Differential Equation Solution

Suppose I have the differential equation p6 d(d(d(d(d(d(x)/dt)/dt)/dt)/dt)/dt)/dt + p4 d(d(d(d(x)/dt)/dt)/dt)/dt + p2 d(d(x)/dt)/dt + p0 x = 0 ,p6 = constant ,p4 = constant ,p2 = constant ,p0 = constant and I state the solution is x = A0 cos(w0 t) . Calculate w0 and p6, p4 and p2.

Let's go carefully, step by step. You have a 6th-order linear homogeneous differential equation with constant coefficients:

p_6 \frac{d^6 x}{dt^6} + p_4 \frac{d^4 x}{dt^4} + p_2 \frac{d^2 x}{dt^2} + p_0 x = 0

and you assume a solution of the form:

x(t) = A_0 \cos(\omega_0 t)

Step 1: Compute derivatives

The derivatives of $x(t)$ are periodic:

$\frac{d^2 x}{dt^2} = -\omega_0^2 A_0 \cos(\omega_0 t) = -\omega_0^2 x$
$\frac{d^4 x}{dt^4} = (\frac{d^2}{dt^2})^2 x = (\!-\omega_0^2)^2 x = \omega_0^4 x$
$\frac{d^6 x}{dt^6} = (\frac{d^2}{dt^2})^3 x = (-\omega_0^2)^3 x = -\omega_0^6 x$

Notice the pattern:

\frac{d^{2n}x}{dt^{2n}} = (-1)^n \omega_0^{2n} x

Step 2: Substitute into the differential equation

p_6 (-\omega_0^6 x) + p_4 (\omega_0^4 x) + p_2 (-\omega_0^2 x) + p_0 x = 0

Factor out $x$ :

(-p_6 \omega_0^6 + p_4 \omega_0^4 - p_2 \omega_0^2 + p_0)x = 0

Since $x \neq 0$ , we have the characteristic equation:

-p_6 \omega_0^6 + p_4 \omega_0^4 - p_2 \omega_0^2 + p_0 = 0

Or equivalently:

p_6 \omega_0^6 - p_4 \omega_0^4 + p_2 \omega_0^2 - p_0 = 0

Step 3: Solve for $\omega_0^2$

Let

y = \omega_0^2

Then the equation becomes a cubic in $y$ :

p_6 y^3 - p_4 y^2 + p_2 y - p_0 = 0

Step 4: Relation between $p_6, p_4, p_2, p_0$ and $\omega_0$

If you want a single cosine solution $x = A_0 \cos(\omega_0 t)$ (i.e., pure harmonic oscillation), then the cubic must have a triple root:

y = \omega_0^2

So we need:

p_6 y^3 - p_4 y^2 + p_2 y - p_0 = p_6 (y - \omega_0^2)^3

Expanding the cube:

p_6 (y^3 - 3 \omega_0^2 y^2 + 3 \omega_0^4 y - \omega_0^6) = p_6 y^3 - 3 p_6 \omega_0^2 y^2 + 3 p_6 \omega_0^4 y - p_6 \omega_0^6

Compare coefficients with the general cubic:

p_6 y^3 - p_4 y^2 + p_2 y - p_0

We get:

$p_6 = p_6$ ✅
$p_4 = 3 p_6 \omega_0^2$
$p_2 = 3 p_6 \omega_0^4$
$p_0 = p_6 \omega_0^6$

✅ Step 5: Summary of relations

\boxed{ \begin{aligned} p_6 &= \text{arbitrary constant (sets overall scale)} \\ p_4 &= 3 p_6 \, \omega_0^2 \\ p_2 &= 3 p_6 \, \omega_0^4 \\ p_0 &= p_6 \, \omega_0^6 \end{aligned} }

And the corresponding frequency:

\boxed{\omega_0 = \sqrt[6]{\frac{p_0}{p_6}}} \quad \text{or equivalently from any coefficient relation.}

From $p_4$ : $\omega_0 = \sqrt{\frac{p_4}{3 p_6}}$
From $p_2$ : $\omega_0 = \sqrt[4]{\frac{p_2}{3 p_6}}$
From $p_0$ : $\omega_0 = \sqrt[6]{\frac{p_0}{p_6}}$

All are consistent if the triple-root condition holds.

If you want, I can also draw the general solution for the 6th-order equation, showing how a single cosine emerges as a triple root scenario. This is helpful to visualize why the triple root is needed.

Do you want me to do that?

yes

Perfect — let’s visualize this carefully. We have a 6th-order linear differential equation:

p_6 x^{(6)} + p_4 x^{(4)} + p_2 x^{(2)} + p_0 x = 0

and the characteristic equation (from assuming $x = e^{\lambda t}$ ) is:

p_6 \lambda^6 - p_4 \lambda^4 + p_2 \lambda^2 - p_0 = 0

Step 1: Factor as quadratic in $\lambda^2$

Let $y = \lambda^2$ . Then the characteristic equation becomes:

p_6 y^3 - p_4 y^2 + p_2 y - p_0 = 0

This is a cubic in $y$ . The roots $y_1, y_2, y_3$ correspond to $\lambda^2$ , so the solutions of the DE are:

x(t) = C_1 e^{\sqrt{y_1} t} + C_2 e^{-\sqrt{y_1} t} + C_3 e^{\sqrt{y_2} t} + \dots

If $y_i < 0$ , the exponentials become cosines/sines, i.e., oscillations.

Step 2: Single cosine solution condition

You want a single harmonic:

x(t) = A_0 \cos(\omega_0 t)

This corresponds to $\lambda = \pm i \omega_0$ , triple root (because it’s 6th order):

y_1 = y_2 = y_3 = -\omega_0^2

So the cubic factors as:

p_6 (y + \omega_0^2)^3 = 0

Step 3: Expand and match coefficients

p_6 (y^3 + 3 \omega_0^2 y^2 + 3 \omega_0^4 y + \omega_0^6) = p_6 y^3 + 3 p_6 \omega_0^2 y^2 + 3 p_6 \omega_0^4 y + p_6 \omega_0^6

Compare to:

p_6 y^3 - p_4 y^2 + p_2 y - p_0

We get exact relations (note the sign because our $y = \lambda^2$ , $\lambda = i \omega_0$ ):

\begin{aligned} p_4 &= 3 p_6 \omega_0^2 \\ p_2 &= 3 p_6 \omega_0^4 \\ p_0 &= p_6 \omega_0^6 \end{aligned}

This is exactly what we derived earlier.

Step 4: General solution for 6th-order DE

If you allow complex roots, the general solution is:

x(t) = (A_0 + A_1 t + A_2 t^2) \cos(\omega_0 t) + (B_0 + B_1 t + B_2 t^2) \sin(\omega_0 t)

Here we see the triple root leads to polynomial factors multiplying sine and cosine.
If you want only $x = A_0 \cos(\omega_0 t)$ , then you must choose initial conditions to eliminate the extra terms (coefficients of $t \cos$ , $t^2 \cos$ , etc., must be zero).

✅ Step 5: Visualization of roots

Think of the cubic in $y$ on the real axis:


diff












y-axis (λ^2)
|       
|        * <- triple root at y = -ω0^2
|       
|       
------------------> x-axis

Single cosine solution comes from the triple root at negative y, giving λ = ±i ω0, multiplicity 3.
Multiplicity 3 ⇒ polynomial terms $t, t^2$ multiplying sin/cos in the general solution.

So in short:

The triple root ensures all derivatives combine to give a pure harmonic.
Coefficients $p_6, p_4, p_2, p_0$ are related as:

\begin{aligned} p_4 &= 3 p_6 \omega_0^2 \\ p_2 &= 3 p_6 \omega_0^4 \\ p_0 &= p_6 \omega_0^6 \end{aligned}

General solution has sine and cosine multiplied by polynomials, but a single cosine is obtained with proper initial conditions.

If you want, I can draw the actual graph showing $x(t) = \cos(\omega_0 t)$ and the effect of triple root multiplicity, which makes it very intuitive.

Do you want me to make that plot?

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