Second order homogenous linear ODE How can I solve this ODE:
$y(x)+Ay'(x)+Bxy'(x)+Cy''(x)+Dx^{2}y''(x)=0$
Can you please also show the derivation.
 A: I don't think you'll find an "elementary" solution in general.  Maple finds a rather complicated solution involving hypergeometric functions:
$$\displaystyle S\, := \,y \left( x \right) ={\it \_C1}\,{\mbox{$_2$F$_1$}(1/2\,{\frac {-d+B+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}},1/2\,{\frac {-d+B- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}}}{d}};\,-1/2\\
\mbox{}\, \left( A-B \sqrt{-{\frac {C}{d}}} \right) {d}^{-1} \left(  \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})}+{\it \_C2}\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) ^{1/2\, \left(  \left( -B+2\,d \right)  \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left(  \sqrt{-{\frac {C}{d}}} \right) ^{-1}}{\mbox{$_2$F$_1$}(1/2\, \left( d \sqrt{-{\frac {C}{d}}}- \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left(  \sqrt{-{\frac {C}{d}}}\\
\mbox{} \right) ^{-1},1/2\, \left( d \sqrt{-{\frac {C}{d}}}+ \sqrt{{d}^{2}+ \left( -2\,B-4 \right) d+{B}^{2}} \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1\\
\mbox{}} \left(  \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left(  \left( 4\,d-B \right)  \sqrt{-{\frac {C}{d}}}+A \right) {d}^{-1} \left(  \sqrt{-{\frac {C}{d}}} \right) ^{-1};\,1/2\, \left( C- \sqrt{-{\frac {C}{d}}}xd \right) {C}^{-1})}
$$
(I used $d$ instead of $D$ because $D$ has a special meaning in Maple)
A: Frobenius method is the most general method I know for this case. Assume your solution is of the form $y = x^r\sum_{n=0}^\infty a_n x^n$, plug it in, and solve for $r$ and $a_n$. You should get two $r$, say $r_1$ and $r_2$. If $r_1 - r_2$ is not an integer, you already have two linearly independent solutions. If $r_1 - r_2$ is an integer, then there are several ways to get two solutions. The simplest theoretical method would be reduction of order. There are, however, special ways tailored for Frobenius method. This seems like a good reference.
