By solving a few examples, I found the pattern that, given a differential equation in $y(t)$ and $x(t)$, where $y(t)$ can be called the input and $x(t)$ the output, if we make the condition that $y(t) = x(t)$, and find the solution $y(t)$, then the poles of $Y(s)$ (the Laplace transform of $y(t)$), occur at the value of $s$ where the transfer function $H(s) = \frac {Y(s)} {X(s)}$ equals one. The transfer equals one because $y(t) = x(t)$ implies $Y(s) = X(s)$.
Any explanations why?
(NOTE: $s$ is complex)
As an example, try the differential equation $$y''(t) + 2y'(t) + 5y(t) = 3u'(t) + 12u(t)$$, where $u(t)$ is the input function.
Now substituting $u(t) = y(t)$, we get the differential equation whose solution is $$C_1 e^{\frac {1} {2}(1 - \sqrt{29})t} + C_2 e^{\frac {1} {2}(1 + \sqrt{29})t}$$
The poles of the Laplace transform are evidently at $$s = \frac {1} {2} (1 - \sqrt{29})$$ and $$s = \frac {1} {2} (1 + \sqrt{29})$$
But if we solved for $s$ such that $$H(s) = \frac {3s + 12} {s^2 + 2s + 5} = 1$$ we would have gotten the same values of $s$.