Differential Equation Solution

Can you give me a differential equation that has x(t) = Constant_0+ A0 cos( wo t + a0) cos( w1 t +a1) as one of its solutions?

Yes. A clean way to do this is to exploit the frequency content of your solution.

Step 1 — Decompose the solution

Using the product-to-sum identity,

\cos(\omega_0 t + a_0)\cos(\omega_1 t + a_1) =\tfrac12\Big[ \cos\big((\omega_0+\omega_1)t + a_0+a_1\big) +\cos\big((\omega_0-\omega_1)t + a_0-a_1\big) \Big].

So your solution

x(t)=\text{Constant}_0 + A_0\cos(\omega_0 t + a_0)\cos(\omega_1 t + a_1)

is a linear combination of:

A cosine with frequency $\Omega$ satisfies:

(D^2 + \Omega^2)x = 0, \quad D=\frac{d}{dt}.

A constant satisfies:

Dx = 0.

So an operator that annihilates all components of $x(t)$ is:

D\,(D^2+\Omega_+^2)(D^2+\Omega_-^2).

A linear differential equation that has

x(t)=\text{Constant}_0 + A_0\cos(\omega_0 t + a_0)\cos(\omega_1 t + a_1)

as one of its solutions is:

\boxed{ \frac{d}{dt} \Bigg[ \left(\frac{d^2}{dt^2}+(\omega_0+\omega_1)^2\right) \left(\frac{d^2}{dt^2}+(\omega_0-\omega_1)^2\right) x(t) \Bigg] =0 }

This is a 5th-order linear homogeneous ODE with constant coefficients.

The equation has many solutions; your $x(t)$ is one specific member.
The constant offset is handled by the $D$ factor.
This is the lowest-order linear ODE with constant coefficients that works without adding forcing terms.

If you want:

I can construct those as well.

Write the differential equation with x, x', x'' , ect.

Certainly. Start from the operator form

\frac{d}{dt} \Big[ \left(\frac{d^2}{dt^2}+(\omega_0+\omega_1)^2\right) \left(\frac{d^2}{dt^2}+(\omega_0-\omega_1)^2\right) x \Big]=0 .

(\,D^2+\Omega_+^2)(\,D^2+\Omega_-^2)x = x^{(4)} + (\Omega_+^2+\Omega_-^2)x'' + \Omega_+^2\Omega_-^2 x

with

\Omega_\pm = \omega_0 \pm \omega_1 .

\frac{d}{dt}(\cdot) \Rightarrow

x^{(5)} +(\Omega_+^2+\Omega_-^2)x^{(3)} +\Omega_+^2\Omega_-^2 x' = 0 .

\Omega_+^2+\Omega_-^2 = (\omega_0+\omega_1)^2+(\omega_0-\omega_1)^2 = 2(\omega_0^2+\omega_1^2)

\Omega_+^2\Omega_-^2 = (\omega_0^2-\omega_1^2)^2 .

\boxed{ x^{(5)} +2(\omega_0^2+\omega_1^2)\,x^{(3)} +(\omega_0^2-\omega_1^2)^2\,x' = 0 }

or, written with primes:

\boxed{ x^{(5)}(t) +2(\omega_0^2+\omega_1^2)\,x^{(3)}(t) +(\omega_0^2-\omega_1^2)^2\,x'(t) =0 }

If you want the expanded form including all integration constants or a forced 4th-order version, say so.

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