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Initial Value systems.

  1. Dec 15, 2004 #1
    Solve the intial value problem using integrals:
    [tex]dy/dx[/tex] = (cube root of x) , y(1) = 2

    The solution:
    if f(x) = F(x), and y(x) is equal to F(x), then,

    y(x) = ([tex]x sub 0[/tex], [tex]y sub 0[/tex])

    Here's what I did, and failed:

    y(1) = 2. This simply means (and I understand this) that if x is 1, then the function y(x) is 2 from the input of x equal to 1.

    So, in order to find the slope of the curve at the cube root of x, when y(1) = 2. We must integrate.

    this is the problem I am having.

    I get:
    y = [tex]integral[/tex] cube root of x [tex]dx[/tex] + C

    [tex]dy/dx[/tex] = cube root of 1 (substitued the intial value of x) + C

    At the coordinates (1,2)

    What am I doing wrong here?

    I dont' know how to integral the cube root of 3. Amd what is C? 2?

    Thank you.
     
  2. jcsd
  3. Dec 15, 2004 #2

    shmoe

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    You should have something general that tells you how to find the antiderivatice of [tex]x^a[/tex] where a is any real number not equal to -1. Think 'power rule' in reverse. In this case, you want an antiderivative of [tex]x^{1/3}[/tex].

    The +C constant in the antiderivative will be determined by your initial condition, y(1)=2.
     
  4. Dec 15, 2004 #3
    Thank you. I am a high school student taking geometry,11th grade, but I always feel under-par and very depressed if someone in my high school knows more calculus than I do. It' slike a strive for perfection, and its annoying. Along with that, I do have depression, suicidal behavioral, and OCD in real life.

    Anyway, here are my thoughts:
    If y(x) = F(x) of some function.

    then y(x) = 2 when (x) is 1.

    The function is an exponetal function curve, therefore it can be put in the slope form of y(x) = mx + b where m is the slope, x is the x intercept, and b is the constant of integration where the curve intercepts the y axis and where this equation is a deriviative of the family of derivatives of F(x) = mx + C.

    I have a trouble with differentation, and an even harder time figuring it out. If differentation is just the rate of change over time in a curve, then, how do they get that wacky formula for it? I'm dying to know.

    Since integration is backwards differentation, is that simliar to saying subtraction is backwards addition??


    In other words, if you can't figure something out with differentation, you can integrate?

    I guess I still dont' get the relationship.

    I forget the power rule. Since [tex]x^3[/tex] is also under the square root of 2, I can't find a calc that will do cube roots.

    And I have to differentate that, and then integrate (backwards differentate) the result of the derivative of it.

    please help, or do it. this isn't for school, it's for figuring it out myselff for fun so I can either work in economics in real life, or be a profession in physics when I graduate.

    thank you
     
  5. Dec 16, 2004 #4

    dextercioby

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    Aha...So that explains the post(s) in the QM forum. :wink: You have no idea about QM because you're in high school.That figures.

    Your problem is:
    Find the function [itex] y(x) [/itex],knowing that [itex]y(1)=2. [/itex] and that:
    [tex] \frac{dy}{dx} =x^{\frac{1}{3}} [/tex].

    Do you know how to integrate a function [itex] y(x)=x^{a} [/itex],when [itex] a\neq 0 [/itex],because i think you know that
    [tex] \int \frac{dy}{dx} dx = y+C [/tex]
    ,right??
    So i don't see what prevents you from integrating [tex] x^{\frac{1}{3}} [/tex]...?????

    Daniel.
     
    Last edited: Dec 16, 2004
  6. Dec 16, 2004 #5

    Gza

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    Everyone else got the math part of it right, but this part of your post concerns me. Please contact your school's counciling department immediately.
     
  7. Dec 16, 2004 #6

    shmoe

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    There will always be someone who knows more than you do in some subject, it's just a fact of life. You're in grade 11, there's no rush to learn calculus now. Do it for enjoyment not competition and don't be afraid to go slowly and learn concepts properly before moving on.

    Please take Gza's advice if you haven't already.
     
  8. Dec 16, 2004 #7
    No, I don't.
     
  9. Dec 16, 2004 #8

    dextercioby

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    In that case,why bother??Why do you want to know how to integrate that differential equation,if u have no idea how to do it all by yourself??Just to put us to work for you,so that it wouldn't make you any use...???

    Daniel.
     
  10. Dec 16, 2004 #9
    I was just looking for some help..

    I do know how to differentiate, I don't understand WHY the formula of finding a deriviative . (like where it comes from).

    I don't like being told, "it just works. use it", not that you are saying that, but that is doing math blind. I like to figure out how it works.

    What I didn't understand was the way you put your equation, which apperead to be 2 derivatives at once:

    [tex] \int \frac{dy}{dx} dx = y+C [/tex]

    This is the function of the equation obviously , the (input) to figure out the result.

    This simply says find the derivative (or rate of change at some instant moment of dx) of the function..

    I don't understand why there is need for dy/dx AND the dx since I already see dx once in dy/dx?
     
    Last edited: Dec 16, 2004
  11. Dec 16, 2004 #10

    dextercioby

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    1.Soory for being a little harsh on you,but i couln't see any cooperation from u.
    2.Almost everything (except definitons and axioms) in mathematics can be proven through (sometimes tedious) calculations.Even to the simple formula
    [tex] (x^{a})'=a x^{a-1} [/tex]
    for arbitrary "a" it can be given a thorough proof.If u want it,well please be my guest and search for it.In the books,the internet...U got where to chose from.
    3.Correspondind to the operation of differentiation,one defines the operation of anti-differentiation.That is,if u're given a function,u have to specify which function should u differentiate in order to get the original/given function.
    In the simple case stressed above,it reads:
    [tex] \int x^{a} dx=\frac{x^{a+1}}{a+1} +C [/tex]
    ,for all "a"'s except "-1".In that case:
    [tex] \int \frac{1}{x} dx=\ln x +C [/tex].

    4.I addopted a notation for the derivative of the function using differentials:
    [tex] f'(x)=:\frac{df(x)}{dx} [/tex]
    ,hence the fundamental formula which links differentiation and anti-differentiation gets the form:
    [tex] \int \frac{df(x)}{dx} dx = f(x)+C [/tex].

    5.Your problem is basically finding the antiderivative to the function [itex] f(x)=:y(x)=x^{\frac{1}{3}} [/itex],knowing that it (the unknown function=the antiderivative) verifies the supplemetary condition:antider(1)=2.

    I believe i've given u enough to get u started and not only.Show me the results and your work as well.

    Daniel.

    PS.And stop complaining if others are better than u at math.So what??U could be atop of them,if u wanted to,and if u'd be gettin your ass to work. :devil:
     
    Last edited: Dec 16, 2004
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