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