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

Solution:

Given:

Equation is xy + 2e^y = 2e

Taking derivative on both sides, we get

⇒ d(xy + 2e^y) / dx = d(2e)/dx

⇒ d(xy)/dx + d(2ey)/dx = d(2e)/dx

On solving,

⇒ (1 + xdy/dx) + 2e dy/dx = 0

On grouping, 

⇒ dy(x + 2e)/dx = -1

⇒ dy/dx = -1/(x + 2e)

Now,

y’’ = d²y/dx²

Finding second order derivative.

⇒ d²y/dx² = - d/dx (1/(x + 2e)

⇒ d²y/dx² = 1/(x + 2e)²

y'' at x = 0 is found to be :

⇒ d²y/dx² = 1/(2e)²

⇒ d²y/dx² = 1/4e²

Therefore, the value of y” at x = 0 is 1/4e².

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

Summary:

For xy + 2e^y = 2e, the value of y'' at the point where x = 0 is 1/4e².