# If xy + ey = e, find the value of y" at the point where x = 0.

If xy + ey = e, find the value of y" at the point where x = 0.
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Demi-Leigh Barrera

$0+{e}^{y}=e$ plug in x=0 to the original equation this would then mean that y=1 where x=0 making the point be (0,1)
$x{y}^{\prime }+y\cdot 1+{e}^{y}\cdot {y}^{\prime }=0$ use implicit differentiation in terms of x to get this equation
0+1+e*y'=0 plug in 0 for X and 1 for Y and solve as far as you can.
y'=-1/e solve the previous equation for y'
$\left(xy{}^{″}+{y}^{\prime }\cdot 1+{y}^{\prime }\right)+\left({e}^{y}y{}^{″}+{y}^{\prime }\cdot {e}^{y}{y}^{\prime }\right)=0$ use implicit differentiation again with respect to x. Hint: You need to use the product rule to get this equation.
$0+\frac{-1}{e}+\frac{-1}{e}+e{y}^{″}+\frac{-1}{e}×\left(e\right)\left(\frac{-1}{e}\right)=0$ plug in the values you got before where x=0, y=1, and ${y}^{\prime }=\frac{-1}{e}$
$\frac{-2}{e}+e{y}^{″}+\frac{1}{e}=0$ simplify the equation to get the final result.
Resut:
$y{}^{″}=\frac{1}{{e}^{2}}$