If xy + 3e^y = 3e, find the value of y'' at the point where x = 0.

Solution:

Given, xy + 3e^y = 3e

Taking derivative,

d(xy + 3e^y)/dx = d/dx (3e)

d(xy)/dx + d(3e^y)/dx = d/dx (3e)

On solving,

(1 + xdy/dx) + 3e dy/dx = 0

On grouping,

dy(x+3e)/dx = -1

dy/dx = -1/(x+3e)

Now, calculating second order derivative

y^'' = d²y/dx²+3e))/dx

d²y/dx² = 1/(x+3e)²

Y’’ at x = 0 is found to be

d²y/dx² = 1/(3e)²

d²y/dx² = 1/9e²

Therefore, the value of y” at x = 0 is 1/9 e².

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If xy + 3e^y = 3e, find the value of y'' at the point where x = 0.

Summary:

For xy + 3e^y = 3e, the value of y'' at the point where x = 0 is 1/9e².