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Math Help - solving an intergal without knowing the funtion

  1. #1
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    solving an intergal without knowing the funtion

    \int_{0}^{^{X}}(x-t)y(t)=2x+\int_{0}^{^{X}}y(t)
    iknow that i should differentiate by x

    \frac{d\int_{0}^{^{X}}(x-t)y(t)}{dx}=\frac{2x+\int_{0}^{^{X}}y(t)}{dx}
    but i dont know if it the correct way.
    i know that differentiation cancels integration but here i have intervals with variables on the integrals
    and i dont know what is y(x)
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  2. #2
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    Assuming your variable of integration is t, you would do this:

    \displaystyle\int_{0}^{x}(x-t)y(t)\,dt=2x+\int_{0}^{x}y(t)\,dt

    \displaystyle x\int_{0}^{x}y(t)\,dt-\int_{0}^{x}t y(t)\,dt=2x+\int_{0}^{x}y(t)\,dt.

    Differentiating yields

    \displaystyle \int_{0}^{x}y(t)\,dt+xy(x)-x y(x)=2+y(x).

    Can you go from here?
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  3. #3
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    why you dissregarded the zero in the lower interval when you differentiate the integral.
    and why the integra of ty(t) turns into xy(x)-xy(x) and not xy(x)-xy(0)
    ?
    about the going further ,i tried is it ok

     \int_{0}^{x}y(t)\,dt+xy(x)-x y(x)=2+y(x)
    turns into
     \int_{0}^{x}y(t)\,dt+0=2+y(x)
    turns to
     \int_{0}^{x}y(t)\,dt+y(x)=2
    after another differentiation
    turns to
     y'(x)+y(x)=0
    correct?
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  4. #4
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    why you disregarded the zero in the lower interval when you differentiate the integral
    Because, when you plug in the zero, the result is a constant. The derivative of a constant is zero.

    and why the integra of ty(t) turns into xy(x)-xy(x) and not xy(x)-xy(0)
    .

    It didn't come from that. You have to use the product rule on

    \displaystyle\frac{d}{dx}\left[x\int_{0}^{x}y(t)\,dt\right]=\int_{0}^{x}y(t)\,dt+xy(x).

    y'(x)+y(x)=0
    You have a sign error there, as well as a losing of the constant. It should be

    y'(x)-y(x)=2.

    [EDIT]: See below for a correction.
    Last edited by Ackbeet; October 21st 2010 at 03:32 PM. Reason: Correction.
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  5. #5
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    regarding the first clame:
    when we do a deriavtive to an integral we dont do the derivative of the inside function we just put one interval function input minus other interval function input
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  6. #6
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    Ah, but what you're talking about is only evaluating the integral in the first place. The next step in our problem is to differentiate. So yes, you start with

    \displaystyle\int_{0}^{x}f(t)\,dt=F(x)-F(0), by the fundamental theorem of calculus, part 2. But then you differentiate thus:

    \displaystyle\frac{d}{dx}\int_{0}^{x}f(t)\,dt=\fra  c{d}{dx}F(x)-0=f(x), by the fundamental theorem of calculus, part 1.
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  7. #7
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    ahhh thanks now i get it
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  8. #8
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    You're welcome. All set?
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  9. #9
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    how did you differentiatied \int_{0}^{x}t y(t)
    ?
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  10. #10
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    It's just the fundamental theorem of the calculus. You get

    \displaystyle\frac{d}{dx}\int_{0}^{x}ty(t)\,dt=xy(  x).
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  11. #11
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    i got it
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  12. #12
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    Ok, great. Have a good one!
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  13. #13
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    Sorry from post #2 I get

    <br />
y'(x)-y(x)=0.<br />
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  14. #14
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    Oh, right. You have to differentiate once more. My mistake.
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  15. #15
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    Quote Originally Posted by transgalactic View Post
    \int_{0}^{^{X}}(x-t)y(t)=2x+\int_{0}^{^{X}}y(t)
    iknow that i should differentiate by x

    \frac{d\int_{0}^{^{X}}(x-t)y(t)}{dx}=\frac{2x+\int_{0}^{^{X}}y(t)}{dx}
    but i dont know if it the correct way.
    i know that differentiation cancels integration but here i have intervals with variables on the integrals
    and i dont know what is y(x)
    I know it is a bit late but there is a different approch using the LaPlace transform.

    The left hand side is the convolution of two functions so its transform is the product of the 2.

    Taking the transform gives

    \frac{1}{s^2}Y(s)=\frac{2}{s^2}+\frac{1}{s}Y(s)

    Solving this for Y(s) gives

    Y(s)=\frac{2}{s+1} and inverting the transform gives

    y(x)=2e^{-x}
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