The solution of the differential equation [(cosx)dx–dy](1+x2)+ex[(1+x2)tan–1x+1]dx=0 is
(where C is arbitrary constant)
A
y=extan−1x+sinx+C
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B
y=tan−1x+exsinx+C
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C
y=e−xtan−1x+sinx+C
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D
y=tan−1x+e−xsinx+C
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Solution
The correct option is Ay=extan−1x+sinx+C [(cosx)dx–dy](1+x2)+ex[(1+x2)tan–1x+1]dx=0
Dividing the equation by (1+x2)dx, we get ⇒cosx−dydx+ex(tan−1x+11+x2)=0⇒dydx=cosx+ex(tan−1x+11+x2)⇒∫dy=∫cosx+ex(tan−1x+11+x2)dx+C⇒y=sinx+∫ex(tan−1x+11+x2)dx+C∴y=sinx+extan−1x+C