Answer to Question #198204 in Calculus for desmond

Question

2. (a) Express 5 sinh x + cosh x in the form Ae^x+Be^{-x}, where A and B are integers.

(b) Solve the equation 5 sinh x + cosh x + 5 = 0, giving your answer in the form ln a, where a ∈ R.

Answer 1

(a)  5 sinh x + cosh x

$=5.\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}$

$=\frac{5}{2}e^x-\frac{5}{2}e^{-x}+\frac{1}{2}e^x+\frac{1}{2}e^{-x} \\ =\frac{6}{2}e^x+\frac{-4}{2}e^{-x}$

$\boxed{A=3 , B=-2}$

(b) 5 sinh x + cosh x + 5 = 0

$=5.\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}+5=0$

let $e^x$ = t

than

$\frac{5}{2}(t-\frac{1}{t})+\frac{1}{2}(t+\frac{1}{t})+5=0$

$\implies3t^2+5t-2=0$

$t=\frac{-5\pm\sqrt{49}}{6}$

$e^x=\frac{-5+7}{6}$ ​$\because$$e^x>0$ (always)

$\boxed{x=ln(\frac{1}{3})}$

here a=$\frac{1}{3}$