If  ey+xy=ex, then  y2(0)=

# If  ${e}^{y}+xy={e}^{x}$, then  ${y}_{2}\left(0\right)=$

1. A

1

2. B

$-2$

3. C

0

4. D

$-1$

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### Solution:

${e}^{y}+xy={e}^{x}\cdots \cdots \left(1\right)$

$x=0⇒{e}^{y}=1⇒y=0$

differentiate with respect to  $x$

${e}^{y}.{y}^{1}+\left(y+x{y}^{1}\right)={e}^{x}\cdots \cdots \left(2\right)$

$x=0,y=0$

${e}^{0}.{y}^{1}\left(0\right)+\left(0+0\right)={e}^{0}=1$

${y}^{1}\left(0\right)=1$

${e}^{y}.{y}^{11}+{e}^{y}.{y}^{1}.{y}^{1}+{y}^{1}+x{y}^{11}+{y}^{1}={e}^{x}$  (differentiate with respect to $x$ in (2))

$\text{put\hspace{0.17em}\hspace{0.17em}}x=0,y=0,{y}^{1}\left(0\right)=1$

$1.{y}^{11}\left(0\right)+{e}^{0}.1.1+1+0+1={e}^{0}$

${y}^{11}\left(0\right)+3=1$

${y}^{11}\left(0\right)=-2$

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