The general solution of the equation (d y)/(d x)=1+x y is :

$\frac{d y}{d x}-x y=1 \Rightarrow P=-x, Q = 1$

I.F. $= e^{-\int x d x}=e^{-\frac{x^2}{2}}$

∴ Solution is $y . e^{\frac{-x^2}{2}}=\int e^{-\frac{x^2}{2}} . 1 . d x+c$

Hence (4) is the correct answer.