Lesson A Bernoulli Equation

MarkFL

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Suppose we are given to solve:

\(\displaystyle \d{y}{x}+\frac{y}{x}=y^2\)

A first order ODE that can be written in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n\)

where \(P(x)\) and \(Q(x)\) are continuous on an interval \((a,b)\) and \(n\) is a real number, is called a Bernoulli equation.

This equation was proposed for solution by James Bernoulli in 1695. It was solved by his brother John Bernoulli. James and John were two of eight mathematicians in the Bernoulli family. In 1696 Gottfried Leibniz showed that the Bernoulli equation can be reduced to a linear equation by making the substitution \(v=y^{1-n}\).

We are given to solve:

(1) \(\displaystyle \frac{dy}{dx}+\frac{1}{x}y=y^2\)

Dividing through by \(y^2\) (observing we are losing the trivial solution \(y\equiv0\)), we obtain:

(2) \(\displaystyle y^{-2}\frac{dy}{dx}+\frac{1}{x}y^{-1}=1\)

Using the substitution of Leibniz, i.e., \(v=y^{-1}\), we find via the chain rule that:

\(\displaystyle \frac{dv}{dx}=-y^{-2}\frac{dy}{dx}\)

and (2) becomes:

(3) \(\displaystyle \frac{dv}{dx}-\frac{1}{x}v=-1\)

Now we have a linear equation in \(v\). Computing the integrating factor, we find:

\(\displaystyle \mu(x)=e^{-\int\frac{dx}{x}}=\frac{1}{x}\)

Multiplying (3) by this integrating factor, we obtain:

\(\displaystyle \frac{1}{x}\frac{dv}{dx}-\frac{1}{x^2}v=-\frac{1}{x}\)

Rewriting the left hand side as the differentiation of a product, we have:

\(\displaystyle \frac{d}{dx}\left(\frac{v}{x} \right)=-\frac{1}{x}\)

Integrating with respect to \(x\), there results:

\(\displaystyle \int\,d\left(\frac{v}{x} \right)=-\int\frac{1}{x}\,dx\)

\(\displaystyle \frac{v}{x}=-\ln|x|+C\)

\(\displaystyle v=x\left(C-\ln|x| \right)\)

Back-substituting for \(v\), we have:

\(\displaystyle \frac{1}{y}=x\left(C-\ln|x| \right)\)

Hence:

\(\displaystyle y(x)=\frac{1}{x\left(C-\ln|x| \right)}\)
 
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harpazo

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What on earth is this?
 

MarkFL

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Let's see if we can solve the general equation:

\(\displaystyle \d{y}{x}+P(x)y=Q(x)y^n\)

Divide through by \(y^n\) to obtain:

\(\displaystyle y^{-n}\d{y}{x}+P(x)y^{1-n}=Q(x)\)

Let:

\(\displaystyle v=y^{1-n}\implies \d{v}{x}=(1-n)y^{-n}\d{y}{x}\implies \d{y}{x}=\frac{1}{1-n}y^n\d{v}{x}\)

And the ODE becomes:

\(\displaystyle y^{-n}\left(\frac{1}{1-n}y^n\d{v}{x}\right)+P(x)v=Q(x)\)

\(\displaystyle \d{v}{x}+(1-n)P(x)v=(1-n)Q(x)\)

Now we have a linear first order ODE, which can be solved according to the method used in this thread:

Lesson - First order linear ODEs and the integrating factor
 
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harpazo

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This is too advanced for me.
 

MarkFL

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I posted this for those studying first order ODEs. This topic, Bernoulli Equations, is introduced fairly early in a course on ODEs. But, if you aren't familiar with the topic or never took a course on ODEs, then it will likely be too advanced.
 
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