# 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 other than 0 or 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

When $n=0$ , the differential equation is linear. When $n=1$ , it is separable. In these cases, standard techniques for solving equations of those forms can be applied. 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, in the case $n=2$ , making the substitution $u=y^{-1}$ in the differential equation ${\frac {dy}{dx}}+{\frac {1}{x}}y=xy^{2}$ produces the equation ${\frac {du}{dx}}-{\frac {1}{x}}u=-x$ , 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

$y'-{\frac {2y}{x}}=-x^{2}y^{2}$ (in this case, more specifically Riccati's equation). The constant function $y=0$ 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 $M(x)$ ,

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

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