Particular solution of y1 + 3xy = x which passes through (0, 4) is :

$\frac{d y}{d x}+(3 x) y=x $

I.F = $e^{\int 3 x d x}=e^{\frac{3}{2} x^2}$

∴ Solution of given equation is

$ye^{\frac{3}{2} x^2}=\int x . e^{\frac{3}{2} x^2} d x+c=\frac{1}{3} e^{\frac{3}{2} x^2}+c$

If curve passes through (0, 4), then

$4 \frac{1}{3}=c \Rightarrow c=\frac{11}{3}$

$y=\frac{1}{3}+\frac{11}{3} e^{-\frac{3}{2} x^2} \Rightarrow 3 y=1+11 e^{-\frac{3}{2} x^2}$

Hence (1) is the correct answer.