Consider the function \(a{e^x} + b{e^{ - x}} = \alpha \cosh \left( {x + \beta } \right)\).
Therefore,
\(\begin{aligned}a{e^x} + b{e^{ - x}} & = \frac{\alpha }{2}\left( {{e^{x + \beta }} \pm {e^{x - \beta }}} \right)\\ & = \left( {\frac{\alpha }{2}{e^\beta }} \right) \pm \left( {\frac{\alpha }{2}{e^{ - \beta }}} \right){e^{ - x}}\end{aligned}\)
On comparing the coefficients we have,
\(\begin{aligned}{l}a & = \left( {\frac{\alpha }{2}{e^\beta }} \right).........\left( 1 \right)\\b & = \pm \left( {\frac{\alpha }{2}{e^{ - \beta }}} \right){e^{ - x}}...........\left( 2 \right)\end{aligned}\)
Divide equation \(\left( 1 \right)\) by \(\left( 2 \right)\) we get,
\(\begin{aligned}\frac{a}{b} & = \pm {e^{2\beta }}\\2\beta & = \ln \left( { \pm \frac{a}{b}} \right)\\\beta & = \frac{1}{2}\ln \left( { \pm \frac{a}{b}} \right)\end{aligned}\)
On solving equations \(\left( 1 \right)\) by \(\left( 2 \right)\) we get,
\({e^\beta } = \frac{{2\alpha }}{\alpha }\)
\(\begin{aligned}{l}{e^\beta } & = \pm \frac{\alpha }{{2b}}\\{\alpha ^2} & = \pm 4ab\\\alpha & = 2\sqrt { \pm ab} \end{aligned}\)
If \(a\) and \(b\) have the same sign we have \(a{e^x} + b{e^{ - x}} = 2\sqrt {ab} \cosh \left( {x + \ln \left( {\frac{a}{b}} \right)} \right)\) whereas if \(a\) and \(b\) have the opposite sign we have\(a{e^x} + b{e^{ - x}} = 2\sqrt { - ab} \sinh \left( {x + \ln \left( { - \frac{a}{b}} \right)} \right)\).
