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Math Help - a general Differential Equations question

  1. #1
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    a general Differential Equations question

    Hello all ,
    i have the equation
    Code:
    (1+x^2)y'+y=arctgx
    now i managed to find the formula for it and got to:
    Code:
     
    y=e^-arctanx[~(arctanx*e^tanx*dx)/(1+x^2)]
    where ~ is the sign for integral
    now i'm supposed to do t=arctanX
    but don't understand where i get the dt from ?
    how do i find the DT is what i'm asking here

    thanks to all the helpers !!
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: a general Differential Equations question

    Quote Originally Posted by smacker View Post
    Hello all ,
    i have the equation
    Code:
    (1+x^2)y'+y=arctgx
    now i managed to find the formula for it and got to:
    Code:
     
    y=e^-arctanx[~(arctanx*e^tanx*dx)/(1+x^2)]
    where ~ is the sign for integral
    now i'm supposed to do t=arctanX
    but don't understand where i get the dt from ?
    how do i find the DT is what i'm asking here

    thanks to all the helpers !!
    It looks like a standard substitution.
    t = arctan(x)

    What is the derivative of arctan(x)? Then compare that form with your integrand. (You'll have to find dx in terms of dt so as a hint you might want to do this by x = tan(t) etc.)

    -Dan
    Last edited by topsquark; November 4th 2012 at 01:59 PM.
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  3. #3
    Newbie
    Joined
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    Israel
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    Re: a general Differential Equations question

    well the answer will be 1/1+x^2
    but the question is - is it always the derrative ?
    like for example :
    lnx\x
    and i want to call t=lnx
    so it will come out tdt ?
    the 1/x becomes the dt ?
    if so math is awsome
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