Transforming an equation to a differential equation We have seen transforming a differential equation to a system of equations.
I ask about the reverse procedure, specifically transforming an equation to a differential equation. Here I have an equation (assume a polynomial  $ p(x)$ of degree n with all roots in real domain) that is structured in a specific way so can be rewritten as a differential equation, for example a first order:
$$ p(x) := u(x) + \alpha u'(x) \tag{*}\label{*}  $$
and to find the roots $p(x) = 0 \to u(x) + \alpha u'(x) = 0$: 
$$ ln (u) = -x/\alpha + c  \to u(x) = e^{(-x/\alpha)+c}  \tag{**}\label{**} $$
The goal is to numerically solve the and analyze the simpler $(**)$ equation (assume $u(x)$ is more well behaved and informative than $p(x)$).
My questions are:


*

*Is this strategy valid and does it recovers all real roots of the original equation (polynomial)?

*If so, how should I chose the constant c ?

*Does this approach obscures root multiplicities, if there is any?

*Is there any publication (book, paper, course note, ...) that addresses such approach in a more general sense


Thanks
 A: You cannot. You have a polynomial $p(x) = u(x) + \alpha u'(x)$. There are two cases to consider (1) This is the zero polynomial (2) This is not the zero polynomial. Lets explore (1).
Lets just assume your polynomial is over $\mathbb{R}$ for simplicity. We can assume a general field, however, then the differential equation would have to find solutions over that field as well. If $p$ is the zero polynomial, then:
$$
p(x) = 0, \forall x\in\mathbb{R}\implies u(x) + \alpha u'(x) = 0, \forall x\in\mathbb{R}
$$
Then, your equation with a derivative is zero for all values of $x$. Thus, it is a differential equation, then you have its solution: $u(x) = Ae^{-x/\alpha}$ for some $A\in\mathbb{R}$. Since $p(0) = 0$ then $A = 0$, then a solution to the differential equation exists: $u(x) = 0$. Which is the zero polynomial. Then, $p(x) = u(x) + \alpha u'(x) = 0$, as expected.

Now we will see why it won't work with case (2). Have a nonzero polynomial $p(x)$ and define it to be $p(x) = u(x) + \alpha u'(x)$ for some function $u$ (which turns out to be polynomials as well. If you make $p(x) = 0$, then you are not saying this is the zero polynomial, you are saying you want $x_k$ such that $p(x_k) = 0$. That is, you wanna solve by finding $x_k$ such that $
u(x_k) + \alpha u'(x_k) = 0$. This is not a differential equation. You can only construct the solution if and only if:
$$
u(x) = Ae^{x/\alpha} \quad\Longleftrightarrow\quad u(x) + \alpha u'(x) = 0,\forall x\in\mathbb{R}
$$
Which is only true if $p(x)$ is the zero polynomial.

I am not saying that: What you asked is impossible. I am saying that it cannot be done in the way you suggested. There might exist a way of doing so. And one can just look at the reverse procedure: How to transform an ODE into a polynomial? The answer is simple: Transforms. Laplace transform, Fourier transform, etc. Then how to transform a polynomial into ODE? The key might be inverse laplace transform, etc.
