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Math Help - variable substitution question(diff)

  1. #1
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    variable substitution question(diff)

    \frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\
    t=\arctan y\\
    t'=\frac{1}{1+y^2}y'\\
    \frac{x}{\ln (x)t-1}=\frac{t}{t'}\\

    \frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\
    xdt=(\ln (x)t-1)tdx\\
    \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\
    still cant beak it as one type of variable on each side
    so i substitute by another variable
    z=\ln x\\
    dz=\frac{dx}{x}\\
    \frac{dt}{zt-1}=dz\\
    i dont know how to separate each variable type on one side
    so i could integrate

    ??
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  2. #2
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    Hi transgalactic

    Do you know integrating factor ?
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  3. #3
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    yes i know that.but how did you know that that i should solve it like this in here?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by transgalactic View Post
    [snip]

    \frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\
    t=\arctan y\\
    t'=\frac{1}{1+y^2}y'\\
    \frac{x}{\ln (x)t-1}=\frac{t}{t'}\\

    \frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\
    xdt=(\ln (x)t-1)tdx\\
    \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\

    [/snip]
    From here,

    \frac{\,dt}{t\ln x-1}=\frac{\,dx}{x}\implies \frac{\,dt}{\,dx}=\frac{\ln x}{x}t-\frac{1}{x} \implies \frac{\,dt}{\,dx}-\frac{\ln x}{x}t=-\frac{1}{x}\sim \frac{\,dt}{\,dx}+P(x)t=Q(x).

    Can you take it from here and apply the integrating factor technique (as songoku mentioned)?
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  5. #5
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    ok i rearanged it like i used to in order to find integration factor
    xdt=(tlnx-1)dx
    i got that the integration factor is:
    y=e^{\frac{ln^2x}{2}-lnx}
    so i mltiply both sides by y
    xe^{\frac{ln^2x}{2}-lnx}dt=(tlnx-1)e^{\frac{ln^2x}{2}-lnx}dx
    now its even more complicated than before
    ??
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  6. #6
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    Hi transgalactic

    Sorry, I have tried to solve this and ended up finding the integration that can't be stated in elementary function.
    For now, maybe we can wait for Chris L T521's help
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  7. #7
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    this is a fll solution:
    http://i30.tinypic.com/2lllrpk.gif

    its really frustrating that i did all the things by the book and i didnt manage to solve its
    what is the mistake in diagnozing the wa of solution

    what should step there in my equation is a red light to solve it like
    the solution above?
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  8. #8
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    Hi transgalactic

    I re-check your work and find out the mistake.

    xdt=(\ln (x)t-1)tdx\\

    \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\ -----------> you missed the term "t"

    So, that's not linear form anymore but Bernoulli form
    Last edited by songoku; August 17th 2009 at 12:18 AM.
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  9. #9
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    <br />
\frac{dt}{(\ln (x)t-1)t}=\frac{dx}{x}\\<br />
    <br />
z=\ln x<br />
    <br />
\frac{dt}{(zt-1)t}=dz\\<br />

    i dont have here a bernuly equation ..
    what to do?
    bernule needs to have f'() form
    i dont know how transform it to this form
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  10. #10
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    Hi transgalactic

    \frac{dt}{(zt-1)t}=dz\\

    \frac{dt}{dz}=zt^2-t

    \frac{dt}{dz}+t=zt^2
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  11. #11
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    here is another problem:
    all those variable are confusing
    how did you know to t it in this way
    dt/dz

    or

    dz/dt

    why you put the derivative part in that way

    what is the logic behind it
    ?
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  12. #12
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    Hi transgalactic

    It's not matter whether you solve \frac{dt}{dz} or \frac{dz}{dt} because it will give you the same result.

    If you state it in \frac{dz}{dt} , it will be :

    \frac{dz}{dt}=\frac{1}{zt^2-t}

    Can you solve it ?
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  13. #13
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    the problem is that i cant see what variable is a function of what variable
    if i had dy and dx
    then i whould do
    dy/dx

    but here dz and dt
    i dont know who is more superior
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  14. #14
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    From here,

    \frac{\,dt}{t\ln x-1}=\frac{\,dx}{x}\implies \frac{\,dt}{\,dx}=\frac{\ln x}{x}t-\frac{1}{x} \implies \frac{\,dt}{\,dx}-\frac{\ln x}{x}t=-\frac{1}{x}\sim \frac{\,dt}{\,dx}+P(x)t=Q(x).

    Can you take it from here and apply the integrating factor technique (as songoku mentioned)?
    The integrating factor is \varrho\left(x\right)=e^{-\int\frac{\ln x}{x}\,dx}=e^{-\frac{\ln^2\left(x\right)}{2}}.

    So it follows now that \frac{\,d}{\,dx}\left[te^{-\frac{\ln^2\left(x\right)}{2}}\right]=-\frac{1}{x}e^{\frac{-\ln^2\left(x\right)}{2}}

    Thus, te^{-\frac{\ln^2\left(x\right)}{2}}=-\int\frac{1}{x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\implies t=-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1}  {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx.

    Therefore, \arctan y=-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1}  {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\implies y=\tan\left[-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1}  {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\right]

    It looks weird, but I guess that's the solution you're looking for...
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  15. #15
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    Hi Chris L T521 and transgalactic

    @ Chris L T521:
    There is error in OP's work.

    @transgalactic :
    Example : \frac{dy}{dx}=x

    dy=x\; dx

    y=\frac{1}{2}\; x^2 + c

    If you re-arrange the question : \frac{dx}{dy}=\frac{1}{x}

    x\; dx = dy

    \frac{1}{2}x^2+c=y

    You will have same result. So, it doesn't matter if you solve \frac{dt}{dz} or \frac{dz}{dt}
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