For an equation with constant coefficients there is a natural guess: an exponential, because differentiating an exponential only multiplies it by its rate. Substituting it turns the differential equation into an algebraic one, a quadratic for the rate, and the next three lessons are about what its roots say.

A tuning fork rings at frequencies set by its shape, not by how it is struck. The roots of the characteristic equation are the rates the equation rings at, fixed by the coefficients alone.

The characteristic polynomial of $y''-3y'+2y=0$ is $r^{2}-3r+2$. Its roots, $1$ and $2$, are the rates of the two exponential solutions $e^{x}$ and $e^{2x}$.

Write $L[y]=ay''+by'+cy$. Substituting gives . An exponential is never zero, so is a solution exactly when is a root of the characteristic equation . Its discriminant sorts every such equation into three cases: two real roots, one repeated root, or a complex pair.

• $y''-3y'+2y=0$: $$, so and are solutions.

Differentiate the exponential once and twice: each derivative multiplies it by r. Substitute and take the exponential out as a common factor. What is left is the characteristic polynomial evaluated at r, times an exponential that is never zero, so the whole is zero exactly when r is a root.

Each derivative of the exponential multiplies it by r.

Substitute and take out the common factor.

An exponential never vanishes.

So the exponential solves the equation exactly when r is a root of the characteristic polynomial.

Replace $y''$ by $r^{2}$, $y'$ by and by .

The discriminant of the characteristic equation decides the shape of every solution.

Great work! You now know how to:

• write the characteristic equation of a constant-coefficient equation
• prove that an exponential solves it exactly when its rate is a root
• sort an equation into one of the three cases by its discriminant
• say why the method needs constant coefficients