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Math Help - Hyperbolic Inverse Equation

  1. #1
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    Hyperbolic Inverse Equation

    Hi, I'm trying to work out the following:

    Find non-trivial values of A and B for which u=Asech^2xBx is a solution of (d^2u/dx^2)-u+3u^2.

    Any help would be really appreciated!
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  2. #2
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    (d^2u/dx^2)-u+3u^2
    That's not an equation. Where does the equals sign go?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    That's not an equation. Where does the equals sign go?
    Apologies, it's =0

    ((d^2u/dx^2)-u+3u^2=0)
    Last edited by AlanC877; January 22nd 2011 at 03:33 AM.
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  4. #4
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    If \displaystyle u = AB\,x\,\textrm{sech}^2\,{x} evaluate \displaystyle \frac{d^2u}{dx^2}.

    Then substitute this all into \displaystyle \frac{d^2u}{dx^2} - u + 3u^2 = 0.

    You should be able to solve for \displaystyle A,B.
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  5. #5
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    I would also point out that you might not be able to uniquely solve for A and B. But you just need to find nonzero values that work. Hence, if you find an equation for A in terms of B, just pick a value for B and determine the corresponding A. You're not asked to find unique nontrivial values, just values.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    If \displaystyle u = AB\,x\,\textrm{sech}^2\,{x} evaluate \displaystyle \frac{d^2u}{dx^2}.

    Then substitute this all into \displaystyle \frac{d^2u}{dx^2} - u + 3u^2 = 0.

    You should be able to solve for \displaystyle A,B.
    Sorry again, I made a stupid mistake when entering the question. u=Asech^2(Bx). Would I just follow the same method as you outlined?
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  7. #7
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    Quote Originally Posted by AlanC877 View Post
    Sorry again, I made a stupid mistake when entering the question. u=Asech^2(Bx). Would I just follow the same method as you outlined?
    Yes, and it should simplify the problem greatly...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Yes, and it should simplify the problem greatly...
    Thank you so much for all your help so far but I'm still a little stuck.

    So, f(x)=Asech^2(Bx), then f'(x)=-2ABsech^2(Bx)tanh(Bx), so f''(x)=4AB^2sech^2(Bx)tanh^2(Bx)-2AB^2sech^4(Bx)

    And 3u^2=3A^2sech^4(Bx)

    So the whole equation is 4AB^2sech^2(Bx)tanh^2(Bx)-2AB^2sech^4(Bx)-Asech^2(Bx)+3A^2sech^4(Bx)

    But I'm stuck on where to go, I've tried various relationships but can't get anything to work. Any more help would be extremely appreciated!
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  9. #9
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    Nevermind, I looked at it and realised I could take out Asech^2(Bx) as a common factor, then use 4B^2tanh^2(Bx)=1-4B^2sech^2(Bx) which allowed me to take out a further sech^2(Bx) common factor to get: Asech^4(Bx)(-6B^2+3A) and so -6b^2+3A=0 so A=2, B=1 is a solution. Thanks for your help.
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