Is it possible to obtain the closed-form expression of this differential equation? As a part of my research, I am stuck in solving the analytical solution of the following 6th order differential equation:
$u_t^{VI}(\theta) + 2 u_t^{IV}(\theta) + A_3 u_t^{II}(\theta) + A_2 u_t (\theta) = 0$,
where the superscript greek letters represent the derivative order and $A_2$ and $A_3$ are the integration constants. The following boundary conditions are given:
$ u_t^V(0) + u_t^{III}(0) = A_4$,
$ u_t^V(\frac{\pi}{2}) + u_t^{III}(\frac{\pi}{2}) = A_4$,
$u_t^{II}(0) + u_t(0) = 0$
$u_t^{II}(\frac{\pi}{2}) + u_t(\frac{\pi}{2}) = 0$
$u_t^{I}(0) = 0$
$u_t^{I}(\frac{\pi}{2}) = 0$
where $A_4$ is another constant.
It should be noted that I have successfully solved this differential equation using Mathematica, however, the solution is huge and can not be shortened. I was wondering if you can suggest another method for solving the above differential equation?
Thanks in advance
 A: $$u_t^{VI}(\theta) + 2 u_t^{IV}(\theta) + A_3 u_t^{II}(\theta) + A_2 u_t (\theta) = 0 \tag 1$$
The usual method to solve this kind of ODE is first to find particular solutions on the form : $\quad u=e^{r\:t}$. Putting it into Eq.$(1)$ leads to :
$$r^6+2r^4+A_3r^2+A_2=0 \tag 2$$
Let : $\quad R=r^2$
$$R^3+2R^2+A_3R+A_2=0$$
Of course, the cubic equation is analytically solvable, but the three roots $R_1,R_2,R_3$ are not nice.
The best way to make understand what "not nice" means is to show them :

It's up to you to use those ugly formulas in the next equations.
The six roots of Eq.$(2)$ are :
$r_1=\sqrt{R_1}\:,\quad r_2=\sqrt{R_2}\:,\quad r_3=\sqrt{R_3}\:,\quad r_4=-\sqrt{R_1}\;,\quad r_5=-\sqrt{R_2}\:,\quad r_6=-\sqrt{R_3} $ 
The general solution of Eq.$(1)$ is :
$$u(t)=c_1e^{r_1t}+c_2e^{r_2t}+c_3e^{r_3t}+c_4e^{r_4t}+c_5e^{r_5t}+c_6e^{r_6t}$$
$$u(t)=\sum_{k=1}^{k=6}c_ke^{r_k t} \tag 4$$
The coefficients $c_1,c_2,c_3,c_4,c_5,c_6$ have to be determined according to the conditions :
$$ u_t^V(0) + u_t^{III}(0) = A_4 \quad\to\quad 
\sum_{k=1}^{k=6}(1+r_k^3)c_k=A_4$$
$$ u_t^V(\frac{\pi}{2}) + u_t^{III}(\frac{\pi}{2}) = A_4 \quad\to\quad 
\sum_{k=1}^{k=6}(r_k^5+r_k^3)\exp(r_k\frac{\pi}{2})c_k =A_4$$
$$u_t^{II}(0) + u_t(0) = 0 \quad\to\quad 
\sum_{k=1}^{k=6}(r_k^2+1)c_k =0$$
$$u_t^{II}(\frac{\pi}{2}) + u_t(\frac{\pi}{2}) = 0 \quad\to\quad 
\sum_{k=1}^{k=6}(r_k^2+1)\exp(r_k\frac{\pi}{2})c_k =0$$
$$u_t^{I}(0) = 0 \quad\to\quad 
\sum_{k=1}^{k=6}r_k c_k =0$$
$$u_t^{I}(\frac{\pi}{2}) = 0 \quad\to\quad 
\sum_{k=1}^{k=6}r_k\exp(r_k\frac{\pi}{2})c_k =0$$
This is a linear system of six equations that you can classically solve for the six unknown $c_k$. All the coefficients have been previously computed.
Finally, put the coefficients $c_k$ into Eq.$(4)$ for the solution $u(t)$.
NOTE : 
Of course, all the above calculus has to be carried out in the complex domain. At end, if among the coefficients $r_k$ some are complex, transform the exponential of complex number into the product of real exponential and sinusoid functions. 
