Hello. Please can you help me integrate with respect to u:
$\displaystyle q_{uu} = (\frac{u-3}{2u})q_{u}$
Let $\displaystyle \displaystyle y(u) = q_u$
Then you have
$\displaystyle \displaystyle y' = \left( \frac {u - 3}{2u} \right)y$ .......................where $\displaystyle \displaystyle y' = \frac {dy}{du}$
$\displaystyle \displaystyle \Rightarrow \frac {y'}y = \frac {u - 3}{2u}$
Solve this differential equation for $\displaystyle \displaystyle y$, and then you can solve for $\displaystyle \displaystyle q$
See if you can finish up