When ${\displaystyle n=0}$ , the differential equation is linear. When ${\displaystyle n=1}$ , it is separable. In these cases, standard techniques for solving equations of those forms can be applied. For ${\displaystyle n\neq 0}$  and ${\displaystyle n\neq 1}$ , the substitution ${\displaystyle u=y^{1-n}}$  reduces any Bernoulli equation to a linear differential equation

${\displaystyle {\frac {du}{dx}}-(n-1)P(x)u=-(n-1)Q(x).}$ 

For example, in the case ${\displaystyle n=2}$ , making the substitution ${\displaystyle u=y^{-1}}$  in the differential equation ${\displaystyle {\frac {dy}{dx}}+{\frac {1}{x}}y=xy^{2}}$  produces the equation ${\displaystyle {\frac {du}{dx}}-{\frac {1}{x}}u=-x}$ , which is a linear differential equation.

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

${\displaystyle {\begin{cases}z:(a,b)\rightarrow (0,\infty ),&{\text{if }}\alpha \in \mathbb {R} \smallsetminus \{1,2\},\\[4pt]z:(a,b)\rightarrow \mathbb {R} \smallsetminus \{0\},&{\text{if }}\alpha =2,\end{cases}}}$ 

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 ${\displaystyle y(x):=[z(x)]^{1/(1-\alpha )}}$  is a solution of

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

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

Consider the Bernoulli equation

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

(in this case, more specifically a Riccati equation). The constant function ${\displaystyle y=0}$  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

${\displaystyle {\begin{aligned}u={\frac {1}{y}}\;&,~u'={\frac {-y'}{y^{2}}}\\[5pt]-u'-{\frac {2}{x}}u&=-x^{2}\\[5pt]u'+{\frac {2}{x}}u&=x^{2}\end{aligned}}}$ 

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 ${\displaystyle M(x)}$ ,

${\displaystyle u'x^{2}+2xu=x^{4}.}$ 

The left side can be represented as the derivative of ${\displaystyle ux^{2}}$  by reversing the product rule. Applying the chain rule and integrating both sides with respect to ${\displaystyle x}$  results in the equations

${\displaystyle {\begin{aligned}\int \left(ux^{2}\right)'dx&=\int x^{4}\,dx\\[5pt]ux^{2}&={\frac {1}{5}}x^{5}+C\\[5pt]{\frac {1}{y}}x^{2}&={\frac {1}{5}}x^{5}+C\end{aligned}}}$ 

The solution for ${\displaystyle y}$  is

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

• 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.

• Index of differential equations