Solution: Let the differential equation be \begin{equation}\label{eq:04Aug2023-1} y^{\prime\prime} + a y^\prime + by = R(x), \end{equation} where $a,b$ are real constants and $R(x)$ is a function of $x$. Given that $x, x + e^x$ and $1 + x + e^x$ are solution of \eqref{eq:04Aug2023-1}, we get \begin{align*} & a + bx = R(x) \\ & e^x + a \left( e^x + 1 \right) + b \left( x + e^x \right) = R(x) \\ & e^x + a \left( e^x + 1 \right) + b \left(1 + x + e^x \right) = R(x) . \end{align*} Using the last two equations, we get \[ b + R(x) = R(x) \implies b = 0 \implies a = R(x). \] Substituting the value of $a$, in \begin{align*} e^x + a \left( e^x + 1 \right) + b \left( x + e^x \right) = R(x) & \implies e^x + R(x)\left( e^x + 1 \right) = R(x) \\ & \implies R(x) = -1. \end{align*} Therefore, the differential equation \eqref{eq:04Aug2023-1} will be \begin{equation}\label{eq:04Aug2023-2} y^{\prime\prime} - y = -1. \end{equation} We need to find the solution of \eqref{eq:04Aug2023-2}. The characteristic equation for the above differential equation will be \[ m^2 - m = 0 \implies m = 0, 1. \] Thus, solution of the corresponding homogeneous equation will be \begin{equation}\label{eq:04Aug2023-3} y_c(x) = c_1 + c_2 e^x, \end{equation} where $c_1$ and $c_2$ are arbitrary constants.

Now we need to find a general solution. If $D$ is the differential operator, then we have \begin{align*} \frac{1}{D^2 - D} (-1) & = \frac{1}{D} (1- D)^{-1} (1) \\ & = \frac{1}{D} \left( 1 + D + D^2 + \cdots \right) (1) \\ & = \frac{1}{D} (1) \\ & = x. \end{align*} Therefore, we got \[ y(x) = c_1 = c_2 e^x +x . \] Now we will use the conditions to determine constants $c_1$ and $c_2$. We have \begin{align*} y(0) = 3 & \implies c_1 + c_2 = 3 \\ y^\prime (0) = 4 & \implies c_2 + 1 = 4 . \end{align*} Therefore, we have $c_1 = 0$ and $c_2 = 3$. Hence, the solution of \eqref{eq:04Aug2023-2} is \[ y(x) = 3e^x + x, \] and $y(1) = 3e + 1$.