How do I show this hyperbolic identity?

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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Could you please state more clearly what your constants c1,c2,k1 and k2 mean? – Dominik Nov 24 '12 at 20:40
They are real numbers, I suspect that during the transformation, the $c_1, c_2$ will become new constants $k_1, K_2$. – Hawk Nov 24 '12 at 20:44
I don't understahd what you want: there are two $\,y'$s there: what relation do you want to get?? – DonAntonio Nov 24 '12 at 20:51
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. – Will Jagy Nov 24 '12 at 21:02

\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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you have about three $\sin$ that should be $\sinh$ – Will Jagy Nov 24 '12 at 21:06
Oh yeah. Thanks. updated it. – user17762 Nov 24 '12 at 21:10

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: $$\exp \left(\dfrac{(a-b)t}2 \right) = \exp \left(\dfrac{at}2 \right) \exp \left(\dfrac{-bt}2 \right) \, .$$ $$\exp \left(\dfrac{(b-a)t}2 \right) = \exp \left(\dfrac{bt}2 \right) \exp \left(\dfrac{-at}2 \right) \, .$$

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