characteristic equation
How do you solve a constant-coefficient linear ODE without guessing wildly? You make one inspired guess — that the solution is an exponential e^(r x) — and let the equation tell you which exponents r are allowed. Substituting that guess collapses the whole differential equation into an ordinary polynomial equation in r. That polynomial is the characteristic equation, and it converts calculus into algebra.
For a y'' + b y' + c y = 0, the trial solution e^(r x) gives (a r^2 + b r + c) e^(r x) = 0, and since the exponential is never zero you need a r^2 + b r + c = 0. The three cases of the roots produce the three solution shapes: two distinct real roots r_1, r_2 give c_1 e^(r_1 x) + c_2 e^(r_2 x); a repeated real root r gives (c_1 + c_2 x) e^(r x), the extra x supplying the second independent solution; a complex pair alpha plus-or-minus i beta gives e^(alpha x)(c_1 cos beta x + c_2 sin beta x), where the imaginary part becomes oscillation via Euler's formula.
This is the workhorse for the whole field of mechanical and electrical vibrations, because those systems are exactly constant-coefficient linear ODEs. The sign of the real part of the roots decides whether a system decays or blows up (stability); the imaginary part sets the oscillation frequency. The same characteristic-polynomial idea reappears as the eigenvalue problem for systems of ODEs — there the roots are eigenvalues of the coefficient matrix.
y'' - 4y' + 13y = 0 gives r^2 - 4r + 13 = 0, roots r = 2 plus-or-minus 3i. So y = e^(2x)(c_1 cos 3x + c_2 sin 3x): an oscillation of frequency 3 growing in envelope like e^(2x).
Complex roots alpha plus-or-minus i beta become growth/decay e^(alpha x) times oscillation at frequency beta.
The exponential guess works ONLY because the coefficients are constant. For variable coefficients (e.g. Cauchy-Euler or Bessel's equation) there is no characteristic equation in this form — you need other methods.