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

[Mechdude]

Homework Statement

\frac{dy}{dx}= - \frac{c}{n} y^{n}

Homework Equations

y' + p(x)y=q(x) y^n

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

 

[rock.freak667]

If c and n are constants then you can just divide by yⁿ and you'd have a separable ODE

 

[Mechdude]

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

 

[LCKurtz]

If c and n are constants then you can just divide by yⁿ and you'd have a separable ODE

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.

 

[Mechdude]

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