Solve: $(1-x)^2y''-4xy'-(1+x^2)y=x$ To solve the differential equation: 
$$(1-x)^2y''-4xy'-(1+x^2)y=x$$
I tried to solve the question by using the normal form of the equation but got stuck. Please help.

There are no Liouvillian solutions (solutions that can be expressed using exp, algebraic functions, and integration).  
There is an irregular point at $x=1$.
You might try a series solution in powers of $x$: 
$$ y = \sum_{j=0}^\infty a_j x^j$$
where $a_j$ satisfy the recurrence
$$ \eqalign{-a_n &+ \left( {n}^{2}-n-7 \right) a_{n+2} +
 \left( -2\,{n}^{2}-10\,n-12 \right) a_{n+3} + \left( {n}
^{2}+7\,n+12 \right) a_{ n+4} = 0\cr
a_0 &= y(0)\cr
a_1 &= y'(0)\cr
a_2 &= y''(0)/2 = a_0/2\cr
a_3 &= y'''(0)/6 = 1/6 - a_0/3 + 5 a_1/6\cr}$$
$$y \left( x \right) =a_{{0}}+a_{{1}}x+{\frac {a_{{0}}}{2}}{x}^{2}+
 \left( {\frac {a_{{0}}}{3}}+{\frac {5\,a_{{1}}}{6}}+{\frac {1}{6}}
 \right) {x}^{3}+ \left( {\frac {17\,a_{{0}}}{24}}+{\frac {5\,a_{{1}}
}{6}}+{\frac {1}{6}} \right) {x}^{4}+ \left( {\frac {29\,a_{{0}}}{30}}
+{\frac {161\,a_{{1}}}{120}}+{\frac {31}{120}} \right) {x}^{5}+
 \left( {\frac {205\,a_{{0}}}{144}}+{\frac {347\,a_{{1}}}{180}}+{
\frac {67}{180}} \right) {x}^{6}+ \left( {\frac {5203\,a_{{0}}}{2520}}
+{\frac {14141\,a_{{1}}}{5040}}+{\frac {2731}{5040}} \right) {x}^{7}+
 \left( {\frac {120257\,a_{{0}}}{40320}}+{\frac {2917\,a_{{1}}}{720}}+
{\frac {493}{630}} \right) {x}^{8}+ \left( {\frac {194149\,a_{{0}}}{
45360}}+{\frac {2109857\,a_{{1}}}{362880}}+{\frac {407527}{362880}}
 \right) {x}^{9}
+\ldots 
$$
