Bernoulli and useful substitutions · Differential Equations

Intuition

Some nonlinear equations are linear equations in disguise, and the right change of variable takes the disguise off. The most useful case is Bernoulli's: a linear equation with a power of the unknown on its right-hand side, which a power of the unknown taken as the new variable turns into a linear equation that the integrating factor solves. The general lesson matters more than the formula — look for the substitution that makes the equation one you already know how to solve.

A message in a simple cipher. Nothing about its content is hard; it only looks foreign. The substitution is the key, and once it is applied the equation reads as one of the kinds this chapter already solves.

plane diagram: a-linear-equation-in-disguise

Solutions of $y'+y=y^{2}$ , a Bernoulli equation. With $v=\tfrac1y$ it becomes the linear $v'-v=-1$ , so $y=\tfrac{1}{1+Ce^{x}}$ : curves starting below one slide down to zero, and a curve starting above one runs off to infinity at a finite input.

Bernoulli equations, and substitutions in general

A Bernoulli equation $y'+p(x)y=q(x)y^{n}$ with becomes linear under . Since , dividing the equation by and multiplying by gives a linear equation for . Solve it, return to , and check separately: the division by loses it when .

v=y^{1-n}:\qquad v'+(1-n)\,p\,v=(1-n)\,q

Cases worth knowing

For $y'+y=y^{2}$ : $n=2$ , , and gives , so , together with .

The Bernoulli substitution makes the equation linear

\begin{aligned}&y'+py=qy^{n},\quad v=y^{1-n}\\&\Longrightarrow\ v'+(1-n)\,p\,v=(1-n)\,q\end{aligned}

The substitution is chosen so that its derivative contains the combination that appears after dividing by y to the n. Differentiate v by the chain rule: one minus n, times y to the minus n, times y prime. Divide the equation by y to the n: the first term becomes y to the minus n times y prime, which is v prime over one minus n, and the second becomes p times y to the one minus n, which is p times v. Multiplying through by one minus n leaves an equation linear in v.

Proof steps

v=y^{1-n}\ \Longrightarrow\ v'=(1-n)\,y^{-n}y'

Differentiate the substitution by the chain rule.

y^{-n}y'+p\,y^{1-n}=q

Divide the equation by y to the n, where y is not zero.

\frac{v'}{1-n}+p\,v=q

Both terms on the left are now written through v.

v'+(1-n)\,p\,v=(1-n)\,q

Multiplying by one minus n gives a linear equation for v, ready for an integrating factor.

Applications

Practice

A Power of y on the Right

A Bernoulli equation is linear except for a power $y^{n}$ multiplying the forcing.

y'+p(x)\,y=q(x)\,y^{n},\qquad n\ne0,1

Try it

Which of these is a Bernoulli equation with $n\ne0,1$ ?