The general solution of the differential equation $\frac{dy}{dx}=1+x+ y +xy $ is given by :
Answer & explanation
Correct answer: option 1
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 $\int\frac{1}{1+y}dy=\int (1+x)dx$
$⇒\log(1+y)=x+\frac{x^2}{2}+C$