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I am trying to derive

$y = \exp \left(\dfrac{a+b}2t \right) \left(k_1 \cosh \left(\dfrac{(a-b)t}2 \right) + k_2 \sinh \left(\dfrac{(a-b)t}2 \right) \right)$

From $y = c_1 \exp(at) + c_2 \exp(bt)$

I've managed to get the $ \exp \left ( \dfrac{a+b}2t \right)\cosh \left ( \dfrac{(a-b)t}{2} \right) $ part, but I cannot the hyperbolic sine part because of the minus sign. Remember, I am going from $y = c_1 \exp(at) + c_2 \exp(bt)$ to $y = \exp \left(\dfrac{a+b}2t \right) \left(k_1 \cosh \left(\dfrac{(a-b)t}2 \right) + k_2 \sinh \left(\dfrac{(a-b)t}2 \right) \right)$, not the other way around

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    I've left all the details you might need. I recommend beginning with your first item with the hyperbolic trig functions, reduce to the simplest form, then carefully write out how to reverse the steps.2012-11-24

2 Answers 2

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\begin{align} y & = c_1 \exp(at) + c_2 \exp(bt) = c_1 \exp\left(\left(\dfrac{a+b}2 + \dfrac{a-b}2 \right)t \right) + c_2 \exp\left(\left(\dfrac{a+b}2 - \dfrac{a-b}2 \right)t \right)\\ & = c_1 \exp\left(\left(\dfrac{a+b}2\right)t \right) \exp \left( \left(\dfrac{a-b}2 \right)t \right) + c_2 \exp\left(\left(\dfrac{a+b}2\right)t \right) \exp \left( -\left(\dfrac{a-b}2 \right)t \right)\\ & = \exp\left(\left(\dfrac{a+b}2\right)t \right) \left(c_1 \exp \left( \left(\dfrac{a-b}2 \right)t \right) + c_2 \exp \left(- \left(\dfrac{a-b}2 \right)t \right) \right)\\ \end{align} Now recall that $\cosh(x) = \dfrac{\exp(x) + \exp(-x)}2$ and $\sinh(x) = \dfrac{\exp(x) - \exp(-x)}2$ Hence, we get that $\exp(x) = \cosh(x) + \sinh(x)$ and $\exp(-x) = \cosh(x) - \sinh(x)$ Hence, \begin{align} y& = \exp\left(\left(\dfrac{a+b}2\right)t \right) \left(c_1 \exp \left( \left(\dfrac{a-b}2 \right)t \right) + c_2 \exp \left(- \left(\dfrac{a-b}2 \right)t \right) \right)\\ & = \exp\left(\left(\dfrac{a+b}2\right)t \right) \left((c_1+c_2) \cosh \left( \left(\dfrac{a-b}2 \right)t \right) + (c_1 - c_2) \sinh \left( \left(\dfrac{a-b}2 \right)t \right) \right)\\ \end{align}

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    Oh yeah. Thanks. updated it.2012-11-24
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The direction does not matter. From the definitions, $ w_1 \cosh ut + w_2 \sinh u t = h_1 e^{ut} + h_2 e^{-ut} $

Let's see, $ (w_1 + w_2)/ 2 = h_1, \; \; (w_1 - w_2)/ 2 = h_2, $ so that $ h_1 + h_2 = w_1, \; \; h_1 - h_2 = w_2. $

For you $ u = (a - b)/2.$

You still need to be confident about this sort of thing: $ \exp \left(\dfrac{(a-b)t}2 \right) = \exp \left(\dfrac{at}2 \right) \exp \left(\dfrac{-bt}2 \right) \, . $ $ \exp \left(\dfrac{(b-a)t}2 \right) = \exp \left(\dfrac{bt}2 \right) \exp \left(\dfrac{-at}2 \right) \, . $