$(1-t^2)\frac{\mathrm{d}^2y}{\mathrm{d}t^2}-t\frac{\mathrm{d}y}{\mathrm{d}t}+(a+2q (1- 2t^2))y=0$ So I have to solve $$(1-t^2)\frac{\mathrm{d}^2y}{\mathrm{d}t^2} -t\frac{\mathrm{d}y}{\mathrm{d}t}+(a+2q (1-2t^2))y=0$$
All substitutions seem to fail, some trigonometric ones fail less than the rest, but they still don't reduce the DE to something solvable. I would also prefer to avoid series solution as I'm not good at it and I probably won't understand the answer, but if you can do it with series, be my guest.
 A: I would try solving these differential equations first.
$$\left(D-\frac{t \pm\sqrt{t^2-4(1-t^2)(a+2q(1-2t^2))}}{2(1-t^2)}\right)y=0$$
Using an integrating factor $F(t)$
$$\left(F(t)D-F(t)\frac{t\pm\sqrt{t^2-4(1-t^2)(a+2q(1-2t^2))}}{2(1-t^2)}\right)y=0$$
such that 
$$DF(t)=F(t)\frac{t\pm\sqrt{t^2-4(1-t^2)(a+2q(1-2t^2))}}{2(1-t^2)}$$
i.e.
$$D\ln(F(t))=\frac{t\pm\sqrt{t^2-4(1-t^2)(a+2q(1-2t^2))}}{2(1-t^2)}$$
or what is the same
$$F(t)=e^{\int\frac{t\pm\sqrt{t^2-4(1-t^2)(a+2q(1-2t^2))}}{2(1-t^2)}dt}$$
There is no much hope to put the integral above nicer with that quartic polynomial inside the square root. But well. Not always we can write down solutions of DEs in nice forms.
Such an $F$ would turn the first equations I wrote above into
$$D(F(t)y)=0$$
from where 
$$y=C/F(t)$$
gives solutions. Solution which would be also solutions to your equation, I think.
Notice that the pair of equations at the beginning are just the factors of the differential operator of your original equation. So,what were doing is finding functions that make these factors vanish. If one of the factors vanish the whole differential operator in your equation would vanish as well.
