9a2 4ab

α e−β (2). We need to find α and β. Dividing equation (1) by equation (2) 2 ⇒ β= 1 2 = ±e2β ⇒ ( ) 2β = ln ± a b ln ± a . Solving equations (1) and (2) for eβ gives us b √ ±ab. a b 2a α 2a α and eβ = ± , so =± α 2b α 2b a b ⇒ α2 = ±4ab ⇒ α = 2 ( ) If function. > 0, we use the + sign and obtain a cosh function, whereas if < 0, we use the − sign and obtain a sinh 1 2 In summary, if a and b have the same sign, we have aex + be−x = 2 opposite sign, then aex…

= ± , so =± α 2b α 2b ( ) If √ 2 , by Equation 3. [or α sinh(x + β )], then ex+β ± e−x−β = and e−x , we have a = ⇒ x = sinh−1 1 = ln 1 + ⇒ β= 1 2 ln ± a . Solving equations (1) and (2) for eβ gives us b ⇒ α2 = ±4ab ⇒ α = 2 > 0, we use the + sign and obtain a cosh function, whereas if √ ±ab. a b < 0, we use the − sign and obtain a sinh function. In summary, if a and b have the same sign, we have aex + be−x = 2 opposite sign,…

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