Solution of the differential equation $\frac{dy}{dx}=1+x+y+x y$ are :
Where C is an arbitrary constant
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\log (1+y)=x+\frac{x^2}{2}+C$
$\frac{dy}{dx}=(1+x)+y(1+x)$
$⇒\frac{1}{1+y}dy=1+xdx$
$⇒\log|1+y|=x+\frac{x^2}{2}+C$