How do I show this hyperbolic identity?

Question

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

Answer 1

\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}

Answer 2

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