$\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 $
Follow Math Help Forum on Facebook and Google+
Originally Posted by transgalactic $\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 $ I don't know about anwone else but I can't see a question here, what is it you want to do? CB
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} $
Originally Posted by CaptainBlack I don't know about anwone else but I can't see a question here, what is it you want to do? CB I believe they have been given the 1st derivative, and have been asked to find the second?
View Tag Cloud