Is it possible to get a frequency equation from limited power expansion of differential equation solution?

Question

I have a system of coupled differential equations (an example )in the form of,

$ x''+ax'+bx-cy=0$

$ y''-ay'+by-cx=0$

The solution to the above system looks like,

$x=Ae^{w_1t}+Be^{w_2t}+Ce^{w_3t}+De^{w_4t}$

$y=Ee^{w_1t}+Fe^{w_2t}+Ge^{w_3t}+He^{w_4t}$

the frequencies $w_1$,$w_2$,$w_3$ and $w_4$ are functions of $a$,$b$ and $c$ My goal is to determine that values of $a$,$b$ and $c$ such that all $w$ are $Real$. However, my original system of differential equations has nonlinear terms. I managed to derive the solution but in power expansion form or,

$x=\sum_{i=0}^{n}{\frac{a_i}{i!}t^i}$

$y=\sum_{i=0}^{n}{\frac{a_i}{i!}t^i}$

I need to determine the frequencies so that I can find the values of $a$,$b$ and $c$ such that all $w$ are $Real$. Is it possible to derive $w$ from limited power expansion (say 8th term) of $x$ and $y$?

My first attempt:

I did a test of my method by taking Fourier Sine Transform of analytical solution which give me the answer in form of $\frac{f(w)}{g(w)}$ where w is the frequency. Then solved the $g(w)$ for $w$ which gives me four answers for $w$ which are the frequencies $w_1$,$w_2$,$w_3$ and $w_4$. However when I attempted with Fourier Sine Transform of series solutions, the answer is different. This is due to the limited power expansion. Is here a way to improve this?

My second attempt:

I linearize the nonlinear terms of differential equations and used the matrix to calculate Determinant of (A-wI) where A is the matrix of the system of differential equation. I managed to calculate the values of $a$,$b$ and $c$ but they were incorrect because of linear terms I made.

A last method that I am considering is Monte Carlo. Is it possible to get a frequency equation from limited power expansion of differential equation solution? Any ideas for other methods that I missed?

Answer 1

Regarding the linear system of ode's, applying the Laplace transform on

$$ \cases{ x''+ax'+bx-cy=0\\ y''-ay'+by-cx=0} $$

we have

$$ \cases{ (s^2+a s+b)X(s)=cY(s)+c_1 x_0+c_2\dot x_0\\ (s^2-as+b)Y(s)=cX(s)+c_3y_0+c_4\dot y_0 } $$

and solving for $X(s)$ we have

$$ X(s)=-\frac{(s (s-a)+b) (c_1 x_0+c_2 \dot x_0)+c (c_3 y_0+c_4 \dot y_0)}{a^2 s^2-\left(b+s^2\right)^2+c^2} $$

now focusing on the denominator, the condition to have only real roots is

$$ a^2-2 b\pm\sqrt{a^4-4 a^2 b+4 c^2}\ge 0 $$