Method of Cauchy for particular solution of second order inhomogeneous differential equation 
Consider
  $$
- u''(t) + p(t) u'(t) + q(t) u(t) = f(t), \quad t \in (a,b).
$$
  For $s \in (a,b)$ find coefficient functions $c_1$ and $c_2$ from the general solution of the homogenous equation
  $$
u_{\text{hom}}(t) = c_1 u_1(t) + c_2(t)
$$
  such that $u_{\text{hom}}(s) = 0$ and $u'_{\text{hom}}(s) = - f(s)$.
  Then
  $$
u_p(t)
:= \int_{t_0}^{t} c_1(s) u_1(t) + c_2(s) u_2(t)
$$
  is a particular solution of the inhomogeneous equation.

And now I am supposed to find a particular solution to
$$
t^2(1 - t) u''(t) + 2t (2 - t) u'(t) + 2(1 + t) u(t) = \frac{1}{t - 1}, \quad t \in (0,1).
$$
I have found the solution for the homogeneous equation to be
$$
u_{\text{hom}}(t) = c_1 t^{-2} + c_2 \cdot \frac{t^2 - 3t + 3}{3t}
$$
But the instructions confuse me in particular because in one equation $c_i$ seem to be constants and the other functions and I con't know how they are connected or if that is the case at all. Can somebody please give me a hint on how to continue?

Also, in the first equation of the yellow block, the highest derivate doesn't have a coefficient function, but since $p$ and $q$ are not relevant to the instructions that follow I don't have to divide my equation by $t^2(t - 1)$, right?

 
So taking the hints from the answer below I got:
So my "Wronski-Matrix" is $$\begin{pmatrix} t^{-2} & \frac{t^2 -3t + 3}{3t} \\ -2 t^{-3} & \frac{1}{3} - t^{-2}\end{pmatrix},$$ solving I get $$C_1'(s) = \frac{t^3(t^2 - 3t + 3)}{(t - 3)^2(t - 1)}$$ and $$C_2'(t) = \frac{3t^2}{(1 - t)(t - 3)^2}.$$
Integrating those gives horrible terms, so what have I done wrong?
 A: Do it literally as described: For any $s$ solve $u_{\rm hom}(s,c_1,c_2)=u_{\rm hom}'(s,c_1,c_2)=0$ and use these coefficients for this $s$ to set the values in the "variating constant coefficient functions" $C_1'(s)=c_1$ and $C_2'(s)=c_2$. For a different $s$ you have a different system to solve, thus get different constants.
Or in a less mystified way, solve
\begin{align}
\pmatrix{u_1(s)&u_2(s)\\u_1'(s)&u_2'(s)}\pmatrix{C_1'(s)\\C_2'(s)}
=\pmatrix{0\\-f(s)}
\end{align}
and then integrate $C_1',C_2'$. This is commonly known an the method of variation of constants.

For that method to work you need to bring the equation into the normal form given in the theorem, so
$$f(t)=\frac1{t^2(1-t)^2}.$$
This gives less horrible terms 
$$
  C_1(t) = - \frac{\log{(1 - t)}}{3} + \frac{1}{9 (t-1)^{3} }, 
\quad
C_2(t) = - \frac{1}{3 (t-1)^{3}}
$$
that also partially cancel in the complete solution
$$
y_p(t) = \frac{3 \log{(1 - t)} + 1}{9 t^{2}}
$$
A: Given  the homogeneous solution for
$$
t^2(1 - t) u''(t) + 2t (2 - t) u'(t) + 2(1 + t) u(t) = \frac{1}{t - 1}, \quad t \in (0,1).
$$
as
$$
u_h = c_1 t^{-2} + c_2 \cdot \frac{t^2 - 3t + 3}{3t}
$$
and considering $c_i = c_i(t)$ after substitution into the full DE we have
$$
3 (t-1)^2 c_1''(t)-6 (t-1) c_1'(t)+t ((t-3) t+3) (t-1)^2 c_2''(t)+2 (2 t ((t-3) t+3)-3) (t-1)
   c_2'(t)+3= 0
$$
now solving for
$$
3 (t-1)^2 c_1''(t)-6 (t-1) c_1'(t) = 0\\
t ((t-3) t+3) (t-1)^2 c_2''(t)+2 (2 t ((t-3) t+3)-3) (t-1)
   c_2'(t)+3= 0
$$
or after simplifications
$$
3 (t-1) c_1''(t)-6  c_1'(t) = 0\\
t ((t-3) t+3) (t-1) c_2''(t)+2 (2 t ((t-3) t+3)-3)
   c_2'(t)+\frac{3}{t-1}= 0
$$
we obtain the $c_1(t), c_2(t)$. As we can see the $c_2(t)$ DE is also complicated. We could also consider as well the set of DEs
$$
3 (t-1) c_1''(t)-6  c_1'(t) +\frac{3}{t-1} = 0\\
t ((t-3) t+3) (t-1) c_2''(t)+2 (2 t ((t-3) t+3)-3)
   c_2'(t)= 0
$$
perhaps a little simpler.
NOTE
The variation of constants method (Lagrange) give us to solve DEs a degree lower than the original. This normally is worth a lot.
