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Question

An integrating factor of the differential equation sinxdydx+2ycosx=1 is

A
sin2x
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B
2sinx
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C
log|sinx|
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D
1sin2x
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E
2sinx
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Solution

The correct option is B sin2x
sinxdydx+2ycosx=1
dydx+2ycotx=cscx
Here, P=2cotx,Q=cscx
Integrating factor I=ePdx
=e2cotxdx=e2cosxsinxdx
=e2lnsinx Integration of cotx=ln|sinx|+C
=sin2x

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