The fluid has a uniform velocity, which means that all components of \(\nabla\bu\) vanish. The Euler equation simplifies to
\begin{equation*}
\rho\pd{\bu}{t}=-\nabla p-\rho g \bk,
\end{equation*}
where
\begin{equation*}
\pd{\bu}{t}=(b, \; 3t^2, \; \e^t).
\end{equation*}
Hence, the pressure is given by solving the equation,
\begin{equation*}
\nabla p = -\rho \pd{\bu}{t} - \rho g \bk
= -\rho (\, b, \; 3t^2, \; \e^t + g \,),
\end{equation*}
which is a vector equation. Splitting into components, we obtain
\begin{equation*}
\frac{\partial p}{\partial x} = -\rho b, \quad
\frac{\partial p}{\partial y} = -3\rho t^2, \quad
\frac{\partial p}{\partial z} = -\rho (\e^t + g).
\end{equation*}
Solving sequentially, the first equation gives
\begin{equation*}
p = -\rho b x + f_1(y,z,t).
\end{equation*}
Substituting into the second equation,
\begin{equation*}
\frac{\partial f_1}{\partial y} = -3\rho t^2
\;\;\Rightarrow\;\;
f_1 = -3\rho t^2 y + f_2(z,t),
\end{equation*}
\begin{equation*}
p = -\rho b x - 3\rho t^2 y + f_2(z,t).
\end{equation*}
The third equation then implies
\begin{equation*}
\frac{\partial f_2}{\partial z} = -\rho(\e^t+g)
\;\;\Rightarrow\;\;
f_2 = -\rho(\e^t+g)z + f_3(t),
\end{equation*}
\begin{equation*}
p = -\rho b x - 3\rho t^2 y - \rho(\e^t+g)z + f_3(t).
\end{equation*}
Using \(p
= p_0\) at the origin gives \(f_3(t) = p_0\) for all \(t\text{.}\) Hence
\begin{equation*}
p = p_0 - \rho \big(bx + 3t^2 y + (\e^t + g)z \big).
\end{equation*}
