Finding the value of the second derivative at an x value? (Implicit Diff)

Question

The question is If $xy + 2e^y = 2e$, find the value of $y''$ at the point where $x = 0$.

I can't seem to figure it out. Any help would be appreciated!

Answer 1

for the first derivative we get $$y+xy'+2e^yy'=0$$ thus $$y'=-\frac{y}{x+2e^y}$$ can you proceed? after this we get $$y''=-\frac{y'(x+2e^y)-y(1+2e^yy')}{(x+2e^y)^2}$$ you must insert $$y'=-\frac{y}{x+2e^y}$$ finally we obtain $$y''=\frac{2e^yy^2}{(2e^y+x)^3}$$ to find $y(0)$ you can plöug in $x=0$ in your first equation: $e^{y(0)}=e$ thus $y(0)=1$

Answer 2

Having, as Dr. Graubner has, $y+ xy'+ 2e^yy'= 0$, I would not solve for y'. Instead, use "implicit differentiation" again. $2y'+ xy''+ 2e^y(y')^2+ 2e^y y''= 0$.

$xy+ 2e^y= 2e$, when x= 0, $2e^y= 2e$, $e^y= e$ so y= 1, Then $y+ xy'+ 2e^yy'= 0$ becomes $1+ 2ey'= 0$, $2ey'= -1$, $y'= -1/(2e)$. Then $2y'+ xy''+ 2e^y(y')+ 2e^yy''= 0$ becomes $2(-1/(2e)+ 2e^{-1/(2e)}y''= 0$ and $y''= (-1/e)2e^{1/2e}= -e^{1/(2e)- 1}$.