A) \[\log \left[ 1+\tan \left( \frac{x+y}{2} \right) \right]+c=0\]
B) \[\log \left[ 1+\tan \left( \frac{x+y}{2} \right) \right]=x+c\]
C) \[\log \left[ 1-\tan \left( \frac{x+y}{2} \right) \right]=x+c\]
D) None of these
Correct Answer: B
Solution :
Put \[x+y=v\] and \[1+\frac{dy}{dx}=\frac{dv}{dx}\] Therefore, the differential equation reduces to \[\frac{dv}{dx}=(1+\cos v)+\sin v\] \[=2{{\cos }^{2}}\frac{v}{2}+2\sin \frac{v}{2}\cos \frac{v}{2}=2{{\cos }^{2}}\frac{v}{2}\left( 1+\tan \frac{v}{2} \right)\] Þ \[\int_{{}}^{{}}{\frac{{{\sec }^{2}}(v/2)dv}{2[1+\tan (v/2)]}}=\int_{{}}^{{}}{dx}\] Þ \[\log \left[ 1+\tan \left( \frac{x+y}{2} \right) \right]=x+c\].
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