Question

If xy + 9e^y = 9e, find the value of y'' at the point where x = 0.

Answer 1

Explanation: When x = 0 and, xy+9e^y = 9e:    

                       0 x y = 0   So,

                       0 + 9e^y = 9e

                       9e^y = 9e

                       Any number raised to 1 is equal to the number itself. So,

                        9e^1 = 9e

                        Since 9e^y = 9e

                        y = 1

Answer 2

The value of y'' at x = 0 is 0.

Given:

xy + 9e^y = 9e

To Find:

y'' at the point where x = 0

Solution:

To find the value of y'' at x = 0, we need to differentiate the given equation twice with respect to x.

Differentiating once, we get:

y' (x) + 9e^y * y' (x) = 0 ... (1)

Differentiating again, we get:

y'' (x) + 9e^y * y'' (x) + 81e^y * y'^2 (x) = 0 ... (2)

Substituting x = 0 in equation (1), we get:

y' (0) + 9e^y * y' (0) = 0

y' (0) * (1 + 9e^y) = 0

Since e^y is always positive, we get y' (0) = 0.

Substituting x = 0 and y' (0) = 0 in equation (2), we get:

y'' (0) + 0 + 81e^y * 0 = 0

y'' (0) = 0

Therefore, the value of y'' at x = 0 is 0.

#SPJ3