If ey+xy=e, then the value of d2ydx2 for x=0, is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{1}{e}$
B
$\frac{1}{e^{2}}$
C
$\frac{1}{e^{3}}$
D
0

Solution:

We have $e^{y} + x y = e$ . Differentiating w.r.t.x, we get $e^{y} \frac{d y}{d x} + y + x \frac{d y}{d x} = 0 . . . . . (i)$

Differentiating w.r.t.x, we get $e^{y} \frac{d^{2} y}{d x^{2}} + e^{y} {(\frac{d y}{d x})}^{2} + 2 \frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}} = 0 . . . . . (i i)$

$Putting x = 0 in e^{y} + x y = e, we get y = 1$

$Putting x = 0, y = 1 in (i), we get e \frac{d y}{d x} + 1 = 0 \Rightarrow \frac{d y}{d x} = - \frac{1}{e}$

$Putting x = 0, y = 1, \frac{d y}{d x} = - \frac{1}{e} in (ii), we get e \frac{d^{2} y}{d x^{2}} + e \cdot \frac{1}{e^{2}} - \frac{2}{e} + 0$ $\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{1}{e^{2}}$

If $e^{y} + x y = e,$ then the value of $\frac{d^{2} y}{d x^{2}}$ for $x = 0,$ is

Solution:

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