91Ó°ÊÓ

\[ y^{(6)}+3 y^{(4)}+3 y^{\rm \prime \prime}+y=0 \]
We assume a solution based on exponentials, which leads us to replace each derivative with a power of a variable, typically denoted as r.
This transforms our differential equation into a polynomial equation:

\[ r^6 + 3r^4 + 3r^2 + 1 = 0 \]
This is our characteristic equation. Solving this equation will give us the roots we need to form our general solution.

This reduces the polynomial to:
\[ u^3 + 3u^2 + 3u + 1 = 0 \]
By factoring, we get:
\[ (u+1)^3 = 0 \]
So, \( u = -1 \).

Reverting back to our original variable r, since \(u = r^2\), we have:
\[ r^2 = -1 \]
Therefore,
\( r = \text{±}i \). Each \(r\) represents one of the roots, meaning in this specific case, we have the repeated roots \( r = i, -i, i, -i, i, -i \), reflecting the polynomial's nature.

Because these involve complex numbers, our general solution will be a combination of exponential and trigonometric functions. Each root contributes pairs of sin and cos terms:

\[ y(t) = C_1 \text{cos}(t) + C_2 \text{sin}(t) + C_3t \text{cos}(t) + C_4t \text{sin}(t) + C_5t^2 \text{cos}(t) + C_6t^2 \text{sin}(t) \]
The constants \(C_1,C_2,C_3,C_4,C_5,\), and \(C_6\) are determined by initial conditions or additional constraints.

In this solution, each term represents part of the general solution for this 6th order differential equation. The higher-order terms \(t \text{cos}(t)\) and \(t^2 \text{cos}(t)\) arise due to the repeated nature of the roots.