Bernoulli Equations

1Introduction

This lecture explains how an appropriate substitution reduces a Bernoulli equation to a first-order linear equation.

In the first-order linear equation

the unknown $$y$$y occurs to the first power. A Bernoulli equation also contains a term $$y^n$$y^n and is therefore generally nonlinear. Its power-law structure nevertheless permits linearization through the substitution $$u=y^{1-n}$$u=y^{1-n}.

2Standard Form

A Bernoulli equation is a first-order differential equation that can be written as

This lecture first treats integer values of $$n$$n. For a general real exponent $$n$$n, the analysis must be restricted to an interval on which $$y^{1-n}$$y^{1-n} is real-valued and differentiable; the region $$y>0$$y>0 is a standard choice.

The cases $$n=0$$n=0 and $$n=1$$n=1 are exceptional. If $$n=0$$n=0, the equation is

which is already first-order linear. If $$n=1$$n=1, then

or equivalently

which is also first-order linear. Thus the Bernoulli substitution is essential only when $$n\ne 0,1$$n\ne0,1.

3Proposition: Reduction of a Bernoulli Equation to First-Order Linear Form

Assume that $$n\ne 0,1$$n\ne0,1 and consider

on a region where $$u=y^{1-n}$$u=y^{1-n} is defined and differentiable. For integer $$n$$n, it is sufficient to work where $$y\ne 0$$y\ne0. Under the substitution $$u=y^{1-n}$$u=y^{1-n}, the function $$u$$u satisfies

This is a first-order linear differential equation in $$u$$u.

3.1Proof

Set $$u=y^{1-n}$$u=y^{1-n}. The chain rule gives

Dividing the original equation by $$y^n$$y^n gives

Since $$y^{1-n}=u$$y^{1-n}=u and $$y^{-n}{y}^{\prime }=u^{\prime }/(1-n)$$y^{-n}y'=u'/(1-n), this equation becomes

Multiplication by $$1-n$$1-n yields

Therefore, the substitution reduces the Bernoulli equation to a first-order linear equation.

4Solution Procedure

1. Rewrite the equation as $$y^{\prime }+p\left(x\right)y=q\left(x\right)y^n$$y'+p(x)y=q(x)y^n.
2. Check whether $$n=0$$n=0 or $$n=1$$n=1.
3. Before division, determine whether $$y=0$$y=0 satisfies the original equation.
4. On a region where $$u=y^{1-n}$$u=y^{1-n} is defined and differentiable, set $$u=y^{1-n}$$u=y^{1-n}.
5. Solve the resulting first-order linear equation in $$u$$u by an integrating factor.
6. On each interval of validity, select every real-valued differentiable branch of the inverse transformation from $$u=y^{1-n}$$u=y^{1-n} to $$y$$y, and verify it in the original equation.

Because the derivation divides by $$y^n$$y^n, the zero solution must be checked beforehand so that it is not discarded.

5Example: $$y^{\prime }+y=x{y}^2$$y'+y=xy^2

The equation

is a Bernoulli equation with $$n=2$$n=2, $$p\left(x\right)=1$$p(x)=1, and $$q\left(x\right)=x$$q(x)=x. The function $$y=0$$y=0 satisfies the original equation and is therefore a constant solution.

For $$y\ne 0$$y\ne0, set $$u=y^{1-2}=y^{-1}$$u=y^{1-2}=y^{-1}. The proposition gives

and hence

This is first-order linear. Its integrating factor is

Multiplication by $$e^{-x}$$e^{-x} gives

Integration yields

so

Since $$u=1/y$$u=1/y,

Together with this family, the previously identified function $$y=0$$y=0 is also a solution.

For each value of $$C$$C, the displayed formula defines a solution on every connected interval on which the denominator $$x+1+C{e}^x$$x+1+Ce^x is nonzero. The solution cannot be extended across a zero of the denominator.

6Connection with the Logistic Equation

Assume $$r>0$$r>0 and $$K>0$$K>0. The logistic equation

can be rewritten as

so it is a special Bernoulli equation.

The logistic equation is also autonomous. Before deriving an explicit solution, the sign of its right-hand side can therefore be used to determine monotonicity of solutions and their direction near equilibrium solutions.

7Scope

Bernoulli equations form a special class of nonlinear equations that can be linearized. A differential equation is not of Bernoulli type merely because it is nonlinear; the power $$y^n$$y^n must occur in the standard form specified above.

Moreover, the substitution is performed where $$y\ne 0$$y\ne0. If the original equation is defined at $$y=0$$y=0 and $$y=0$$y=0 is actually a solution, that constant solution must be treated separately.

8Exercises