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Question

The equation of the curve that passes through the point (1,2) and satisfies the differential equation dydx=2xy(x2+1) is

A
y(x2+1)=4
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B
y(x2+1)+4=0
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C
y(x21)=4
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D
None of these
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Solution

The correct option is A y(x2+1)=4
dydx=2xy(x2+1)dyy=2xx2+1dx
On integrating, we get
logy=log(1+x2)+logcy(1+x2)=c
Since curve passes through (1, 2), we have
c=2(1+12)c=4
Hence solution is y(x2+1)=4

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