If e^y + xy = e then find the value of y'' at y = 1

Question

vishweshshrimali5 · Answer

y'' means second derivative of y w.r.t. x

anonymous · Answer

Clarification needed: Is $e$ approximately $2.71$ or a variable?

vishweshshrimali5 · Answer

Well sorry , I don't have any idea regarding that.
but the answer says 1/e^2

vishweshshrimali5 · Answer

hope that this may help u

anonymous · Answer

I unfortunately cannot see how to solve this. Someone other than me needs to give it a try; it seems like a differential equation.

anonymous · Answer

Maybe implicit differentiation?

anonymous · Answer

Hmm, I still have an x in the equation at the end..

anonymous · Answer

$$\frac{d}{dx}[e^y+xy=e] \rightarrow y'e^y+xy'+y=0.$$

anonymous · Answer

$$\frac{d}{dx}[y'e^y+xy'+y=0] \rightarrow y'y'e^y+y''e^y+xy''+y'+y'=0.$$
$$y'=\frac{-y}{(e^y+x)}$$
Can anyone check my work up to this point?
I think from here, it's $$y''=\frac{-2y'-(y')^2e^y}{e^y+x}$$
then substitute in y' and y..

vishweshshrimali5 · Answer

y' = ?

anonymous · Answer

Oh, duh! I forgot to simply solve for x. x=0 when y =1, so yeah all that's left is y''=1/e^2.