What is the ordinary differential equation for double exponential summation?

$$y=c_{1}e^{a_{1}x}+c_{2}e^{a_{2}x}$$ The characteristic polynomial is: $$(r-a_1)(r-a_2)=0$$ $$r^2-(a_1+a_2)r+a_1a_2=0$$ Then you deduce the differential equation: $$y''-(a_1+a_2)y'+a_1a_2y=0$$

General method: From an expression with two parameters $c_1, c_2$, to find the second-order DE with that solution.

Step 1, solve for $c_1$
Step 2, differentiate to eliminate $c_1$
Step 3, solve for $c_2$
Step 4, differentiate to eliminate $c_2$.
The result is an equation involving $y, y', y''$.

If $y''(x) + ay'(x) + by(x) = 0$, then it is well known that the general solution is:

$$c_1 e^{alpha x} + c_2 e^{beta x}$$

where $alpha, beta$ are the roots of the quadratic $x^2+ax + b = 0$.

This result can be obtained by letting $y = e^{rx}$, finding $y'(x)$ and $y''(x)$, and then substituting those values into the differential equation.

So the original problem has been transformed into: given the roots of a quadratic equation, how can I find the original quadratic? This is easy if you just expand $(x + alpha)(x + beta) = 0$, or use Vieta's formulas.