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Question

The solution of the differential equation [(cosx)dxdy](1+x2)+ex[(1+x2)tan1x+1]dx=0 is
(where C is arbitrary constant)

A
y=extan1x+sinx+C
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B
y=tan1x+exsinx+C
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C
y=extan1x+sinx+C
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D
y=tan1x+exsinx+C
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Solution

The correct option is A y=extan1x+sinx+C
[(cosx)dxdy](1+x2)+ex[(1+x2)tan1x+1]dx=0
Dividing the equation by (1+x2)dx, we get
cosxdydx+ex(tan1x+11+x2)=0dydx=cosx+ex(tan1x+11+x2)dy=cosx+ex(tan1x+11+x2) dx+Cy=sinx+ex(tan1x+11+x2) dx+Cy=sinx+extan1x+C

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