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Question

Consider the differential equation
y2dx+(x1y)dy=0.
If y(1)=1, then x is given by :



A
42ye1ye
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B
31ye1ye
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C
1+1ye1ye
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D
11ye1ye
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Solution

The correct option is C 1+1ye1ye
dxdy+xy2=1y3
I.F.=e1y2dy=e1y
So x.e1y=1y3e1ydy
Let 1y=t1y2dy=dt
I=tetdt=ettet=e1y+1ye1y+c
x=1+1y+c.e1y
Since y(1)=1
c=1e
x=1+1y1e.e1y

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