Understanding the Bernoulli Differential Equation

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In summary: This is usually not the case, and can easily be checked by substituting known values for k for different values of k. But, even if it wasn't, you should still be able to arrive at the correct result by using the solution strategy given in your textbook.
  • #1
Susanne217
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Homework Statement



This diff.eqn here.

[tex]\frac{dy}{dx} = \frac{y}{x} - y^2[/tex]


which in my mind can be re-written to

[tex]\frac{dy}{dx} -\frac{1}{x}y = -y^2[/tex]

does this mean that the above is a socalled bernoulli diff.eqn? And should solved as such? I may be dumb, but that the only way I can deduce is to solve that eqn..

So please light me up..

Sincerely
Susanne
 
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  • #2
Sure, looks Bernoulliriffic to me. Just trust yourself, go through the solution strategy for the Bernoulli equations, and you will arrive at a result. It is always good to try things out before asking questions. Maybe you can solve that with exact differentials + integrating factor as well, but at least using solution methods a la Bernoulli works out just fine too.
 
  • #3
LawlQuals said:
Sure, looks Bernoulliriffic to me. Just trust yourself, go through the solution strategy for the Bernoulli equations, and you will arrive at a result. It is always good to try things out before asking questions.


Okay thanks Lawl,

The reason I'm asking is that I get the right solution using Maple and the wrong one by hand :(

But now I know that it is a Bernoulli diff.eqn thanks :)

But why do I keep getting the wrong result by hand??:uhh:

Maple says to me Susanne the solution is

[tex]y(x) = \frac{2x}{x^2+2\cdotC}[/tex]

But then I use my own brain I get a totally different result. Look.

First I choose [tex]w = y^{-1}[/tex]

Why by the solution method for Bernoulli diff.eqn from my textbook( Zigs First Course in differential eqn p. 62-63).

[tex]\frac{dw}{dx}+ \frac{1}{x}w = 1[/tex]

I find the integration factor to be x^-1.

thus I get that [tex]x^{-1} \cdot w = \int(\frac{1}{x})dx = ln(x)[/tex]


and by replacing w with y^-1

I get

[tex]y(x) = \frac{1}{x\cdot ln(x)+kx}[/tex]

Which as you can see is lightyears away from what Maple says the solution. What am I doing wrong? Please point out where I in my calc are doing wrong :(
 
Last edited:
  • #4
The incongruence between your work and the solution quoted by Maple stems from the integrating factor. Please examine that again. And, note that Maple has (for some reason) chosen the constant of integration in your solution to be equal to unity.
 

1) What is a Bernoulli differential equation?

A Bernoulli differential equation is a type of nonlinear differential equation in the form of dy/dx + P(x)y = Q(x)y^n, where n is a constant. It is named after the Swiss mathematician Daniel Bernoulli and is commonly used in physics and engineering to model various phenomena.

2) How is a Bernoulli differential equation different from a linear differential equation?

A Bernoulli differential equation is different from a linear differential equation because it involves a variable raised to a power instead of a constant coefficient. This makes it a nonlinear equation, and it requires a different approach to solve compared to linear equations.

3) What are some applications of Bernoulli differential equations?

Bernoulli differential equations are commonly used in physics and engineering to model various phenomena such as population growth, chemical reactions, and fluid dynamics. They are also used in economics to model supply and demand, and in biology to model population growth and predator-prey relationships.

4) How do you solve a Bernoulli differential equation?

The general approach to solving a Bernoulli differential equation involves making a substitution of y = u^n, which transforms the equation into a linear differential equation in terms of u. The equation can then be solved using standard techniques such as separation of variables or integrating factors.

5) Are there any special cases of Bernoulli differential equations?

Yes, there are two special cases of Bernoulli differential equations that have simpler solutions. The first is when n = 0, which reduces the equation to a linear differential equation. The second is when n = 1, which is known as a Riccati equation and can be solved using a substitution and subsequent linearization.

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