the characteristic equation
When the coefficients of a linear equation are constant — a y'' + b y' + c y = 0 with fixed numbers a, b, c — there is a wonderfully direct route to the answer that turns calculus into algebra. You guess that the solution is an exponential, and that single guess collapses the differential equation into an ordinary polynomial equation. That polynomial equation is the characteristic equation.
Here is how the magic works. Try y = e^(r x) for some unknown number r. Then y' = r e^(r x) and y'' = r^2 e^(r x); every derivative just multiplies by another r. Substitute into a y'' + b y' + c y = 0 and factor out the never-zero e^(r x): you are left with a r^2 + b r + c = 0. That quadratic in r is the characteristic equation, and any root r makes e^(r x) a genuine solution. The differentiation has been traded for multiplication, and a differential equation for a high-school quadratic — solve it with the quadratic formula and read off the exponents.
The character of the roots dictates the form of the solution, in three cases. Two distinct real roots r1, r2 give y = c1 e^(r1 x) + c2 e^(r2 x). A repeated real root r gives y = (c1 + c2 x) e^(r x). A complex-conjugate pair a ± b i gives y = e^(a x)(c1 cos(b x) + c2 sin(b x)). Each case fills out the two-dimensional solution space; together they solve every constant-coefficient linear second-order equation completely. This shortcut is one of the most reliable tools in all of differential equations — but it works only because the coefficients are constant; with variable coefficients the exponential guess fails.
For y'' + 5 y' + 6 y = 0 the characteristic equation is r^2 + 5 r + 6 = 0, which factors as (r + 2)(r + 3) = 0, so r = -2 or r = -3. Hence y = c1 e^(-2x) + c2 e^(-3x). The whole solution dropped out of one quadratic.
The exponential guess turns the ODE into a quadratic; its roots become the exponents of the solution.
The characteristic-equation shortcut applies only to constant-coefficient equations. For variable coefficients p(x), q(x) the trick fails — e^(rx) does not solve them — and you must turn to reduction of order, series methods, or special techniques.