {-y'}{y^{2}}}.}$
${\displaystyle w'+{\frac {2}{x}}w=x^{2}}$

which can be solved using the integrating factor

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

Multiplying by ${\displaystyle M(x)}$,

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

Note that left side is the derivative of ${\displaystyle wx^{2}}$. Integrating both sides results in the equations

${\displaystyle \int d[wx^{2}]=\int x^{4}dx}$
${\displaystyle wx^{2}={\frac {1}{5}}x^{5}+C}$
${\displaystyle {\frac {1}{y}}x^{2}={\frac {1}{5}}x^{5}+C}$

The solution for ${\displaystyle y}$ is

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

## References

• {{#invoke:citation/CS1|citation

|CitationClass=citation }}. Cited in Template:Harvtxt.

• {{#invoke:citation/CS1|citation

|CitationClass=citation }}.

1. y'-2*y/x=-x^2*y^2, Wolfram Alpha, 01-06-2013