The general solution of the differential equation $\frac{dy}{dx}=1+x+ y +xy $ is given by :

$log(1+y)=x+\frac{x^2}{2}+C,$ where C is a constant $log\, y =x+\frac{x^2}{2}+C,$ where C is a constant $log (y-1)=x+\frac{x^2}{2}+C,$ where C is a constant $log (1+y)=x-\frac{x^2}{2}+C,$ where C is a constant

$log(1+y)=x+\frac{x^2}{2}+C,$ where C is a constant

The correct answer is Option (1) → $log(1+y)=x+\frac{x^2}{2}+C,$ where C is a constant $\frac{dy}{dx}=1+x+ y +xy=(1+x)+y(1+x)$ $\frac{dy}{dx}=(1+x)+(1+x)$ so $\int\frac{1}{1+y}dy=\int (1+x)dx$ $⇒\log(1+y)=x+\frac{x^2}{2}+C$