Bernoulli Equation

The Bernoulli differential equation is an equation of the form $y'+ p(x) y=q(x) y^n$. This is a non-linear differential equation that can be reduced to a linear one by a clever substitution. The new equation is a first order linear differential equation, and can be solved explicitly. The Bernoulli equation was one of the first differential equations to be solved, and is still one of very few non-linear differential equations that can be solved explicitly. Most other such equations either have no solutions, or solutions that cannot be written in a closed form, but the Bernoulli equation is an exception.

The idea behind the Bernoulli equation is to substitute $v=y^{1-n}$, and work with the resulting equation, as shown in the example below.

Solve the differential equation $y'+y=xy^2.$

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We start by dividing through by $y^2$, to get $$y' y^{-2} +y^{-1}=x.$$ Now, if we let $v=y^{-1}$, we have $v'=-y^{-2}y'$, so our equation becomes $$-v'+v=x\implies v'-v=-x$$ Now, we multiply through by the integrating factor $e^{-x}$, and we have $$e^{-x}v'-e^{-x} v=-xe^{-x}\implies \left[e^{-x}v\right]'=-xe^{-x}\implies e^{-x}v=e^{-x}+xe^{-x}+C\implies v=1+x+Ce^x.$$ Finally, since $v=y^{-1}$, we have the solution $$y=\frac{1}{Ce^x+x+1}.$$

The same procedure can be used to solve the general version of the equation.

Solve the differential equation $y'+p(x) y=q(x) y^n$.

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We divide through by $y^n$, so we have $$y' y^{-n}+p(x)y^{1-n}=q(x).$$ Then, if $v=y^{1-n}$, $v'=(1-n)y^{-n} y'$, so this equation is $$\frac{1}{1-n} v'+p(x) v=q(x)\implies v'+(1-n)p(x) v=q(x)(1-n).$$ Then, the integrating factor will be a function $f(x)$ such that $f(x)=e^{\int (1-n)p(x)\, dx}$. We multiply through by this to get $$f(x)v'+p(x)(1-n)f(x)v=q(x)f(x)(1-n)\implies \left[ f(x)v\right]'=q(x)f(x)(1-n)\implies v(x)=\frac{1-n}{f(x)}\int q(x)f(x)\, dx,$$ which is the solution to the Bernoulli equation.