Before doing any examples, we begin with recalling that the solutions to the differential differential equation
\[ay''+by' +cy = 0\]is intimately connected to the roots of the parabola
\[ar^2 + br+c = 0.\]Consider
\[y''+6y' + 5y = 0.\]The characteristic polynomial is
\[r^2 + 6r + 5 = (r+5)(r+1),\]and so the roots are
\[r = -1\qquad \text{and}\qquad r= -5.\]Thus the solution is
\[y = c_1 e^{-t} + c_2 e^{-5t}.\]Consider
\[y''+ 8y' + 25 y = 0,\]which has characteristic equation
\[r^2 + 8r + 25 = 0.\]This terms cannot be factored with real roots but does have complex roots. We use the quadratic formula
\[r = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-8\pm \sqrt{64- 100}}{2},\]which can be rewritten as
\[r = \frac{-8 \pm \sqrt{-36}}{2} = \frac{-8 \pm 6i}{2} = -4\pm 3i.\]Therefore the solution is
\[y = e^{-4t} \left(c_1 \cos(3t) + c_2 \sin(3t) \right).\]Sometimes it is easier to just complete the square when we have complex roots:
\[r^2+ 8r +25 = r^2+8r + 16 + 9 = (r+4)^2 + 3^2\]and so
\[r^2+8r+25 = 0\]means
\[\begin{split} (r+4)^2 &= -3^2\\ r+4 &= \sqrt{-3^2} = \pm 3i\\ r &=-4 \pm 3i. \end{split}\]Consider
\[y'' + 6y' + 9 y = 0.\]The characteristic equation is
\[r^2+6r+9 = (r+3)^2.\]This is a repeated root which is $r= -3$.
This means the general solution to the ODE is
\[y = (c_1+c_2t)e^{-3t}.\]