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Question

I.F. of (1+x2)dydx+2xy=4x2 is:

A
2x
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B
11+x2
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C
1+x2
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D
x2
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Solution

The correct option is C 1+x2
y1+2x1+x2y=4x 2
I.F=eP(n)dx=e2xdx1+x2
t=1+x2
dt=2x dx
edtt=elogt=t
I.F=(1+x2)

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