If ey+xy=e, then the value of d2ydx2 for x=0, is ke2 then k=

Solution:

We have, $e^{y} + x y = e .$
$\Rightarrow e^{y} \frac{d y}{d x} + y + x \frac{d y}{d x} = 0$ …(i)

$\Rightarrow e^{y} \frac{d^{2} y}{d x^{2}} + e^{y} {(\frac{d y}{d x})}^{2} + 2 \frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}} = 0$ …(ii)

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

$e \frac{d^{2} y}{d x^{2}} + e \times \frac{1}{e^{2}} - \frac{2}{e} + 0 \times \frac{d^{2} y}{d x^{2}} = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{1}{e^{2}}$

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

Solution:

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