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Math Help - Tough ODE problem

  1. #1
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    Tough ODE problem

    Solve the ODE:

    \frac{dy}{dx}=-\frac{y}{\sqrt{a^2-y^2}}

    3. The attempt at a solution

    My teacher hinted that the substitution z^2=a^2-y^2 could be helpful, but once I make the substitution, I can't seem to take the next step.

    z^2=a^2-y^2

    Differentiating WRT x, we have:

    2z\frac{dz}{dx}=-2y\frac{dy}{dx}

    Substituting back in we have:

    \frac{dz}{dx}=\frac{a^2-z^2}{z^2}

    But I can't take it from here... much less obtain an equation for y(x). Any thoughts?
    Last edited by justaboy; September 21st 2010 at 03:38 PM.
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  2. #2
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    Quote Originally Posted by justaboy View Post
    Solve the ODE:

    \frac{dy}{dx}=-\frac{-y}{\sqrt{a^2-y^2}}

    3. The attempt at a solution

    My teacher hinted that the substitution z^2=a^2-y^2 could be helpful, but once I make the substitution, I can't seem to take the next step.

    z^2=a^2-y^2

    Integrating WRT x, we have:



    Substituting back into the equation, we have:

    \frac{dz}{dx}=\frac{a^2-z^2}{z^2}

    But I can't take it from here... much less obtain an equation for y(x). Any thoughts?
    \displaystyle \frac{dx}{dy} = \frac{- \sqrt{a^2 - y^2}}{y} \Rightarrow x = - \int \frac{\sqrt{a^2 - y^2}}{y} \, dy.

    Now use the substitution u = a^2 - y^2 etc.
    Last edited by mr fantastic; September 21st 2010 at 04:03 PM. Reason: Edited due to edit in original post.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \displaystyle \frac{dx}{dy} = \frac{\sqrt{a^2 - y^2}}{y} \Rightarrow x = \int \frac{\sqrt{a^2 - y^2}}{y} \, dy.

    Now use the substitution u = a^2 - y^2 etc.
    Ok, making the sub z^2=a^2-y^2 and plugging that into x = -\int \frac{\sqrt{a^2 - y^2}}{y} \, dy I get

    x = \int \frac{z}{a^2 - z^2} \, dz. Sort of the same as I ended up with in the OP, only the inverse. How do I take this from here?
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  4. #4
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    Quote Originally Posted by justaboy View Post
    Ok, making the sub z^2=a^2-y^2 and plugging that into x = -\int \frac{\sqrt{a^2 - y^2}}{y} \, dy I get

    x = \int \frac{z}{\sqrt{a^2 - z^2}} \, dz. Sort of the same as I ended up with in the OP, only inversed. How do I take this from here?
    Please read my reply more carefully. That is NOT the substitution I advised.
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  5. #5
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    You're the boss. If u=a^2-y^2 then

     x = \int \frac{du}{2u\sqrt{a^2-u^2}}
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  6. #6
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    Quote Originally Posted by justaboy View Post
    You're the boss. If u=a^2-y^2 then

     x = \int \frac{du}{2u\sqrt{a^2-u^2}}
    No, you don't get that. Nevertheless, what you do get is not directly helpful. So forget my initial suggestion. The given hint of u^2 = a^2 - y^2 is better.

    You get \displaystyle \int \frac{u^2}{a^2 - u^2} \, du = - \int \frac{u^2}{u^2 - a^2} \, du = - \int 1 + \frac{a^2}{u^2 - a^2} \, du.

    Partial fraction decomposition should be used to integrate the second term.
    Last edited by mr fantastic; September 21st 2010 at 04:04 PM. Reason: Edited due to edit in original post.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post

    You get \displaystyle \int \frac{u^2}{a^2 - u^2} \, du = - \int \frac{u^2}{u^2 - a^2} \, du = - \int 1 + \frac{a^2}{u^2 - a^2} \, du.
    I had accidently added a negative sign to the initial equation in my first post... anyways, with that in mind, I get:

     \int \frac{u^2}{a^2-u^2} \, du

    so x=\frac{1}{2}(-a\log(u-a)+a\log(a+u)-2u)+c
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    Quote Originally Posted by justaboy View Post
    I had accidently added a negative sign to the initial equation in my first post... anyways, with that in mind, I get:

     \int \frac{u^2}{a^2-u^2} \, du

    so x=\frac{1}{2}(-a\log(u-a)+a\log(a+u)-2u)+c
    That looks OK: http://www.wolframalpha.com/input/?i...-+y%5E2%5D%2Fy

    (Note that the form given by Wolfram is not the simplest)
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  9. #9
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    Wolfram also returns this solution:

    x = a*arctanh(\frac{\sqrt{a^2-y^2}}{a})-\sqrt{a^2-y^2}+c

    In either case is it possible to solve for y(x)?



    Thanks for you time.
    Last edited by justaboy; September 21st 2010 at 06:53 PM.
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  10. #10
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    Quote Originally Posted by justaboy View Post
    Wolfram also returns this solution:

    x = a*arctanh(\frac{\sqrt{a^2-y^2}}{a})-\sqrt{a^2-y^2}+c

    In either case is it possible to solve for y(x)?



    Thanks for you time.
    No.
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  11. #11
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    I just want to point out that there are integrals of the following type
    \displaystyle\int x^m(a+bx^n)^p dx
    that can be solved only in three cases:

    1. p\in\mathbb{Z}
    2. \frac{m+1}{n}\in\mathbb{Z} substitute a+bx^n=z^s where  s is the denominator of the fraction p
    3. \frac{m+1}{n}+p\in\mathbb{Z} substitute ax^{-n}+b=z^s where again  s is the denominator of the fraction p

    The integral here is one such integral \displaystyle\int\frac{\sqrt{a^2-y^2}}{y}dy=\int y^{-1}(a^2-y^2)^{\frac{1}{2}}dy in which m=-1, n=2 and p=\frac{1}{2}. Notice that \frac{m+1}{n}=0\in\mathbb{Z} hence it can be solved using the substitution a^2-y^2=z^2.

    Of course, your teacher gave you that as a hint, and you and the rest of the crew here solved the problem. I think it is worth noting that this integral is nothing special.
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  12. #12
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    I don't think my prof would have assigned an impossible question to me though... there's no way to find y(x)?
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  13. #13
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    Quote Originally Posted by justaboy View Post
    I don't think my prof would have assigned an impossible question to me though... there's no way to find y(x)?
    There is nothing wrong with an implicit solution like the one that has been found. I suggest you ask your professor what his/her expectation is.
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