Solution: The given differential equation is \begin{equation}\label{eq:24Feb2023-1} y^\prime +6y=3 \end{equation} which is a linear differential equation. In order to solve this, we will multiply this by the integrating factor.

In general, if the differential equation is \[ y^\prime + P(x)y = Q(x), \] then the integrating factor is \[ I = e^{\int P(x)\mathrm{d} x}. \] Note that \[ I^\prime = P(x) e^{\int P(x)\mathrm{d} x} = P(x)\cdot I. \] In order to solve this differential equation, we multiply by the integrating factor both sides to obtain: \begin{align*} & y^\prime I + P(x)y\cdot I = Q(x)\cdot I \\ \implies & \left( y \cdot I\right)^\prime = Q(x) \cdot I \\ \implies & y\cdot I = \int Q(x)\cdot I \mathrm{d} x + c, \end{align*} where $c$ is the constant of integration.

We can apply the same method explain above to obtain the solution of the given differential equation. Here we have \[ P(x) = 6,~Q(x) = 3. \] So the integrating factor will be \[ I(x) = e^{\int 6\mathrm{d} x} = e^{6x}. \] So the \eqref{eq:24Feb2023-1} is equivalent to \[ \left( ye^{6x} \right)^\prime = 3e^{6x}. \] Thus, we have \begin{align*} ye^{6x} = \int 3e^{6x} + c & \implies ye^{6x} = \frac{e^{6x}}{2} + c \\ & \implies y(x) = \frac{1}{2}+ ce^{-6x}. \end{align*} Thus, the solution of the differential equation \eqref{eq:24Feb2023-1} will be \[ \textcolor{blue}{\boxed{ y(x) = \frac{1}{2}+ ce^{-6x}}} \]