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I Differential equations and numerical methods questions

  1. Jul 22, 2016 #1
    i have few doubts about differential equations and numerical methods ...
    in a differential equation question ... you are given an instantaneous rate of change...
    and you are supposed to find the function that gives , this instantaneous rate of change

    maths_5.png

    is this same as ...

    f(0)=1,
    f(0.01)=1.01,
    f(0.02)=1.0201,
    f(0.03)= 1.030301

    ??
     
  2. jcsd
  3. Jul 22, 2016 #2

    BvU

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    Yes. Note that replacing ##\ dy\over dx\ ## by ##\ \Delta y\over \Delta x\ ## is an approximation....
     
  4. Jul 22, 2016 #3
    thanks ...

    maths_3.png

    does anyone know how to re write this equation for numerical approximations ??
     
  5. Jul 22, 2016 #4

    BvU

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    No approximation needed: you have solved the equation....
     
  6. Jul 22, 2016 #5
    i am trying to understand some things in plain english ...
    in a differential equation question,you are given an instantaneous rate of change of an object
    The gradient of the curve dy/dx is the instantaneous change of y with respect to x ,the rate at which y is changing as x is changing
    and you are supposed to find the function that gives , this instantaneous rate of change

    maths_5.png

    and this is same as ...

    f(0)=1,
    f(0.01)=1.01,
    f(0.02)=1.0201,
    f(0.03)= 1.030301

    ??


    wasn't i supposed to find the function that gives ,this instantaneous rate of change ??
    The answer is the function that has that rate of change
    The answer is a function that has the instantaneous rate of change that you were given ...


    now i have four functions and four instantaneous rate of change ??
    what does this mean ??
    does it mean that i have four functions that has the instantaneous rate of change that i was given

    ??
     
  7. Jul 22, 2016 #6

    BvU

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    You have two equations with a different character: one (##{dy\over dx}=x^2-3##) is of the type ## y'= f(x) ## and the other is of the type ##y'= f(y)## (I use the abbreviation ##y'## for the derivative ##{dy\over dx}## ).

    ## y'= f(x) ## is solved if you can find a primitive of f : a function ##F(x)## for which ##F'= f##.

    There is the small matter of the integration constant: If you know your rate of change but you don't know where you start, then you can not determine where you will be next.

    In other words: With every differential eqiuation you need something to deal with the integration constant (one for every "quote") . For example a function value at x = 0 (or a function value at any other point).

    Your first equation isn't accompanied by such a condition, so you can't get further than ## y = F(x) = x^3/3-3x + K##.

    Your second differential equation does have a condition: ##y(0) = 1##. You solve numerically by approaching y(0) as a linear function for a small step:$$y(0+\Delta x) \approx y(0) + \Delta x {dy\over dx}$$ when all you really know is $$ \lim_{h\downarrow 0} { y(0+h) - y(0) \over h } = {dy\over dx} \ .$$
    When you take a few steps you get a sequence of approximated function values, but not a function (in the sense of an exact prescription what to do with x to get y).

    So for ##\ y'= y\ \ \& \ \ y(0) = 1 \ ## you have found four function values, not four functions.
     
    Last edited: Jul 22, 2016
  8. Jul 22, 2016 #7
    BvU ,

    thanks a lot .. there is so much good information in that post ... i have already learned a lot from your post ... now i think its just a matter of googling the proper words from your post ..



    even differential equation is kind of getting interesting ...

    few more basic questions ...

    i dont understand this properly ...

    most of the time , the original question is ...

    find the function that gives ,this instantaneous rate of change


    maths_5.png

    which is same as this ..

    f(0)=1,
    f(0.01)=1.01,
    f(0.02)=1.0201,
    f(0.03)= 1.030301




    for ##y'= y & y(0) = 1 ## you have found four function values, not four functions.

    what is the next thing you do with these function values... you plug it into the equations ??

    i really thought the answer is a function ..that has that rate of change

    i mean this is how a question really starts ...

    find the function that gives ,this instantaneous rate of change

    <Mod note: Irrelevant image deleted>


    if answer is a function that has that rate of change ...

    what are these four function values ...

    ???
     
    Last edited by a moderator: Jul 29, 2016
  9. Jul 22, 2016 #8

    BvU

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    Nice picture; taken from inside the book ? (I recognize Rotterdam, though...)

    Seems to me you still have trouble understanding what you did for ##y'=y## so let's step back and treat ##y'= x^2-3## the same way. Add initial condition y(0) = 0 so it yields a unique solution.

    ##x = 0 \rightarrow y(0) = 0##

    ##x = 0.1 \rightarrow y(0.1) = 0 + (0.01 - 3)* 0.1 = -0.299 \ ## exact solution ##y(0.1) = 0.001 - 3 = -0.299 \ ## hurray.

    etc. etc. we get:
    upload_2016-7-22_16-21-54.png

    upload_2016-7-22_16-44-51.png

    So the exact solution is a function ##y = x^3/3-3x## and the numerical approach is a table of aproximated function values.

    (I took steps of 0.1 so the difference between exact and numerical becomes visible).

    In fact, for your other equation ##y'=y## and ##y(0)=1## an exact solution can also be found. Use the same method (everything with y on one side gives ##{dy\over y}=dx##, then integrate and apply the initial condition). But you'll learn that one soon enough.
     

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  10. Jul 22, 2016 #9
    let me re arrange a few things ...so that i can understand this properly ...


    ok , there are two questions in this one ...

    find the function that gives, this instantaneous rate of change ...


    maths_5.png


    which is same as this ..

    f(0)=1,
    f(0.01)=1.01,
    f(0.02)=1.0201,
    f(0.03)= 1.030301








    for ##y'= y & y(0) = 1 ## you have found four function values, not four functions.

    for ##f(x)=y & f(0) = 1 ## you have found four function values, not four functions.

    f(0)=1,
    f(0.01)=1.01,
    f(0.02)=1.0201,
    f(0.03)= 1.030301







    BvU ,

    thanks for the reply ...

    its from a book called "numerical methods for ordinary differential equations " ...i did a horizontal flip of that image with ms paint ..
    Mod note: Why? How does that add any information to your question? Also, you have posted an image of a bridge twice, which also adds nothing of substance to what you're asking.
    few more re arrangements in a way that i can understand this ...


    i am going to take a little bit of time to read and re read this ...

    there is so much good information again on this one ...

    let me see what i can learn from this , and i don't know where to move to from here ...

    again , thanks a lot ...

    if i have more doubts i will come back and post more ...
     
    Last edited by a moderator: Jul 29, 2016
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