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Question

If dydx=1+x+y+xy and y(-1)=0, then function y is

A
e(1x)2/2
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B
e(1+x)2/21
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C
loge(1+x)1
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D
1+x
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Solution

The correct option is D e(1+x)2/21
dydx=1+x+y+xy
dydx=(1+x)+y(1+x)
dydx=(1+x)(1+y)
dy1+y=(1+x)dx
log(1+y)=x+x22+C.....(1)
y(1)=0
Therefore,
log(1+0)=1+12+C
C=12
Substituting the value of C in equation (1), we get
log(1+y)=x+x22+12
log(1+y)=(x+1)22
y=e(x+1)221

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