How can I solve the following higher order ODE? I am trying to solve a system of two second-order ODEs. After separating them, I obtained a fourth-order independent ODE as illustrated below. I wonder if there is a specific technique to solve it.
$$y^{(4)}+\frac{a_1}{x} y^{(3)}+\frac{a_2}{x^2}y^{(2)}+a_3y^{(2)}+\frac{a_4}{x}y^{(1)}+a_5y=0$$
 A: A power series method might do the trick. Substitute
$$ y = \sum_{k=0}^\infty c_k x^k$$
in the differential equation $$x^2 y^{(4)} + a_1 x y^{(3)} + \cdots + a_5 x^2 y = 0.$$
You will obtain a series in which the coefficients are difference equations in the unknowns $c_k$. For concrete constants $a_n$, this difference equation maybe can be solved. 
A: This is a follow-up to the answer given by doraemonpaul above.Let us express the unknown function through its Laplace transform as follows:
\begin{equation}
y(x)=\int\limits_0^\infty e^{-x s} K(s) ds
\end{equation}
By inserting the above into the ODE and then by integrating by parts in order to eleiminate the powers of $x$ from the integrand in the $s$-domain we obtain the following equation:
\begin{eqnarray}
&&\frac{d^2}{d s^2} \left[s^4 K(s) \right] - a_1 \frac{d}{d s} \left[s^3 K(s)\right]+(a_2+a_3 \frac{d^2}{d s^2}) \left[s^2 K(s) \right] - a_4 \frac{d}{d s}\left[ s K(s) \right] + a_5 \frac{d^2}{d s^2} \left[ K(s) \right] =\\
&& -a_5 x K(0) - a_5 K^{'}(0)
\end{eqnarray}
Now let us for simplicity assume that $K(0)=K^{'}(0)=0$ which means that $y(x) = O(1/x^3)$ when $x\rightarrow \infty$. Then we have:
\begin{eqnarray}
&&\frac{a_2 s^2 K(s) + \frac{d}{d s}\left( \frac{d}{d s}\left((a_5+a_3 s^2+s^4)K(s)\right) - (a_1 s^3 + a_4 s)K(s)\right)}{a_5+a_3 s^2+s^4}=\\
&&
\frac{K[s] \left(s^2 (-3 a_1+a_2+12)+2 a_3-a_4\right)}{a_3 s^2+a_5+s^4}-\frac{s K'(s) \left((a_1-8) s^2-4 a_3+a_4\right)}{a_3 s^2+a_5+s^4}+K''(s) =0
\end{eqnarray}
Then by substituting for $u=s^2$, $d/d s = 2 \sqrt{u} d/d u$ and $d^2/d u^2 = 2 d/d u + 4 u d^2/d u^2$  we get a following ODE satisfied by the Laplace transform:
\begin{equation}
\frac{K[u] (u (-3 a_1+a_2+12)+2 a_3-a_4)}{4 u (u (a_3+u)+a_5)}+\frac{K'(u) (a_5-u ((a_1-9) u-5 a_3+a_4))}{2 u (u (a_3+u)+a_5)}+K''(u)=0
\end{equation}
Now we use the common technique to eliminate the coefficient at the first derivative. We write $K(u) = M(u) \cdot V(u)$ where $M(u)$ is the exponential from minus one half times the antiderivative of the coefficient in question. In this case it reads:
\begin{equation}
M(u):=\frac{(u (a_3+u)+a_5)^{\frac{a_1-8}{8}} \exp \left(-\frac{(a_1 a_3-2 a_4) \tan ^{-1}\left(\frac{a_3+2 u}{\sqrt{4 a_5-a_3^2}}\right)}{4 \sqrt{4
   a_5-a_3^2}}\right)}{\sqrt[4]{u}}
\end{equation}
Now define:
\begin{eqnarray}
P^{(4)}(u)&:=&u^4 \left(-a_1^2+2 a_1+4 a_2+3\right)+\\
&&u^3 (a_3 (-2 a_1+4 a_2+6)-2 (a_1-3)a_4)+\\
&&u^2 \left(2 a_5 (-3 a_1+2 a_2+3)+3 a_3^2+2 a_3 a_4-a_4^2\right)+\\
&&u (6 a_3 a_5-2 a_4 a_5)+\\
&&3 a_5^2
\end{eqnarray}
and then the function $V(u)$ satisfies the following ODE:
\begin{equation}
V^{''}(u)+\frac{P^{(4)}(u)}{16 u^2(a_5+a_3 u + u^2)^2}V(u)=0
\end{equation}
which maps to the Heun equation https://en.wikipedia.org/wiki/Heun_function .
