how to solve a 3rd order differential equation with non-constant coefficients I am trying to solve this third order differential equation with non-constant coefficients
$$(x^3)y'''+6 x^2 y''+[6+(1+a-bx^2)x^2]x y'+[1+3 a-5 b x^2]x^2 y=0$$
where $a$ and $b$ are constants and $y$ is a function of $x$ only. 
The problem originated form a Micropolar fluid flow problem involving longitudinal and torsional oscillations. The original differential equation was 4th order and a combination of the Laplace transform, a change in variables as well as multiple integral transforms were used to bring the form  given below. 
I have tried the standard methods for solving differential equations with variable coefficients, as well as the Frobenius power series method, using x =0 as a regular singular point, however it became too complicated for a pattern and hence solutions to be formed. 
I would be grateful if anyone had any further ideas on how to solve this differential equation.
 A: The indicial roots are $0$, $-1$, $-2$.  There is a series solution 
of the form
$\sum_{k=0}^\infty c_{2k} x^{2k}$ with $c_0 = 1$, $c_2 = -a/8 - 1/24$, and
$$ -(n+5) b c_n + ((a+1)n + 5a + 3) c_{n+2} + 
     + (n+4)(n+5)(n+6) c_{n+4} = 0 $$
and a series solution of the form $\sum_{k=0}^\infty c_{2k-1} x^{2k-1}$ with $c_{-1}=1$, 
$c_1 = -a/3$, and this same recurrence.  A third fundamental solution involves $x^n$ for even $n \ge -2$ and $x^n \ln(x)$ for even $n \ge 0$. 
A: $$(x^3)y'''+6 x^2 y''+[6+(1+a-bx^2)x^2]x y'+[1+3 a-5 b x^2]x^2 y=0$$
This is not a final answer, only a way to simplify the problem.
One observe that the change of $x$ to $-x$ doesn't change the equation. This draw us to a change of variable in order to simplify :
$$X=x^2\quad;\quad X'=2x \quad;\quad y(x)=u(X)$$
$y'= 2xu'\quad;\quad y''=4x^2u''+2u' \quad;\quad y'''=8x^3u'''+12xu''$
The derivatives of $y$ are with respect to $x$. The derivatives of $u$ are with respect to $X$.
$x^3(8x^3u'''+12xu'')+6 x^2 (4x^2u''+2u') +[6+(1+a-bx^2)x^2]x (2xu')+[1+3 a-5 b x^2]x^2 y=0$
After simplification :
$$8X^2u'''+36Xu'' +[24+2(1+a)X-2bX^2]u'+[1+3 a-5 bX] y=0$$
In the general case, I don't think that a closed form solution exists with the available standard functions.
In the case $a=0$ and $b\neq 0$ WolframAlpha cannot find a closed form solution.   
In the case $b=0$ and $a\neq 0$ WolframAlpha gives a very complicated closed form solution involving hypergeometric functions and a Meijer-G function.
In the particular case $a=0$ and $b=0$ WolframAlpha gives the solution on the form $u=c_1\frac{1}{\sqrt{X}}+c_2\:_1F_2(\frac12,\frac32;2;-\frac{X}{4})+c_3$(Meijer-G function). 
A: Hint:
Let $r=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=2x\dfrac{dy}{dr}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{dr}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+2\dfrac{dy}{dr}=2x\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+2\dfrac{dy}{dr}=2x\dfrac{d^2y}{dr^2}2x+2\dfrac{dy}{dr}=4x^2\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}=4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}$
$\dfrac{d^3y}{dx^3}=\dfrac{d}{dx}\left(4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=2x\left(4r\dfrac{d^3y}{dr^3}+6\dfrac{d^2y}{dr^2}\right)$
$\therefore2r^2\left(4r\dfrac{d^3y}{dr^3}+6\dfrac{d^2y}{dr^2}\right)+6r\left(4r\dfrac{d^2y}{dr^2}+2\dfrac{dy}{dr}\right)+(6+(1+a-br)r)2r\dfrac{dy}{dr}+(1+3a-5br)ry=0$
$8r^2\dfrac{d^3y}{dr^3}+12r\dfrac{d^2y}{dr^2}+24r\dfrac{d^2y}{dr^2}+12\dfrac{dy}{dr}+(12+2r(1+a-br))\dfrac{dy}{dr}+(1+3a-5br)y=0$
$8r^2\dfrac{d^3y}{dr^3}+36r\dfrac{d^2y}{dr^2}-(2br^2-(a+1)r-24)\dfrac{dy}{dr}-(5br-3a-1)y=0$
