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Question Number 178415 by Spillover last updated on 16/Oct/22

If acosh x + bsinh x = c

show that .

x = \ln [\frac{c \pm \sqrt{c^{2} + b^{2} - a^{2}}}{a + b}]

Answered by haladu last updated on 16/Oct/22

a \cosh x + b \sinh x = C

\cosh (x) = \frac{e^{x} + e^{- x}}{2}

\sinh (x) = \frac{e^{x} - e^{- x}}{2}

a {\frac{e^{x} + e^{- x}}{2}} + b {\frac{e^{x} - e^{- x}}{2}} = c

mutiply both sides by e^{x}

\frac{a}{2} {{(e^{x})}^{2} + 1} + \frac{b}{2} {{(e^{x})}^{2} - 1} = {ce}^{x}

{(e^{x})}^{2} (\frac{a + b}{2}) + \frac{a - b}{2} = {ce}^{x}

{(e^{x})}^{2} (\frac{a + b}{2}) - {ce}^{x} + \frac{a - b}{2} = 0

mutiply both sides by \frac{2}{a + b}

\Rightarrow {(e^{x})}^{2} - \frac{2 c}{a + b} (e^{x}) + \frac{a - b}{a + b} = 0

\Rightarrow {(e^{x})}^{2} + 2 (- \frac{c}{a + b}) (e^{x}) = \frac{b - a}{a + b}

\Rightarrow {(e^{x} - \frac{c}{a + b})}^{2} = \frac{b - a}{a + b} + \frac{c^{2}}{{(a + b)}^{2}}

e^{x} - \frac{c}{a + b} = \sqrt{\frac{(b - a) (b + a) + c^{2}}{{(a + b)}^{2}}}

\Rightarrow e^{x} = \frac{- c \pm \sqrt{b^{2} - a^{2} + c^{2}}}{(a + b)}

\Rightarrow x = \ln {\frac{- c \pm \sqrt{b^{2} + c^{2} - a^{2}}}{a + b}}

Commented by Spillover last updated on 16/Oct/22

t h a n k y o u

Commented by haladu last updated on 16/Oct/22

You are wellcome

Answered by CElcedricjunior last updated on 16/Oct/22

on acoshx + bsinhx = c

=> acoshx + bsinhx = a (\frac{e^{x} + e^{- x}}{2}) + b (\frac{e^{x} - e^{- x}}{2})

=

=> (a + b) e^{2 x} + (a - b) - 2 {ce}^{x} = 0

=> Δ = 4 c^{2} - 4 (a^{2} - b^{2})

supposons que Δ > 0 ie c^{2} > a^{2} - b^{2} => Δ = 2 \sqrt{c^{2} + b^{2} - a^{2}}

=> e^{x} = \frac{c \mp \sqrt{c^{2} + b^{2} - a^{2}}}{a + b}

=> x = \ln ∣ \frac{c \mp \sqrt{c^{2} + b^{2} - c}}{a + b} ∣

\dots \dots \dots \dots l e c e l e b r e c e d r i c j u n i o r \dots \dots .

Commented by Spillover last updated on 16/Oct/22

t h a n k s