<s>[INST] 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.
 [/INST] 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
$$ \eqalign{-a_n &+ \left( {n}^{2}-n-7 \right) a_{n+2} +
 \left( -2\,{n}^{2}-10\,n-12 \right) a_{n+3} + \left( {n}
^{2}+7\,n+12 \right) a_{ n+4} = 0\cr
a_0 &= y(0)\cr
a_1 &= y'(0)\cr
a_2 &= y''(0)/2 = a_0/2\cr
a_3 &= y'''(0)/6 = 1/6 - a_0/3 + 5 a_1/6\cr}$$
$$y \left( x \right) =a_{{0}}+a_{{1}}x+{\frac {a_{{0}}}{2}}{x}^{2}+
 \left( {\frac {a_{{0}}}{3}}+{\frac {5\,a_{{1}}}{6}}+{\frac {1}{6}}
 \right) {x}^{3}+ \left( {\frac {17\,a_{{0}}}{24}}+{\frac {5\,a_{{1}}
}{6}}+{\frac {1}{6}} \right) {x}^{4}+ \left( {\frac {29\,a_{{0}}}{30}}
+{\frac {161\,a_{{1}}}{120}}+{\frac {31}{120}} \right) {x}^{5}+
 \left( {\frac {205\,a_{{0}}}{144}}+{\frac {347\,a_{{1}}}{180}}+{
\frac {67}{180}} \right) {x}^{6}+ \left( {\frac {5203\,a_{{0}}}{2520}}
+{\frac {14141\,a_{{1}}}{5040}}+{\frac {2731}{5040}} \right) {x}^{7}+
 \left( {\frac {120257\,a_{{0}}}{40320}}+{\frac {2917\,a_{{1}}}{720}}+
{\frac {493}{630}} \right) {x}^{8}+ \left( {\frac {194149\,a_{{0}}}{
45360}}+{\frac {2109857\,a_{{1}}}{362880}}+{\frac {407527}{362880}}
 \right) {x}^{9}
+\ldots 
$$
</s>