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Thread: y = f(y) ?

  1. #1 y = f(y) ? 
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    Hi, I'll start calculus next year but I am trying to self-teach.

    I was wondering what is called a function in which the increment of y (derivative) is not based on x but on y itself.
    If that is not clear, consider a function in which x is time and y is space t*v, if v =1 then the derivative of space is constant = 1.
    If we consider the increment of space with respect to space, then the increment is not constant and is 1/y. so y(t+1)= y(t)+1/y

    Probably this is a mess, I hope you can help me and say it this exists and has a name or how it shoud be formulated.
    Thanks for your help.


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  3. #2  
    KJW
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    An autonomous differential equation is a differential equation which does not explicitly depend on the independent variable.


    There are no paradoxes in relativity, just people's misunderstandings of it.
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  4. #3  
    Forum Junior whizkid's Avatar
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    It is y'/y
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  5. #4  
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    Quote Originally Posted by KJW View Post
    An autonomous differential equation is a differential equation which does not explicitly depend on the independent variable.
    Thanks, I googled that and found it's an autonomous system, can you tell me how to formulate it correctly, is this the only way?

    An autonomous system is a ...of the form
    where x takes values...


    Can I say y(tn+1)= y(tn)+1/y(tn)
    Last edited by Password; September 4th, 2014 at 05:52 AM.
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  6. #5  
    New Member HyD@'s Avatar
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    Can I say y(tn+1)= y(tn)+1/y(tn)

    I think your equation holds true if you replace the with , where is very small and approaching 0.


    The problem with your equation is you have assumed that the rate of change of y at to be constant in the interval between , which need not be the case.


    means that the rate of change of the function x depends on the value of x at t. The moment t changes, the rate of change of the function takes a new value, which need not be same.
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  7. #6  
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    This post is old, but I'd almost like to bargain that Password's password for his account is Username...
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