The particular solution of the differential equation $y\text{'} + 3xy = x$ which passes through $(0,4)$ is

 Given the differential equation $y^{\text{'}}+3xy=x$. 

Rearrange the given equation and consider 

$y^{\text{'}}$ as $\frac{dy}{dx}$,

 we have $\frac{dy}{dx}=x-3xy$.
take Common $x$ from right-hand-side, 

$\frac{dy}{dx}=x(1-3y)$
And, divide by $(1-3 y)$.
$\frac{dy}{(1-3y)}=xdx\text{. }$
Integrating both sides of the above  equation,
$$\begin{gathered} \int \frac{dy}{1-3y}=\int xdx. \\ \frac{\ln |1-3y|}{-3}=\frac{x^2}{2}+C. ...........\left(1\right) \end{gathered}$$
Rearrange the equation $\left(1\right)$.
$$\begin{gathered} \Rightarrow \ln |1-3y|=\frac{-3x^2}{2}-3c \\ 1-3y=\frac{-3x^2}{e^2}-3c \left\{ Ina=b \\ \Rightarrow a=e^b \right\} \\ 3y=I-\frac{-3x^2}{e^2}\left(I\right) ...\left(2\right) \left\{I=e^{-3c}\right\} \end{gathered}$$.
where $I$ is integration $\left\{I=e^{-3c}\right\}$.
According to the given information, Equation passes

through $\left(0,4\right).$ So

$$\begin{gathered} 3\left(4\right)=1-e^{\frac{-3}{2}}{\left(0\right)}^2\left(I\right) \\ 12=1-e^0\left(I\right) \\ 12=1-I \\ I=-11 \end{gathered}$$

So, the particle are solution is:

$3y=1+11e^{\frac{-3x^2}{2}}$              {substitute I in (2)}