A) \[\log (1+y)=x+\frac{{{x}^{2}}}{2}+c\]
B) \[{{(1+y)}^{2}}=x+\frac{{{x}^{2}}}{2}+c\]
C) \[\log (1+y)=\log (1+x)+c\]
D) None of these
Correct Answer: A
Solution :
\[\frac{dy}{dx}=1+x+y+xy\] Þ \[\frac{dy}{dx}=(1+x)(1+y)\] Þ \[\frac{dy}{dx}+\sin \left( \frac{x+y}{2} \right)=\sin \left( \frac{x-y}{2} \right)\] On integrating, we get \[\log (1+y)=\frac{{{x}^{2}}}{2}+x+c\].
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