# Solve the equation ( hyperbolic functions)

• July 28th 2009, 11:35 PM
Steve Mckenna
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?
• July 29th 2009, 12:28 AM
DeMath
Quote:

Originally Posted by Steve Mckenna
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.

$6\sinh \theta - 4\cosh \theta - 3 = 0 \Leftrightarrow 6 \cdot \frac{{{e^\theta } - {e^{ - \theta }}}}
{2} - 4 \cdot \frac{{{e^\theta } + {e^{ - \theta }}}}
{2} - 3 = 0 \Leftrightarrow$

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

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

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

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

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

${e^\theta } = \frac{{3 + \sqrt {29} }}{2} \Leftrightarrow \theta = \ln \frac{{3 + \sqrt {29} }}{2}.$
• July 29th 2009, 12:30 AM
girdav
I have the same equation.
Let $e^{\theta} = t$ and we have to solve $t-5\frac1t -3=0 \Leftrightarrow t^2-3t-5 = 0$.
• July 29th 2009, 12:30 AM
songoku
Hi steve

Yes, that's correct

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