I would like to know if this differential equation can be transformed into the hypergeometric differential equation
$ 4 (u-1) u \left((u-1) u \text{$\varphi $1}''(u)+(u-2) \text{$\varphi $1}'(u)\right)+\text{$\varphi $1}(u) \left((u-1) u \omega ^2-u (u+4)+8\right)=0$
On
Hint:
$4(u-1)u((u-1)u\varphi1''(u)+(u-2)\varphi1'(u))+\varphi1(u)((u-1)u\omega^2-u(u+4)+8)=0$
$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\dfrac{(\omega^2-1)u(u-1)-5u+8}{4u(u-1)}\varphi1(u)=0$
$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\left(\dfrac{\omega^2-1}{4}-\dfrac{5}{4(u-1)}+\dfrac{2}{u(u-1)}\right)\varphi1(u)=0$
$u(u-1)\varphi1''(u)+(u-2)\varphi1'(u)+\left(\dfrac{\omega^2-1}{4}-\dfrac{2}{u}+\dfrac{3}{4(u-1)}\right)\varphi1(u)=0$
$\varphi1''(u)+\dfrac{u-2}{u(u-1)}\varphi1'(u)+\left(\dfrac{\omega^2-1}{4u(u-1)}-\dfrac{2}{u^2(u-1)}+\dfrac{3}{4u(u-1)^2}\right)\varphi1(u)=0$
$\varphi1''(u)+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\varphi1'(u)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)\varphi1(u)=0$
Let $\varphi1=u^a(u-1)^bv$ ,
Then $\dfrac{d\varphi1}{du}=u^a(u-1)^b\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v$
$\dfrac{d^2\varphi1}{du^2}=u^a(u-1)^b\dfrac{d^2v}{du^2}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v=u^a(u-1)^b\dfrac{d^2v}{du^2}+2u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v$
$\therefore u^a(u-1)^b\dfrac{d^2v}{du^2}+2u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(u^a(u-1)^b\dfrac{dv}{du}+u^a(u-1)^b\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v\right)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)u^a(u-1)^bv=0$
$\dfrac{d^2v}{du^2}+2\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(\dfrac{dv}{du}+\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v\right)+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$
$\dfrac{d^2v}{du^2}+\left(\dfrac{2a}{u}+\dfrac{2b}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{2}{u}-\dfrac{1}{u-1}\right)\left(\dfrac{a}{u}+\dfrac{b}{u-1}\right)v+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$
$\dfrac{d^2v}{du^2}+\left(\dfrac{2(a+1)}{u}+\dfrac{2b-1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a(a-1)}{u^2}+\dfrac{2ab}{u(u-1)}+\dfrac{b(b-1)}{(u-1)^2}\right)v+\left(\dfrac{2a}{u^2}-\dfrac{a-2b}{u(u-1)}+\dfrac{b}{(u-1)^2}\right)v+\left(\dfrac{2}{u^2}+\dfrac{\omega^2-12}{4u(u-1)}+\dfrac{3}{4(u-1)^2}\right)v=0$
$\dfrac{d^2v}{du^2}+\left(\dfrac{2(a+1)}{u}+\dfrac{2b-1}{u-1}\right)\dfrac{dv}{du}+\left(\dfrac{a^2+a+2}{u^2}+\dfrac{8ab-4a+8b+\omega^2-12}{4u(u-1)}+\dfrac{4b^2+3}{4(u-1)^2}\right)v=0$
Choose $a^2+a+2=0$ and $4b^2+3=0$ , and the ODE reduces to Gaussian hypergeometric equation.
$$ 4 (u-1) u \left((u-1) u \text{$\varphi $1}''(u)+(u-2) \text{$\varphi $1}'(u)\right)+\text{$\varphi $1}(u) \left((u-1) u \omega ^2-u (u+4)+8\right)=0$$ HINT :
I think that it might be reduced to hypergeometric equation thanks to a change of function of this kind : $$\varphi(u)=u^a(u-1)^bF(u)$$ where $F(u)$ becomes the new unknown function.
$a$ and $b$ being real parameters, to be determined after the transformation, so that the equation becomes simpler.
This attempt supposes a big work and to spent much time without being certain of success. Sorry, I will not do it for you because I am not convinced that the result is worth the effort and even if there is no typo in the equation.