Solving Bernoulli's ODE with y' + p(x)y=q(x)y^n

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In summary, the given equation is a separable ODE that can be solved using Bernoulli's method. Dividing by yn and using the change of variable v = y(1-n) transforms it into a linear first order equation that can be solved with an integrating factor. The source of the equation is a post on Physics Forums about a specific problem.
  • #1
Mechdude
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Homework Statement


[tex] \frac{dy}{dx}= - \frac{c}{n} y^{n} [/tex]


Homework Equations




[tex] y' + p(x)y=q(x) y^n [/tex]

The Attempt at a Solution


im strictly speaking able to do it , i just wanted to kno whether I am on the right track using bernoulli's equation, not that i can see any other methods!
;-)
cheers
 
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  • #2
If c and n are constants then you can just divide by yn and you'd have a separable ODE
 
  • #3
thanks , i try that ( c and n are constants). seems less involving if not actually suggesting i was on the wrong track
 
  • #4
rock.freak667 said:
If c and n are constants then you can just divide by yn and you'd have a separable ODE

Mechdude said:
thanks , i try that ( c and n are constants). seems less involving if not actually suggesting i was on the wrong track

I don't think you will find it to be separable. Bernoulli's method is the way to go. Dividing by yn puts it in the form

y(-n)y' +p(x)y(1-n)= q(x)

and the change variable v = y(1-n) transforms it into a linear first order equation which can be done with an integrating factor.

[edit] Correction -- your specific equation is indeed separable, it is the general Bernoulli equation that isn't.
 
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  • #5
LCKurtz said:
I don't think you will find it to be separable. Bernoulli's method is the way to go. Dividing by yn puts it in the form

y(-n)y' +p(x)y(1-n)= q(x)

and the change variable v = y(1-n) transforms it into a linear first order equation which can be done with an integrating factor.

[edit] Correction -- your specific equation is indeed separable, it is the general Bernoulli equation that isn't.

i did find it to be separable and got an expression for a problem I am working on that agrees with the solution provided, so I am confident about that . thanks though.

here's the source of the d.e. for the curious :
https://www.physicsforums.com/showthread.php?p=2587065#post2587065
 

FAQ: Solving Bernoulli's ODE with y' + p(x)y=q(x)y^n

What is Bernoulli's ODE?

Bernoulli's ODE (ordinary differential equation) is a type of first-order differential equation in the form of y' + p(x)y=q(x)y^n, where p(x) and q(x) are functions of x and n is a constant. It is named after the Swiss mathematician Jacob Bernoulli who first studied and solved this type of equation.

What is the difference between Bernoulli's ODE and other types of ODEs?

The main difference between Bernoulli's ODE and other types of ODEs is the presence of the term y^n, which makes it a non-linear equation. This term makes it more challenging to solve compared to linear or homogeneous ODEs.

How do you solve a Bernoulli's ODE?

To solve a Bernoulli's ODE, you can use the substitution method. By substituting y^(-n) = z, the equation can be transformed into a linear ODE, which can be solved using standard methods such as separation of variables or integrating factor. After obtaining the solution in terms of z, you can then substitute back to get the solution in terms of y.

What are the applications of Bernoulli's ODE?

Bernoulli's ODE is commonly used in various fields, including physics, engineering, and economics. In physics, it is used to model phenomena like population growth, radioactive decay, and chemical reactions. In engineering, it is used in the design of control systems and to study the behavior of fluids. In economics, it is used to analyze the relationship between variables in economic models.

Are there any special cases of Bernoulli's ODE?

Yes, there are two special cases of Bernoulli's ODE: when n=0 and when n=1. When n=0, the equation becomes linear, and when n=1, it becomes separable. These cases have simpler solutions compared to the general case of Bernoulli's ODE and are often used as starting points for solving more complex equations.

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