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Math Help - Solve the equation ( hyperbolic functions)

  1. #1
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    Solve the equation ( hyperbolic functions)

    6 sinh (theta) - 4 cosh (theta) - 3 = 0

    giving the root correct to 3 decimal places.


    I have arrived at the following equation thus far:

    e^theta - 5e^-theta - 3 = 0.

    Is this correct?
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Steve Mckenna View Post
    6 sinh (theta) - 4 cosh (theta) - 3 = 0

    giving the root correct to 3 decimal places.


    I have arrived at the following equation thus far:

    e^theta - 5e^-theta - 3 = 0.

    Is this correct?
    Yes, it's correct.

    6\sinh \theta  - 4\cosh \theta  - 3 = 0 \Leftrightarrow 6 \cdot \frac{{{e^\theta } - {e^{ - \theta }}}}<br />
{2} - 4 \cdot \frac{{{e^\theta } + {e^{ - \theta }}}}<br />
{2} - 3 = 0 \Leftrightarrow

    \Leftrightarrow 3\left( {{e^\theta } - {e^{ - \theta }}} \right) - 2\left( {{e^\theta } + {e^{ - \theta }}} \right) - 3 = 0 \Leftrightarrow {e^\theta } - \frac{5}{{{e^\theta }}} - 3 = 0 \Leftrightarrow

    \Leftrightarrow {e^{2\theta }} - 3{e^\theta } - 5 = 0.

    {\text{Let }}{e^\theta } = t \Rightarrow {e^{2\theta }} = {t^2}.

    {t^2} - 3t - 5 = 0 \Rightarrow {t_{1,2}} = \frac{{3 \pm \sqrt {29} }}{2}.

    \frac{{3 + \sqrt {29} }}{2} > 0,{\text{ }}\frac{{3 - \sqrt {29} }}{2} < 0.

    {e^\theta } = \frac{{3 + \sqrt {29} }}{2} \Leftrightarrow \theta  = \ln \frac{{3 + \sqrt {29} }}{2}.
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  3. #3
    Super Member girdav's Avatar
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    I have the same equation.
    Let e^{\theta} = t and we have to solve t-5\frac1t -3=0 \Leftrightarrow t^2-3t-5 = 0.
    Last edited by girdav; July 29th 2009 at 12:04 AM. Reason: error
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  4. #4
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    Hi steve

    Yes, that's correct

    EDIT :
    @girdav : there's typo, 1 should be 5 ^^
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