= Solution
The <pressure> force in the $x$-direction on a rectangle of dimensions $\delta x\times\delta y$ is
$$
-\frac{\partial p}{\partial x}\,\delta x\delta y+o(\delta x\delta y).
$$
The shear stress is $\tau_{xy}=\mu u_y$, so the net viscous force is
$$
\mu u_{yy}\,\delta x\delta y+o(\delta x\delta y).
$$
There is no convective <acceleration>. Dividing the <momentum> balance by the area and taking the <limit> gives
$$
\boxed{\rho u_t=-\frac{dp}{dx}+\mu u_{yy}}.
$$
The $y$-momentum balance gives $p_y=0$, so the <pressure> depends only on $x$.
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