Determining a component of velocity vector of a fluid flow explicitly

If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a

(a) linear function of y.

(b) quadratic function of y.

(c) cubic function of y.

(d) constant.

I started with Euler dynamical equation as follows:

$\frac{D\overrightarrow{q}}{Dt}=\overrightarrow{F}-\frac{1}{\rho}(\mathrm{\nabla}p)$

where $\overrightarrow{q}=(u(y),0,0)$ and $\overrightarrow{F}$ is the external force, $\mathrm{\nabla}p$ being pressure gradient. Now as per question we have along X-direction that

$\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}t}+u(y)\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}x}+0.\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}y}+0.\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}z}=0-\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}t}=-\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}u(y)=-\int {\textstyle (}\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}{\textstyle )}dt+C$

C being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?

If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a

(a) linear function of y.

(b) quadratic function of y.

(c) cubic function of y.

(d) constant.

I started with Euler dynamical equation as follows:

$\frac{D\overrightarrow{q}}{Dt}=\overrightarrow{F}-\frac{1}{\rho}(\mathrm{\nabla}p)$

where $\overrightarrow{q}=(u(y),0,0)$ and $\overrightarrow{F}$ is the external force, $\mathrm{\nabla}p$ being pressure gradient. Now as per question we have along X-direction that

$\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}t}+u(y)\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}x}+0.\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}y}+0.\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}z}=0-\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{\partial}u(y)}{\mathrm{\partial}t}=-\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}u(y)=-\int {\textstyle (}\frac{1}{\rho}\frac{\mathrm{\partial}p}{\mathrm{\partial}x}{\textstyle )}dt+C$

C being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?