Unidirectional flow and shear rate This is from a previous exam paper that I’m going through for my upcoming exam regarding the modelling of PDE’s.

Parts a) and b) are fine but I’ve been stuck on part c) for a while. I’ve tried googling it and all I’ve come across is something called the power law. I just can’t seem to figure this out, even with the hint. Could someone please start the problem off for me or give me some hints (other than the hint given)?
 A: The velocity is zero at the walls where $y=0,h$ due to the no-slip condition. By symmetry, the maximum velocity must be attained at the centerline $y = h/2$. Hence, the velocity gradient is positive for $0 \leqslant y \leqslant h/2$ and $\left|\frac{du}{dy} \right| = \frac{du}{dy}$ in this region.
Let $q(y) = \frac{du}{dy}$.  The differential equation applicable for $0 \leqslant y \leqslant h/2$ is 
$$\mu_0 \left|\frac{du}{dy} \right|^{n-1} \frac{d^2 u}{dy^2}  = \mu_0q^{n-1} \frac{dq}{dy} = G,$$
where $G = \frac{dp}{dx} < 0$ is the pressure gradient which is constant for fully-developed flow.
This implies 
$$q^{n-1} \frac{dq}{dy} =\frac{1}{n}\frac{dq^n}{dy} = \frac{G}{\mu_0}$$
Integrating we get
$$\left(\frac{du}{dy}\right)^n = q^n = \frac{nG}{\mu_0}y + C.$$
We can evaluate the constant $C$ by applying the condition $\frac{du}{dy} = 0$ at $y = h/2$ where the velocity has a maximum to find
$$\frac{du}{dy} = \left[\frac{nG}{\mu_0}\left( y - \frac{h}{2}\right)\right]^{1/n}.$$
I'm sure you can complete the problem from this point.
