Find compressed form for cumbersome calculation Given the three functions
$u^{\mathrm{(I)}}(t)\;=t \left(t^2\right)^{k}\,e^{2\beta t^2},\\
u^{\mathrm{(II)}}(t)=\sqrt{\left(t^2\right)^{2k}-\left(t^2\right)^{2k+1}}\,e^{2\beta t^2},\\
u^{\mathrm{(III)}}(t)\,=\sqrt{\left(t^2\right)^{2k+1}-\left(t^2\right)^{2k+2}}\,e^{2\beta t^2}$
for $k \in \{0,...,n\}$ and $C,\beta \in \mathbb{R}$ fixed.
such that $$u'''(t)^{\mathrm{(x)}}-\frac{t}{(1-t)}u'(t)^{\mathrm{(x)}}+4\left(A_n^{\mathrm{(x)}}\frac{(2t)}{(1-t^2)}+\ \beta \frac{(2t)^2}{(1-t^2)}+ \frac{C}{(1-t^2)}\right)u(t)^{\mathrm{(x)}}=0$$
where $A_n^{\mathrm{(I)}} = n\\$
$A_n^{\mathrm{(II)}} = 2n\\$
$A_n^{\mathrm{(III)}} = 3n\\$
Now, I was wondering whether it is possible to find an explicit representation for this, such that for every $k \in \{0,...,n\}$ and $(I,II,III)$ you end up with an equation that looks like $c_0 (t^2)^{k-1} + c_1 (t^2)^{k} +c_2 (t^2)^{k+1} = 0$?
 A: As far as I understand you need just to simplify those expressions for different $u$ and $A$.
Here is a Mathematica code that do this:
U1[t_] := t^(2 k + 1) Exp[2 b t^2]
U2[t_] := Sqrt[1 - t^2] t^(2 k) Exp[2 b t^2]
U3[t_] := Sqrt[1 - t^2] t^(2 k + 1) Exp[2 b t^2]

A1 := (2 n + 2) b
A2 := (2 n + 3) b
A3 := (2 n + 4) b

L[u_, A_] := (1 - t^2) D[u[t], {t, 2}] - t D[u[t], t] + 
  4 (A (2 t^2 - 1) + b^2 (2 t^2 - 1)^2 + C) u[t]

(E^(2 t^2 b)) (t^(-1 + 2 k))
Collect[Simplify[L[U1, A1]/((E^(2 t^2 b)) (t^(-1 + 2 k)) )], t]

(E^(2 t^2 b)) (t^(-2 + 2 k)) Sqrt[1 - t^2] 
Collect[Simplify[
  L[U2, A2]/((E^(2 t^2 b)) (t^(-2 + 2 k)) Sqrt[1 - t^2] )], t]

E^(2 t^2 b) t^(-1 + 2 k) Sqrt[1 - t^2]
Collect[Simplify[L[U3, A3]/(E^(2 t^2 b) t^(-1 + 2 k) Sqrt[1 - t^2])],
  t]

After formatting improvements this code gives the following results
$$
\small
\left[
t^4 (16 \beta  n-16 \beta  k)+
t^2 \left(4 \beta ^2+4 \beta+16 \beta k-8 \beta  n +4 C-4 k^2-4k-1\right)+
4 k^2+2 k
\right]
e^{2\beta t^2}t^{2k-1}
$$
$$
\small
\left[
t^4 (16 \beta  n-16 \beta  k+8 \beta )+
t^2 \left(4 \beta ^2-8 \beta + 16 \beta k-8 \beta  n +4 C-4 k^2-4k-1\right)+
4 k^2-2 k
\right]
e^{2\beta t^2}t^{2k-2}\sqrt{1-t^2}
$$
$$
\small
\left[
t^4 (16 \beta n-16 \beta  k+8 \beta)+ 
t^2 \left(4 \beta ^2-4 \beta+16 \beta k- 8 \beta n+4 C-4 k^2 -8 k-4\right)+
4 k^2+2k
\right]
e^{2\beta t^2}t^{2k-1}\sqrt{1-t^2}
$$
