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# If xy + 5e^{y} = 5e, find the value of y'' at the point where x = 0.

**Solution:**

Given, xy + 5e^{y} = 5e

Taking derivative,

d/dx (xy + 5e^{y}) = d/dx (5e)

d/dx(xy) + d/dx (5e^{y}) = d/dx (5e)

On solving,

(1 + xdy/dx) + 5e dy/dx = 0

On grouping,

dy/(x + 5e) dx = -1

dy/dx = -1/(x + 5e)

Now,

y’’ = d^{2}y/dx^{2}

d^{2}y/dx^{2} = - d/dx (1/(x + 5e))

d^{2}y/dx^{2} = 1/(x + 5e)^{2}

Y’’ at x = 0 is found to be

d^{2}y/dx^{2} = 1/(5e)^{2}

d^{2}y/dx^{2} = 1/25 e^{2}

Therefore, the value of y” at x = 0 is 1/25 e^{2}.

## If xy + 5e^{y} = 5e, find the value of y'' at the point where x = 0.

**Summary:**

For xy + 5e^{y} = 5e, the value of y'' at the point where x = 0 is 1/25e^{2}.

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