$\displaystyle

y'=Ae^y+Be^{-y}+2

$

y is a function of x

$\displaystyle

y'=Ae^y+Be^{-y}+2

$

so

$\displaystyle

y''=Ae^yy'+Be^{-y}y'+2

$

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- Aug 17th 2009, 01:06 AMtransgalactichow to make a derivative by x..
$\displaystyle

y'=Ae^y+Be^{-y}+2

$

y is a function of x

$\displaystyle

y'=Ae^y+Be^{-y}+2

$

so

$\displaystyle

y''=Ae^yy'+Be^{-y}y'+2

$ - Aug 17th 2009, 02:18 AMCaptainBlack
- Aug 17th 2009, 02:19 AMMush
If $\displaystyle \frac{dy}{dx} = Ae^y + Be^{-y} + 2 $

then,

$\displaystyle \frac{d^2y}{dx^2} = Ae^y \frac{dy}{dx} - Be^{-y} \frac{dy}{dx} $ - Aug 17th 2009, 02:20 AMMush