[Math] Linear differential equations: when the characteristic polynomial has no roots

ordinary differential equations

The following is from a file of the Differential Equations course of MIT:
enter image description here
enter image description here

This is then generalized for degree $n$.

If the roots of the characteristic polynomial are different, then it's clear that the solution is

$$
x(t) = C_1 e^{r_1t} + C_2 e^{r_2t}
$$

But what happens when we have only one (double) root? Or no roots at all? The solution must exist, right? So I guess this just means that it is not of this (exponential) form.

Best Answer

The characteristic polynomial will have roots. They may not be real though. If you get a characteristic polynomial with complex roots, you use eulers formula to get a solution, remember $e^{i\theta}$. With repeated roots, we multiply by $t$ to get a linearly independent solution. (i.e if $r=2$ then $y = c_1e^{2t} + c_2te^{2t}$)

Related Question