Method to solve this ODE $x^{(6)}+2Ax^{(4)}+A^2x^{(2)}+B^2x = 0$

Question

I have to solve this ODE: $$x^{(6)}+2Ax^{(4)}+A^2x^{(2)}+B^2x = 0$$ where the upper index in brackets () indicates the order of the time derivative, $A = 4(m^2-2eHs_z)$ and $B= 4meH$, both are constant.
I would like to use a perturbative method because $\frac{eH}{m} \ll 2m$.
I already tried to write the characteristic polynomial and note the square of the first three terms, but then I don't know how to go ahead.

Answer 1

$$x^{(6)}+2Ax^{(4)}+A^2x^{(2)}+B^2x = 0$$ In the case of $A$ and $B$ are constants (not nul) and supposing that the symbol of the variable is $t$, i.e. : $x=x(t)$ .

Looking for particular solutions on the form $$x(t)=e^{\lambda t}$$ $$\lambda^6+2A\lambda^4+A^2 \lambda^2+B^2=0$$ This is a cubic equation wrt $(\lambda^2)$ .

This equation can be solved for six roots (real or complex, distinct or not) called $\lambda_1$ , $\lambda_2$ , . . . , $\lambda_k$ , . . . , $\lambda_6$ .

In the general case (six distinct roots) the particular solutions are $x_k(t)=e^{\lambda_k t}$ with $k=1$ to $6$.

The solution of the ODE is $$x(t)=\sum_{k=1}^{k=6}C_k \:e^{\lambda_k t}$$ $C_1$ , $C_2$ , . . . , $C_k$ , . . . , $C_6$ are arbitrary constants.

Depending on some particular values of $A,B$ occasionally two values of $\lambda$ might be equal. Accordingly a particular solution would be on the form $t\,e^{\lambda t}$. This special case has to be studied distinctly.

Note that if some root $\lambda_k$ is complex sinusoidal terms appear in place of exponential. The complex cojugate root appears as well making the final result real anyway (insofar $A,B$ are real).

Answer 2

Taking the characteristic polynomial

$$ s^6+2As^4+A^2s^2+B = (s^4+c_1s^2+c_2)(s^2+c_3) $$

and solving

$$ \cases{ B^2-c_2c_3=0\\ A^2-c_2-c_1c_3=0\\ 2A - c_1-c_3=0} $$

After the substitutions

$$ \cases{ A=4(m^2-e H s_z)\\ B = 4me H\\ s_z= \frac 12\\ e = 2\frac{\mu m^2}{H} } $$

with $\mu$ representing a fraction of the unity, we obtain

$$ \left\{ \begin{array}{ccl} c_1&=& \frac{2}{3} \left(m^2 (8-16 \mu )+2^{2/3} \sqrt[3]{3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu(\mu (32 \mu -21)+24)-4)}-m^6 (\mu (\mu (16 \mu +3)+12)-2)}+\frac{2 \sqrt[3]{2} m^4(1-2 \mu )^2}{\sqrt[3]{3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu-21)+24)-4)}-m^6 (\mu (\mu (16 \mu +3)+12)-2)}}\right) \\ c_2&=& \frac{8}{9} \left(6 m^4 (1-2 \mu )^2-\frac{4 \sqrt[3]{2} m^6 (2 \mu -1)^3}{\sqrt[3]{3\sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu -21)+24)-4)}-m^6 (\mu (\mu (16 \mu+3)+12)-2)}}+\sqrt[3]{2} \left(3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu-21)+24)-4)}-m^6 (\mu (\mu (16 \mu +3)+12)-2)\right)^{2/3}+\frac{2\ 2^{2/3} m^8 (1-2\mu )^4}{\left(3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu -21)+24)-4)}-m^6 (\mu(\mu (16 \mu +3)+12)-2)\right)^{2/3}}+2\ 2^{2/3} m^2 (1-2 \mu ) \sqrt[3]{3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu -21)+24)-4)}-m^6 (\mu (\mu (16 \mu+3)+12)-2)}\right) \\ c_3&=& \frac{1}{3} \left(m^2 (8-16 \mu )-2\ 2^{2/3} \sqrt[3]{3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu -21)+24)-4)}-m^6 (\mu (\mu (16 \mu +3)+12)-2)}-\frac{4 \sqrt[3]{2} m^4 (1-2 \mu )^2}{\sqrt[3]{3 \sqrt{3} \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu -21)+24)-4)}-m^6 (\mu (\mu (16 \mu +3)+12)-2)}}\right) \\ \end{array} \right) $$

all this can be done with the help of a symbolic processor. Important is the analysis of

$$ \sqrt{m^{12} \mu ^2 (\mu (\mu (32 \mu-21)+24)-4)} $$

or

$$ p(\mu) = 32 \mu ^3-21 \mu ^2+24 \mu -4 $$

because according with its signal the solutions should be real (exponential) for $p(\mu) \ge 0$ or complex conjugate (oscilatory) otherwise.

NOTE

Also can be used the Routh-Hurwitz criterion to qualify the roots.