Answer: $\large\frac{d^2x}{dt^2}$$ + w^2x = 0$

Since the given function has two arbitrary constants, a second order differential equation has to be derived.

Given $x = c_1 \cos \;wt + c_2 \sin\;wt$ where $c_1, c_2$ are arbitrary constants that need to be eliminated.

Here $w$ is a parameter that cannot be eliminated.

Differentiating, we get: $\large\frac{dx}{dt}$$ = -w c_1\sin \; wt + wc_2 cos\;wt$

Differentiating again, we get: $\large\frac{d^2x}{dt^2}$$ = -w^2c_1\cos \; wt - w^2 c_2 \sin wt = -w^2 (c_1\cos\;wt + c_2 \sin \;wt)$

Now, from the original equation, we can see that $x = c_1 \cos \;wt + c_2 \sin\;wt$ where $c_1, c_2$. Therefore substituting,

$\Rightarrow \large\frac{d^2x}{dt^2}$$ = -w^2x \rightarrow \large\frac{d^2x}{dt^2}$$ + w^2x = 0$