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Math Help - Help substituting !

  1. #1
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    Help substituting !

    I have

    (D^2theta/ds^2)+((2/s)dtheta/ds)+((k^2)/s)theta=0

    I am asked to make the subtitution x=2k*sqrt(s) in order to get the above in terms of theta and x.

    How do i do this?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by LooNiE View Post
    I have

    (D^2theta/ds^2)+((2/s)dtheta/ds)+((k^2)/s)theta=0

    I am asked to make the subtitution x=2k*sqrt(s) in order to get the above in terms of theta and x.

    How do i do this?
    I'm assuming you mean,

     \frac{D^2 \theta}{dS^2} + \frac{2d \theta}{sdS} + \frac{k^2 \theta}{s}=0

    edit- Mulligen, I think i violated some rules here...1 momento

    I'm not entirely sure off the bat where to go with this. But let us try some manipulation.

     \frac{D^2 \theta}{dS^2} + \frac{k^2 \theta}{s}= - \frac{2d \theta}{sdS}

     \frac{s D^2}{dS} + k^2 dS= - \frac{2d \theta}{ \theta }

    Noting that

    x=2k \sqrt{s}

    s = \frac{x^2}{4k^2}

    Going with

    \frac {dS}{dX}= \frac {2x}{4k^2}

    dS= dX \frac {2x}{4k^2}

    Sub this into our original equation to get

     \frac{s D^2}{ \frac{2xdX}{4k^2}} + k^2 ( \frac{2xdX}{4k^2} )= - \frac{2d \theta}{ \theta }

     \frac{4k^2 s D^2}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }

    Now we sub in

    s = \frac{x^2}{4k^2}

    To get

     \frac{4k^2 D^2 \frac{x^2}{4k^2}}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }

     \frac{D^2 x}{ 2dX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }

     \frac{D^2 x}{ dX} + xdX= - \frac{d \theta}{ \theta }

    And since I had no idea where to begin, I have run out of manipulations that I can think of. Hopefully this will prompt some of your own!
    Last edited by AllanCuz; April 15th 2010 at 11:53 AM.
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  3. #3
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    Yes, that's what I meant.
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  4. #4
    Senior Member AllanCuz's Avatar
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    I'd like to see an answer to this if you guys wouldn't mind I'll work on it more when i decide i need more of a break from exam studying.
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