The general solution of the differential equation dydx+sinx+y2=sinx−y2 is

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logtany2=c−2sinx

logtany4=c−2sinx2

logtany2+π4=c−2sinx

logtany2+π4=c−2sinx2

answer is B.

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Detailed Solution

dydx=sin⁡x−y2−sin⁡x+y2 sinA-B-sinA+B=-2cosAsinB=−2cos⁡x2sin⁡y2⇒∫cos⁡ecy2⋅dy=−2∫cos⁡x2⋅dx⇒ln⁡tan⁡y412=−2⋅sin⁡(x2)12+2c⇒ln⁡|tan⁡(y4)|=c−2sin⁡x2

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