u''+u'/z=(K*g/q^2*z+T*K/q^2*z^2)*u
First let $\displaystyle a = \frac{K\,g}{q^2}$ and $\displaystyle b = \frac{T\,K}{q^2}$ so your ODE is
$\displaystyle
z^2 u'' + z u' -(a z+b)u = 0
$
Under the transformation $\displaystyle z = \frac{x^2}{4a}$ this ODE becomes
$\displaystyle x^2 u'' + x u' - (x^2 + 4b)u= 0$ a modified Bessel equation which has standards solutions.